Collins Cambridge International AS & A Level Further Mathematics Further Mechanics STUDENT'S BOOK: Worked solutions 9780008271893

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WS TITLE PAGE_Further Mechanics
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Cambridge International AS & A Level Further Mathematics

Further Mechanics STUDENT’S BOOK: Worked solutions

Anthony Alonzi, Chris Chisholm Series Editor: Dr Adam Boddison

Pure Mathematics 1 International Students Book Title page.indd 1 57736_Pi_viii.indd 1 WS TITLE PAGE_Further Mechanics.indd 1

14/11/17 10:46 pm 6/18/18 10:36 3:21 PM 31/07/18 AM

1 

WORKED SOLUTIONS

Worked solutions 1 Motion of a projectile Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

Prerequisite knowledge 1 a Horizontal: 10 cos 23° = 9.21

Vertical: 10 sin 23° = 3.91

b Horizontal: Vertical: 12.6 cos 45° = 8.91 12.6 sin 45° = 8.91 c Horizontal: 29 cos 65° = 12.3

Vertical: 29 sin 65° = 26.3

d Horizontal: Vertical: 0.2 cos 29° = 0.175 0.2 sin 29° = 0.0970 2 a v = u + at = 2 + 3 × 4 = 14 m s−1 b v2 = u2 + 2as = 12 + 2 × 2 × 4 = 17, v = 4.12 m s−1 c u2 = v2 − 2as = 252 − 2 × 10 × 16 = 305, u = 17.5 m s−1 1 d s = ut + at 2 = 5 × 5 + 0.5 × 4 × 52 = 75 m 2 e v = u + at = 10 + −2 × 5 = 0 m s−1 3 a v2 = u2 + 2as = 02 + 2 × 10 × 10 = 200, v = 14.1 m s−1 1 b s = ut + at 2, 10 = 0 × t + 0.5 × 10 × t2, 2 5t2 = 10, t = 1.41 s 4 a Maximum height, v = 0 v = u + at, 0 = 2 + −10 × t, t = 0.2 s b v2 = u2 + 2as, 0 = 22 + 2 × −10 × s, s = 0.2 m

Exercise 1.1A 1 a u = 25, q  = 45° 1 y = Ut sin q − 2 gt2 1 0 = 25t sin 45° − 2 × 10 × t2 t(25 sin 45° − 5t) = 0

t = 0 or t = 3.536



Time in air = 3.54 s

b Range, x = ut cos 45° x = 25 × 3.536 × cos 45° x = 62.5 m 2 a u = 40, q  = 45° Time to maximum height



0 = 40 sin 45° − 10t t = 2.83 s



When t = 2.828



y = 40 × 2.828 sin 45° − 1 × 10 × 2.8282 2 Maximum height = 40.0 m



b Time of flight = 2 × 2.828 = 5.656 s Range = 40 × 5.656 × cos 45° Range = 160 m 3 a Time of flight is found by finding t when y = 0 0 = 8.8 sin 42° t − 1 × 10 t2 2 t(8.8 sin 42° − 5t) = 0

t = 0 or t = 1.178



Tiger will be in air for 1.18 s

b Range = Ut cos q = 8.8 × 1.178 × cos 42° = 7.70 m

It will cross the river with 2.70 m to spare.

c  Motion only occurs in two dimensions, air resistance is negligible, tiger is a point mass 4 a x = Ut cos q

x = 30 × 3 × cos 53°



x = 54.2 m

1 b y = Ut sin q − 2 gt2 1 y = 30 × 3 × sin 53° − × 10 × 32 2 y = 26.9 m c vx = U cos q

vx = 30 cos 53° = 18.05 m s−1



vy = U sin q − gt



vy = 30 × sin 53° − 10 × 3 = −6.041 m s−1 i.e. downwards



So speed = 18.052 + (−6.041)2 = 19.0 m s−1

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1 Motion of a projectile

5 First find the time of flight: 1 0 = 22t sin q − × 10 × t2 2 0 = t(22 sin q − 5t) 22sin θ t = 0 or t = 5 Then find the range: x = Ut cos q 12 = 22 × 22sin θ × cos q 5 30 = sin 2θ 121 2q  = 14.36° q  = 7.18° 6 a When t = 4, y = 0 0 = 25 × 4 × sin q − 1 × 10 × 42 2 100 sin q = 80 sin θ = 80 = 4 100 5 4 3 b If sin θ = 5 then cos θ = 5 Range = Ut cos q 3 x = 25 × 4 × 5 = 60 

b Range = 35 × 4.427 × cos 20° = 146 m c  Motion only occurs in two dimensions, air resistance is negligible, stone is a point mass, gravity is constant 2 a u = 20, q  = 0°, h = 20

Time of flight is found by finding t when y = 0 0 = 20t sin 0° − 1 × 10 × t2 + 20 2 t = −2 or t = 2



So time of flight = 2 s



b Range = 20 × 2 × cos 0° = 40 m 3 a u = 29.4, q  = 30°, h = 49

Time of flight is found by finding t when y = 0 0 = 29.4t sin 30° − 1 × 10 × t2 + 49 2 5t2 − 29.4t sin 30° − 49 = 0



Using the quadratic formula:



t = −1.988 or t = 4.928



So particle will hit ground after 4.93 seconds.



b Range = 29.4 × 4.928 × cos 30° = 125 m

7 a Maximum height is when vy = 0

c vx = U cos q



0 = 10 sin 30° − 10t



t = 0.5 s

= 29.4 cos 30°



Maximum height = 10 × 0.5 × sin 30° − 1 × 10 × 0.52 2 = 1.25 m

= 25.46 m s−1





= 29.4 sin 30° − 10 × 4.93

b Range = Ut cos q = 10 × 0.5 × cos 30° = 4.33 m

vy = U sin q − gt = −34.58 m s−1 (i.e. downwards)



Speed = 25.462 + (−34.58)2 = 42.9 m s−1



5t2 − 5t + 1 = 0

d  If q  is an angle between the ground and the direction of the particle as it hits the ground then 34.58 q = tan−1  25.46 = 53.6°



Using the quadratic formula:

4 u = 20, q  = −10°, h = 50



t = 0.2764 s and t = 0.7236 s



So it is above 1 m for 0.447 seconds.

c When y = 1 1 = 10 sin 30°t − 5t2

Exercise 1.2A 1 a u = 35, q = 20°, h = 45

Time of flight is found by finding t when y = 0 1 0 = 35t sin 20° − × 10 × t2 + 45 2 5t2 − (35 sin 20°)t − 45 = 0



Using the quadratic formula:



t = −2.033 or t = 4.427



So time of flight = 4.43 s



(

)

Time of flight is found by finding t when y = 0 0 = 20t sin (−10°) − 1 × 10 × t2 + 50 2 5t2 − 20t sin (−10°) − 50 = 0 Using the quadratic formula: t = −3.529 or t = 2.834 So particle will hit the ground after 2.83 seconds. Range = 20 × 2.834 × cos (−10°) = 55.8 m

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1 

WORKED SOLUTIONS 1 b y = Ut sin q −  gt2 2 1 0 = 40t sin q −  gt2 2 80t sin q −  gt2 = 0

5 a When the stone lands in the sea, y = 0 1 0 = U × 3 × sin 0° − 2 × 10 × 32 + h 0 = h − 45

h = 45 m

b  Distance from base of cliff =  73.52 − 452 = 58.11 m 58.11 = U × 3 × cos 0° U = 19.4 m s−1



t(80 sin q −  gt) = 0



t = 0 or 80 sin q  −  gt = 0



t = 0 or 80 sin q =  gt

c 80 sin q   =  gt so t =

6 Find the time taken for the ball to travel 4 m horizontally:



80sin θ 3 = sec θ g 2 160 sin q cos q = 3g 3g 2 sin q cos q = 80 3g sin 2θ = 80 gx 2(1 + tan 2 θ ) 4 y = x tan q − 2U 2

16 tan2 q  − 500 tan q + 366 = 0

y = 2x −

Using the quadratic formula: tan q  = 0.75 or tan q  = 30.5 So q  = 36.9° or q  = 88.1° Smallest possible angle is 36.9°.

2U sin θ 2U sin 30° 2U × = g = g g

=U g

2 4xU 2 5gx 2 − 2 2U 2U

y=

x(4U 2 − 5gx) 2U 2

5 a R = =

2

2 U 2 sin 60° U 23 U 2 3 b Range = U sin 2θ = = = 2g g g g 2

2

2

=

2

U sin θ U sin 30° c Maximum height = = 2g 2g U ×( 2g 2

=

)

1 2 2

=

2 For Mei, q  = 135° would be launching the particle backwards. Her range is correct but negative. She should have found the smallest value of 2q  for which cos 2q = 0 which would have given an acute value of q . U2 If 2q = 90°, then q  = 45° and the range is g m. For Xing, although the graph of sin q  has a maximum value for 90°, the graph of sin 2q  has a maximum value for 45°, from which the 2 maximum range is U m. g 3 a  x = ut cos q  



V 2 sin 2θ g 2V 2 sin θ cos θ g

( )( )

2V 2 3 4 5 5 g

24V 2 = 25g m

2

U 8g

gx 2(1 + 4) 2U 2

y=

Exercise 1.3A 1 a Time of flight =

Equating this with the expression from part a:



4 = 25t cos q 4 t = 25cos θ The height when at this point must be 2.8 m. 2 4 4  2.8 = 25 × × sin θ − 1 × 10 ×  25cos θ 2  25cos θ  16 2.8 = 4 tan q − sec2 q 125 350 = 500 tan q  − 16 (1 + tan2 q )

1 2

80sin θ g

b Projected at a V 2 sin 2α g



R=



Projected at (90° – a)



2 R = V sin 2(90° − α) g



sin 2(90° – a ) = sin (180° – 2a)



sin (180° – 2a ) = sin (180°) cos (2a) – sin (2a) cos (180°) = sin 2a V 2 sin 2α So R = g These ranges are the same.

60 = 40t cos q   3 t = 2 sec q

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1 Motion of a projectile

6 Starting from the formula y = x tan q − For A, q  = 0, U = 14 m s−1 y = x tan 0° −

(

gx 2 1 + tan 2 0° 2 × 14

2

)

For B, q  =  45°, U = 28 m s−1 y = x tan 45° −

(

gx 2 1 + tan 2 45° 2 × 28 2

=

(

gx 2 1 + tan 2 θ 2U 2

10 t = 23sin θ

):

1 2 gt + h 2 0 = 23t sin q  − 5t 2 + 20 10 10 0 = 23 sin q − 5 23sin θ 23sin θ y = Ut sin q −

−10x 2 392

) = x − 20x

(

1568

Since A is 60 m vertically above B, the particles collide if 2 2 x − 20x = −10x + 60 1568 392

1568x − 20x2 = −40x2 + 94 080

(



5t2 – (23 sin 10.19°)t − 20 = 0

Using the quadratic formula:



Using the quadratic formula:

x = −118 or x = 39.8



t = −1.63 or t = 2.45

So the particles collide after travelling a horizontal distance of 39.8 m.



So the time of flight is 2.45 s.



2v sin 45° 10 v = 21.2

U 2 sin 2θ 2 a R = g

2 2 h = 21.2 sin 45° + 1.5 2 × 10 h = 12.7

b When t = 1

vx = U cos q =  16.31 cos 35° = 13.36



vy = U sin q − gt = 16.31 sin 35° − 10 × 1 = −0.6450 m s−1 2

2

So speed = (13.36) + (−0.6450) = 13.4 m s−1 0.6450 Direction = tan −1 = 2.76° below the 13.36 horizontal

3 a     x = Ut cos q

b speed =

10 = 23t sin q

(18cos 20°)2 + (18sin 20° − 30)2

c  Displacement after 3 s is the same as the height of the starting point. 1   y = Ut sin q − 2 gt2 −h = 18 × 3 × sin 20° − 0.5 × 10 × 32

   h = 26.5 m

1 5 a y = Ut sin q − 2 gt2 2 = (35 sin 38°)t − 5t2

U 2 sin70° 10 250 2 U = sin70° U = 16.3 m s−1



vy = U sin q − gt = 18 sin 20° − 10 × 3 = 18 sin 20° − 30 18sin 20° − 30  Hits ground at angle of q  = tan −1   18cos 20°  = 54.6° = 29.2 m s−1

25 =





3=

2 2 b Maximum height = v sin θ + 1.5 2g



4 a vx = U cos q = 18 cos 20°

1 a If it reaches its maximum height after 1.5 s, it will return to 1.5 m after 3 s. 2v sin θ t= g

2

0 = 30 −

20x2 + 1568x – 94 080 = 0

Exam-style questions

) + 20

500 529sin 2 θ 15 870 sin2 q  = 500 500 sin q =  15 970 q =  10.2° 1 b y = Ut sin q −  gt2+ h 2 0 = 23t sin 10.19° − 5t2 + 20

2

)



5t2 – (35 sin 38°)t + 2 = 0



Using the quadratic formula:



t = 0.0949 or t = 4.215



The ball hits the platform on the downward phase of its motion so this happens after 4.22 seconds.

b When t = 4.215

vx = U cos q =  35 cos 38° = 27.58



vy = U sin q − gt = 35 sin 38° − 10 × 4.215 = −20.60 (i.e. downwards)



speed = 27.582 + (−20.60)2  =  34.42 m s−1



The ball strikes the platform at a speed of 34.4 m s−1.

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1 

WORKED SOLUTIONS

6 a Find t when y = 0: 1 y = Ut sin q − gt2 + h 2 0 = 10 × t × sin 30° − 5t2 + 19



Using the quadratic formula:



t = 0.1467 or t = 1.364



So the ball is above 1 m for 1.22 s.



5t2



Using the quadratic formula:



vx = U cos q =  12 cos 39° = 9.326



t = −1.512 or t = 2.512



vy = U sin q − gt = 12 sin 39° − 10 × 0.1467 = 6.085



vx = U cos q = 10 cos 30° = 8.660



speed = (9.326)2 + (6.085)2 = 11.1 m s−1



vy = U sin q − gt = 10 sin 30° − 10 × 2.512 = −20.12



q =  tan −1



When y = 1, the speed is 11.1 m s−1 at an angle of 33.1° above horizontal.



– (10 sin 30°)t − 19 = 0

2

b When t = 0.1467

2

speed = (8.660) + (−20.12) =

21.9 m s−1

Strikes the ground at an angle of q =  tan −1

(

= 66.7° b x = Ut cos q  = 10 × 2.512 × cos 30° = 21.8 m so d = 21.8 2 7 a Range = u sin 2θ g

u 2 sin 2θ 10 u2 sin 2q =  150 15 =



u2 =

150 sin 2θ

Maximum height =

u 2 sin 2 θ 2g

u 2 sin 2 θ 2 × 10 2 2 u sin q = 300 300 u2 = sin 2 θ Equating: 150 300 = sin 2θ sin 2 θ  15 =



20.12 8.660

)

6.085 ( 9.326 ) = 33.1°

9 a Particle A u 2 sin 2θ u 2 sin 2θ Range = , 60 = , 600 = u2 sin 2q g g Particle B 75 x = Ut cos q, 75 = ut cos q, t = u cos θ Time of flight = 75 when y = 0. u cos θ 1 y = Ut sin q − 2 gt2 + h 1 0 = Ut sin q − gt2 + h 2 2 75 75 0 = u × u cos θ × sin θ − 5  u cos θ  + 25 28125 0 = 75 tan q  − u 2 cos2 θ + 25

600 300 = sin 2θ sin θ cos θ 28 125 + 25 0 = 75 tan q  – 300 cos2 θ sin θ cos θ 0 = 75 tan q  – 28 125 + 25  300cos θ   sin θ  28 125tan θ + 25 0 = 75 tan q  – 300

From above, u 2 =

(

)



150 sin2 q = 300 sin 2q



150 sin2 q − 600 sin q cos q = 0



150 sin q (sin q  − 4 cos q ) = 0



150 sin q = 0 or sin q  − 4 cos q = 0



0 = 22 500 tan q –  28 125 tan q +  7500



q  = 0





or tan q  = 4



5625 tan q  = 7500 7500 q  = tan −1 = 53.13° 5625

q  = 76.0° b u 2 = 150 sin 2θ = 150 sin152° u = 17.9

1 8 a y = Ut sin q −   gt2 2 1 = (12 sin 39°)t − 5t2

5t2 – (12 sin 39°)t + 1 = 0





( )

The angle of projection of the particles is 53.1°.

b 600 = u2 sin 2q, 600 u2 = sin106.26

or, leave it as sin(2 × 53.13) so as to not lose the precision



u = 25 m s−1



The initial speed of the particles is 25 m s−1.

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1 Motion of a projectile

10 a y = x tan q − 



gx 2(1 + tan 2 θ ) 2U 2

gx 2(1 + tan 2 30) 2 × 52 10x 2 1 + 1 3 3 y= x− 3 50

y = x tan 30° −

( )



3 4 2 x− x 3 15  3 4  0 = x − x  3 15  3 4 − x =0 3 15

So x = 0 or



x = 0 or x = 5 3 4 So range = 2.17 m

c y=

x=



Range =



dy For a turning point, dx = 0 So x = 5



So maximum height = 4(5) − 0.4(5)2 + 1 = 11 m.

b When x = 0, y = 1 So height of launch is 1 m. c For range, y = 0

2U 2 sin θ cos θ g U 2 sin 2θ g

b Maximum range when sin 2q  = 1 2q  = 90°

q  = 45° Maximum range occurs when angle of 2 projection is 45°, so max range = U g

13 a R =

4 2 3 x− x +5 3 15

11 a y = 4x − 0.4x2 + 1 dy dx = 4 − 0.8x



0=





x =U ×

4 2 3 y= x− x 3 15

b Range is when y = 0

2U sin θ × cos q g



U 2 sin 2θ g

6.52 sin2θ 10 6.52 sin 2q = 20 20 sin 2θ = 42.25 2q = 28.25 or 2q = 151.8 2=

q = 14.13 or q = 75.9

The two possible angles are 14.1° or 75.9° above the horizontal.

b t=

2U sin θ g

t = 2 × 6.5 × sin14.13 = 0.317 10 The ball will be in the air for 0.317 s. gx 2 sec2 θ 14 y = x tan q − 40 x = 1, y = 0.4 2 0.4 = tan q − 10sec θ 40



0 = 4x − 0.4x2 + 1



0 = 20x − 2x2 + 5



2x2 − 20x − 5 = 0



Using the quadratic formula:

sec2 θ 4 1.6 = 4 tan q − (1 + tan2 q )



x = −0.244 or x = 10.2

1.6 = 4 tan q − 1 − tan2 q



So the range is 10.2 m.

tan2 q − 4 tan q + 0.6 = 0

12 a For time of flight, y = 0

Substitute into y = Ut sin q − 1 gt2: 2 1 0 = Ut sin q  − gt2 2 1 From which 0 = t(U sin q  − gt) 2 1 So t = 0 or U sin q  − gt = 0 2 t = 0 is when the particle is launched so 2U sin θ . t= g x = Ut cos q

0.4 = tan q −

Using the quadratic formula: tan q  = 0.1561 or tan q  = 3.844 q  = 8.87° or q = 75.4° The two possible angles are 8.87° or 75.4° above the horizontal. 15 a Particle P: x = 20t cos 60° = 10t When x = 15 , t = 1.5 s

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1 

WORKED SOLUTIONS

b We need to know the height of P at t = 1.5 s 1 y = 20t sin 60° – (10)t2 2 y = 10 3t − 5t 2

Particles land 10 m apart V 2 36 3 So 10 = 5 +10 V 2 = 224.7

When t = 1.5

V = 15.0 m s–1. 5(1.5)2



y = 10 3(1.5) −



y = 14.73 m



The particles collide at t = 1.5 s at a height of 14.73 m



For Q

18 In order for the ball to pass through the window we need 2 < y < 3.5 when the ball has travelled 12 m. Time taken to travel 12 m? x = 14t cos 35°

 y = Vt − 5t2

12 = 14t cos 35°

14.73 = V(1.5) − 5(1.5)2



t = 1.046 s

 V = 17.3

At this time, y = 14 × 1.046 × sin 35° – 5 × 1.0462 = 2.93 m

c Max height of P H =



2

2

u sin θ 2g

This means the ball will pass through the window.





16 a Max height of P u 2sin 2 θ H= 2g



322sin 2 55° 20 = 34.3557… = 34.4 m (to 3 s.f.)

b When q =  10°, 5 – 2x = tan 10° 1 x = (5 − tan 10°) = 2.412 2

=



1 of max height = 11.45 m (to 4 s.f.) 3 2 of max height = 22.90 m (to 4 s.f.) 3 For P,

b

y = 32t sin  55° − 5t2 11.45 = 32t sin  55° − 5t2 Solving using the quadratic formula gives: t = 0.4809 or t = 4.762 When y = 22.90, 22.90 = 32t sin  55° − 5t2 Solving using the quadratic formula gives: t = 1.108 or t = 4.135 So total time between this range = (1.108 – 0.4809) + (4.762 – 4.135) = 1.25 s (to 3 s.f.) 17 Range of P 122 sin 60° 36 3 R= = 10 5 Range of Q R=



When x = 2.412, y = 5(2.412) – (2.412)2 = 6.24 m

Mathematics in life and work 1 u = 5, q  = 30°, with block, h = 0.6 Without block: 52 sin 60° U 2 sin 2θ = = 2.165 m 10 g With block: R=

When y = 11.45,

2

opposite which is the gradient adjacent of the curve dy Since y = 5x − x2, dx = 5 − 2x = tan q

19 a tan  q = 

20 2sin 2 60° = 20 = 15 m

2

V sin 90° V = 10 10

Time of flight, y = 0 1 y = Ut sin q − gt2 + h 2 0 = (5 sin 30°)t − 5t2 + 0.6 5t2 − 2.5t − 0.6 = 0 Using the quadratic formula: t = −0.1772 or t = 0.6772 So time of flight = 0.677 s. x = ut cos q   = 5 × 0.6772 × cos 30°  = 2.932 m So he can dive 0.767 m further using the starting block.

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1 Motion of a projectile

2 Without the block: 2U sinθ = 2 × 5 × sin30° = 0.5 s g 10 So speed at landing:

Time of flight = vx = U cos q = 5 cos 30° = 4.330 m s−1 vy = U sin q − gt

= 5 sin 30° − 10 × 0.5 = −2.5 m s−1 Enters the water at an angle of tan −1

2.5 = 30.0° 4.33

With the block: Time of flight = 0.6772 s vx = U cos q = 5 cos 30° = 4.330 m s−1 vy = U sin q − gt = 5 sin 30° − 10 × 0.6772 = −4.272 m s−1 Enters the water at an angle of tan

−1 4.272

4.330 = 44.6°

He will enter the water at the smallest angle when diving from the poolside. 3 u = 5.2, q =  30°, h = 0.6 Time of flight, y = 0 1 y = Ut sin q − gt2 + h 2 0 = (5.2 sin 30°)t − 5t2 + 0.6 5t2 − 2.6t − 0.6 = 0 Using the quadratic formula: t = −0.1731 or t = 0.6931 So time of flight = 0.693 s x = ut cos q = 5.2 × 0.6931 × cos 30° = 3.121 m So he will be able to travel an extra 18.9 cm.

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WORKED SOLUTIONS

2 

2 Equilibrium of a rigid body Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

Prerequisite knowledge

5 Both methods are valid. Both methods lead to an answer of the form Fd sin q. Chen’s method is more efficient in this example.

1 W = mg W = 0.125 × 10 = 1.25 N 2 Horizontal component = 9 cos 52° = 5.54 N Vertical component = 9 sin 52° = 7.09 N 3 F = µR 7.2g µ= F = = 0.4 R 18g 4 a Resolving vertically, R + 20 sin 30° = 5g, R = 40 N b F = µR = 0.25 × 40 = 10 N c Resultant force = ma

20 cos 30° − 10 = 5a



a = 1.46 m s−2

Exercise 2.1A 1 a i 6 × 5 = 30 N m b i 28 ÷ 4 = 7 N

ii  3.5 × 12 = 42 N m ii  15 ÷ 4 = 3.75 N

2 Clockwise: 12 × 4 = 48 N m Anticlockwise: 17 × 3 = 51 N m Overall turning moment = 51 − 48 = 3 N m anticlockwise 3 a 3 × 9 = 27 N m clockwise b 5 × 4 + 3 × 5 = 35 N m clockwise c 5 × 9 = 45 N m clockwise d 5 × 15 − 3 × 6 = 57 N m clockwise 4 a Clockwise: 7 × 3 = 21 N m

Anticlockwise: 7 × 4 = 28 N m



Overall turning moment = 28 − 21 = 7 N m anticlockwise

b 7 × 3 + 7 × 4 = 49 N m clockwise c Clockwise: 7 × 3 = 21 N m

Anticlockwise: 7 × 4 = 28 N m



Overall turning moment = 28 − 21 = 7 N m anticlockwise

d 7 × 3 + 7 × 2 sin 45° = 30.9 N m anticlockwise

6 a Clockwise: 8 × 3 sin 30° = 12 N m Anticlockwise: 10 × 3 sin 60° = 25.98 N m Overall turning moment = 25.98 − 12 = 14.0 N m anticlockwise b The turning moment will still be anticlockwise but smaller (11.0 N m). 7 a 13 × 4 = 6 × 5 + 11d d = 2  b (15 − 11) × 2 = 8 N m anticlockwise 8 Let x m be distance of the man from the centre of the seesaw. 60g × x = 36g × 2 x = 1.2 m 9 a Taking moments anticlockwise 7 × 4 + 9 × 6 = 82 N m Turning moment = 82 N m anticlockwise b i Let force = F N 2 × F = 82 F = 41 N F acts vertically downwards. ii R + 9 = 7 + 41 R = 39 N 10 The value of U is correct but the working is incorrect. She has taking moments incorrectly with cos and sin. She has then also evaluated the compound angle formulae incorrectly. a Resolving vertically: T cos a  = U cos a T=U q=U b Taking moments about O: T × 8 sin a  = U × 3 cos a q × 8 sin a  = q × 3 cos a 3 tan a  = 8 a  = tan−1 3 = 20.6° 8

()

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2 EQUILIBRIUM OF A RIGID BODY

c Resolving horizontally:

V = T sin a + U sin a



V = q sin a + q sin a



V = 2q sin a 3 6q 73 V = 2q × = 73 73





Exercise 2.2A

1 a 3.7x = 0.2 × 2 + 0.7 × 1 + 0.8 × 0.5 + 1.2 × 0.2

3.7x = 1.74



x = 0.470 m from A

9x = 16.9

So ( x , y ) = (2.30, 3.83) d 680x = 230 × 2 + 342 × 4 + 108 × 9 x = 4.12 680 y = 230 × 3 + 342 × 1 + 108 × 2 y = 1.84 2 12.5x = 0 × 5 + 4 × 2.5 + 4 × 2 + 0 × 3 x = 1.44 12.5 y = 0 × 5 + 0 × 2.5 + 2.5 × 2 + 2.5 × 3

x = 1.88 m from A

y =1

c 4mx = 0.3 × m + 1 × m + 1.2 × m + 1.3 × m 4mx = 3.8 m

So ( x , y ) = (1.44, 1) 3 4mx = 2 × m + 6 × m + 2 × m + 6 × m

x = 0.95 m from A



d 10.5mx = 0.2 × 2m + 0.7 × 6m + 0.8 × 2m + 1.2 × 0.5m

x =4 4m y = 5 × m + 5 × m + 8 × m + 8 × m

10.5mx = 6.8m

y = 6.5

x = 0.648 cm from A

2 12x = 0 × 2 + 0.4 × 2 + 0.8 × 2 + 1.2 × 2 + 1.6 × 2 + 2 × 2

So ( x , y ) = (4, 6.5) 4 (10 + m) × 4 = 2 × 1 + 3 × 8 + 5 × 4 + 2 × m

12x = 12

40 + 4m = 46 + 2m

x = 1 m from A

m = 3 kg 13 y = 2 × 1 + 3 × 1 + 5 × 5 + 3 × 3

3 (3m + 3.2) × 1.2 = 0.2 × m + 1.3 × 2m + 2.5 × 3.2 3.6m + 3.84 = 2.8m + 8

0.8m = 4.16



8 + 1.6m + 1.6n = 4n + 8

18 000 = 150d + 5000

12.5 + 2.5m + 2.5n = 10 + 5m

d = 86.7 cm from A 5 (3m + 3) × 6 = m × 1 + 2m × 7 + 3 × 9

m = 3 kg

Exercise 2.3A

Exercise 2.2B

1 a 8y = 3 × 2 + 5 × 5 y = 3.875



x = 4.625



So ( x , y ) = (4.63, 3.88)

b 10x = 2 × 5 + 5 × 6 + 3 × 5 10y = 2 × 3 + 5 × 2 + 3 × 0

10

Solving simultaneously: gives n = 2 and m = 3

3m = 9

So ( x , y ) = (5.5, 1.6)

2

2.5n − 2.5m = −2.5 Substituting m = 1.5n into 2

18m + 18 = 15m + 27

x = 5.5

1

2.4n − 1.6m = 0 (5 + m + n) × 2.5 = 3 × 0 + n × 0 + 5 × 2 + 5 × m

150d = 13 000

1 a 8x = 3 × 4 + 5 × 5

y =3 5 (5 + m + n) × 1.6 = 3 × 0 + 4 × n + 4 × 2 + 0 × m

m = 5.2 kg

4 450 × 40 = 200 × 10 + 100 × 30 + 150 × d



y = 3.83

x = 2.30

So ( x , y ) = (4.12, 1.84)

b 9x = 0.9 × 2 + 2.1 × 3 + 2.2 × 4

c 7.6x = 1.4 × 5 + 4.3 × 2 + 1.9 × 1 7.6 y = 1.4 × 1 + 4.3 × 6 + 1.9 × 1

y = 1.6

Shape

Area

x-coordinate of centre of mass

y-coordinate of centre of mass

Large rectangle

54

6.5

5

Small rectangle

12

9

9.5

Lamina

66

x

y

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2 

WORKED SOLUTIONS



66x = 54 × 6.5 + 12 × 9

66 y = 54 × 5 + 12 × 9.5

x = 6.95 (to 3 s.f.)

y = 5.81 (to 3 s.f.)



e Shape

Area

b Shape

x-coordinate of centre of mass

Area

x-coordinate of centre of mass

y-coordinate of centre of mass

Large rectangle

72

9

7

Large rectangle

60

9

4.5

Small rectangle

18

4.5

13

Medium rectangle

21

13.5

10.5

Circle

π

10

8

90 − π

x

y

Small rectangle

15

Lamina

96

Lamina 4.5

9.5 (90 − π) x = 72 × 9 + 18 × 4.5 − 10 × π

x

y

(90 − π) y = 72 × 7 + 18 × 13 − 8 × π x = 8.03 (to 3 s.f.)

96x = 60 × 9 + 21 × 13.5 + 15 × 4.5

Shape

c

Shape

Area

x-coordinate of centre of mass

y-coordinate of centre of mass

Large rectangle

90

9

8.5

Medium rectangle

10

11

8.5

Small rectangle

6

6

10.5

Lamina

74

x

y

74x = 90 × 9 − 10 × 11 − 6 × 6 74 y = 90 × 8.5 − 10 × 8.5 − 6 × 10.5 x = 8.97 (to 3 s.f.)

Area

y = 6.59 (to 3 s.f.)

x = 9.28 (to 3 s.f.)

y = 8.34 (to 3 s.f.)

d

Shape

Area

x-coordinate of centre of mass

y-coordinate of centre of mass

Rectangle

60

8

7

Triangle

15

8

11

Lamina

75

x

y

x = 8

75 y = 60 × 7 + 15 × 11 y = 7.8

x-coordinate of centre of mass

y-coordinate of centre of mass

Rhombus

32

8

10

Rectangle

4

8

10.5

Lamina

28

x

y

28x = 32 × 8 − 4 × 8 x = 8

28 y = 32 × 10 − 4 × 10.5 y = 9.93 (to 3 s.f.)

2 From diagram, x = 6

Using formula with r = 2, a = p 2 p 2 × 2sin 2 = 2.85 y =2+ 3p 2

3 Distance of centre of mass from (4, 2) 2 × 4 × sin p 4 = 2.401 = 3× p 4 p So x = 4 + 2.401 cos  = 5.70 4 p and y = 2 + 2.401 sin  = 3.70 4 4 Using formula with r = 10, a = p 12 Distance from O to centre of mass = = 6.59 cm

75x = 60 × 8 + 15 × 8

y = 8.21 (to 3 s.f.)

f

96 y = 60 × 4.5 + 21 × 10.5 + 15 × 9.5

y-coordinate of centre of mass

2 × 10 sin p 12 3p 12

5 Length of median = 32 − 1.52 = 2.60 2 So centre of mass lies × 2.60 = 1.73 cm along 3 each median from the vertex.

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2 EQUILIBRIUM OF A RIGID BODY

Exercise 2.4A 1 3 r = 3 × 4 = 1.5 8 8 The centre of mass lies 1.5 cm above the centre of the base of the hemisphere.

6 Assuming the origin is at the bottom left of the cross-section.

2 a 3 h = 3 × 16 = 12 so centre of mass lies 12 cm 4 4 below the vertex. This means it is 4 cm above the base. 3 3 b h = × 16 = 12 so centre of mass lies 12 cm 4 4 below the vertex. This means it is 4 cm above the base. c The centre of mass of a solid cone does not depend on its radius, only its height. 3 x and y will lie directly above the centre of the base of the cuboid. Shape

Volume

Cuboid

z-coordinate of centre of mass

128

4

Pyramid

32

8 + 1 × 6 = 9.5 4

Solid

160

z



Shape

Area

x-coordinate of centre of mass

y-coordinate of centre of mass

Large rectangle

1500

15

25

Small rectangle

750

45

12.5

Crosssection

2250

x

y

2250x = 1500 × 15 + 750 × 45 2250 y = 1500 × 25 + 750 × 12.5 x = 25

y = 20.8 (to 3 s.f.)

z = 25 (due to the symmetry of the step) 7 Assuming the centre of the flat base of the cylinder is (y, z) = (0, 0)

160z = 128 × 4 + 32 × 9.5

Shape

Volume

Cylinder

πr2 r = πr3

Hemisphere

2 pr 3 3

Solid

5 3 pr 3

z = 5.1 The centre of mass is 5.1 cm above the base. 4 x = 0 and y = 0 due to the symmetry of the solid. Shape

Volume

z-coordinate of centre of mass

Cone

261.8

3 × 10 = 7.5 4

Hemisphere

261.8

Solid

523.6



10 +

r 2 2r +

19 3 r= r 8 8 x

5 3 r 2 19 pr x = pr 3 × + pr 3 × r 3 2 3 8 5 3 25 4 pr x = pr 3 12 13 x= r 8

The centre of mass is 13 r from the end of the 8 sculpture.

3 × 5 = 11.88 8 z

x-coordinate of centre of mass

8 x = 12.5 and y = 12.5 because of symmetry. Shape

523.6z = 261.8 × 7.5 + 261.8 × 11.88

Area

z = 9.69 5 3 r = 6 so r = 16 cm 8



z-coordinate of centre of mass

Large square

625

12.5

Small square

25

2.5

Cross-section

600

z

600z = 625 × 12.5 − 25 × 22.5 z = 12.1 (to 3 s.f.)

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WORKED SOLUTIONS

Exercise 2.5A

e

R

1 a R

W

25g N 20g N

2m

40g N

30g N

40g N

4m 4m

f

R2

8m b T

T 12 m

A

R1

B C

2.5 m

0.8 m 5m

9g N

2 a It is rigid, straight and has uniform mass. b Moments about C:

6m 15 m c 5g N

3g N

A

B



50g × 3 + 60g × 0.5 = 30g × 2 + RD × 2



RD = 60g = 600 N



Resolving vertically:



RC + RD = 30g + 60g + 50g



RC + 60g = 140g



RC = 80g = 800 N

3 a Let x m be the distance of the centre of mass from A.

W 6m d RC

RD

A

B C

D

6m

4m 12 m 6m



Moments about C:



2g × 4 + 20g(x − 1) = 30g × 1



20(x − 1) = 22



x − 1 = 1.1



x = 2.1 m

b Resolving vertically:

20g N 2m

7g N



RC = 30g + 20g + 2g



RC = 52g = 520 N

4 Let A be the end of the pole that he is holding and let B represent the position of his other hand. Moments about A:

5g × 2 = RB × 0.8



RB = 12.5g = 125 N



Resolving vertically:



RA acts in the same direction as W (also described in the solution to 1f).

RA + 5g = 12.5g RA = 75 N

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2 EQUILIBRIUM OF A RIGID BODY

5 a Let JL = x m

Moments about L:



40g(12 − x) = 24gx + 16g(x − 6)



480 − 40x = 24x + 16x − 96



576 = 80x



x = 7.2 m

b Resolving vertically:

d The boulders are particles with their masses concentrated at a single point so that the perpendicular distances are exact. 8 a Let the weight of the rod be W and the distance of the centre of mass of the rod from C be x m.

For the 10 kg object:

5 Resolving vertically: W + 10g = RA1 + RB1 = 4 RB1 RB1 = 4 (W + 10g) 5 Moments about A:



RL = 24g + 16g + 40g





RL = 80g = 800 N

W(x − 3) + 10g × 7 = RB1 × 5 4 W(x − 3) + 70g = (W + 10g) × 5 5 W(x − 3) + 70g = 4W + 40g

c The masses are concentrated at a single point and the centre of mass of the beam is at its midpoint. 6 a Let string be attached at A.





W(x − 3) = 4W − 30g



For the 5kg object:



Moments about A:





7g(0.375 − d) = 5gd + 3g(0.375 + d)





2.625 − 7d = 5d + 1.125 + 3d





1.5 = 15d d = 1  m or 10 cm 10 b T = 3g + 5g + 7g



7 Resolving vertically: W + 5g = RA2 + RB2 = RB2 3 3 RB2 = (W + 5g) 7 Moments about A:



RA + RB = 64g + 16g



W(x − 3) = RB2 × 5 + 5g × 3 3 W(x − 3) = (W + 5g) × 5 + 15g 7 15 W(x − 3) = (W + 5g) + 15g 7 15 Hence 4W − 30g = (W + 5g) + 15g 7 28W − 210g = 15W + 75g + 105g



RA + 3RA = 80g



13W = 390g

4RA = 80g



W = 30g = 300 N

RA = 20g



The weight of the rod is 300 N.



T = 15g = 150 N

7 a Resolving vertically:

RB = 20g × 3 = 60g = 600 N b Let AB = x



b W(x − 3) = 4W − 30g

30g(x − 3) = 120g − 30g



Moments about A:



30g(x − 3) = 90g



64g × 3 + 16g × 6 = 60gx



x−3=3



288 = 60x x = 4.8 m

c Resolving vertically:

RA + RB = 64g + Mg



RA + 5RA = 64g + Mg



6RA = 64g + Mg 1 RA = (64g + Mg) 6 Moments about B:



Mg × 1.2 + RA × 4.8 = 64g × 1.8 1 Mg × 1.2 + (64g + Mg) × 4.8 = 64g × 1.8 6 7.2M + 307.2 + 4.8M = 691.2



12M = 384



M = 32



x=6

The distance of the centre of mass from C is 6 m.

Exercise 2.5B 1 Since TK = 0, the position of K is not required. Let the position of the particle from A be x m. Moments about J: 6g × d = 9g(d − x) 6gd = 9gd − 9gx 9x = 3d 1 x= d 3 1 The particle is d m from A. 3

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2 

WORKED SOLUTIONS

2 Let the distance of the centre of mass from A be x m.

M(x − 2) = 54

RC = 0

Mx − 2M = 54

Moments about D:

Mx = 2M + 54

4g × 0.7 = 10g (1.4 − x)

After block placed, RE = 0.

2.8 = 14 − 10x

Moments about F:

10x = 11.2

72g × 1.5 = Mg(5 − x)

x = 1.12 The distance of the centre of mass from A is 1.12 m. 3 a Since the rod is on the point of tilting about D, the reaction force at E is 0 N. b Moments about D:

mg × 6 + 70g × 1.5 = 50g × 3



6m + 105 = 150 6m = 45 m = 7.5

4 a Moments about P:

20g × 2 = RQ × 5



RQ = 8g = 80 N



Resolving vertically:



RP + RQ = 20g



RP + 8g = 20g



RP = 12g = 120 N

b RP = 0

Moments about Q:



Mg × 1 = 20g × 3



M = 60

5 a RV = 0

Moments about U:



5g × 8 + W × 3 = 70g × 1



W = 10g = 100

b Let the mass of the added load be M kg.

RU = 0



Moments about V:



(M + 5)g × 3 = 10g × 2 + 70g × 6



3M + 15 = 20 + 420



3M = 425 425 M= 3 The mass of the load that has been added is 425  kg. 3 6 Let x m be the distance of the rod’s centre of mass from A.

108 = M(5 − x) 108 = 5M − Mx Mx = 5M − 108 Hence 2M + 54 = 5M − 108 162 = 3M M = 54 Substituting M = 54 into Mx = 2M + 54: 54x = 108 + 54 54x = 162 x=3 The distance of the rod’s centre of mass from A is  3 m.

Exercise 2.6A 1 Let the reaction force acting between the ladder and the floor be RF, the friction force acting between the ladder and the floor be FF and the reaction force acting between the ladder and the wall be RW. a Resolving vertically:

RF = Mg N

b F = µR

FF = 0.4RF = 0.4Mg N

c Resolving horizontally:

RW = FF = 0.4Mg N

d Let the length of the ladder be 2a m.

Moments about point of contact of ladder with floor:



Mga sin a =  RW × 2a cos a Mg sin a =  0.4Mg × 2 cos a tan a =  0.8 a =  tan−1(0.8) = 38.7°

2 Let distance of the centre of mass of the ladder from B be x.

Before block placed:

Let the reaction force acting between the ladder and the ground be RB, the friction force acting between the ladder and the ground be FB and the reaction force acting between the ladder and the wall be RA.

Moments about E:

Resolving horizontally:

Mg(x − 2) = 18g × 3

RA = FB

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2 EQUILIBRIUM OF A RIGID BODY

Resolving vertically: 50g = RB Moments about B: 50g × x cos 70° = RA × 8 sin 70° 50g × x cos 70° = FB × 8 sin 70° 1 50g × x cos 70° = RB × 8 sin 70° 4 1 50g × x cos 70° = 4 × 50g × 8 sin 70° x = 2 tan 70° x = 5.49 m 3 Let the reaction force acting between the ladder and the ground be RG, the friction force acting between the ladder and the ground be FG and the reaction force acting between the ladder and the wall be RW. Resolving horizontally: RW = FG Resolving vertically: 36g + 45g = RG

4 F = RC 3 C 27g + 4 (324g − 6RC) = RC 3 27g + 432g − 8RC = RC

27g +

459g = 9RC RC = 51g N FC = 324g − 6 × 51g = 324g − 306g = 18g N F 18g 6 µC = C = = RC 51g 17 5 Let the reaction force acting between the ladder and the ground be RB, the friction force acting between the ladder and the ground be FB and the reaction force acting between the ladder and the wall be RA. Let the maximum distance the man can climb safely be x m. Resolving horizontally: FB = RA Resolving vertically: RB = 60g + 70g = 130g Moments about B:

RG = 81g Moments about G: 45g × 2 sin 30° + 36g × 3 sin 30° = RW × 6 cos 30° 198g sin 30° = 6RW cos 30° 33g tan 30° = RW RW = 11g 3 FG = µRG F R 11g 3 11 3 = µ= G = W = 81 RG 81g 81g 4 Let the reaction force acting between the ladder and the ground be RC, the friction force acting between the ladder and the ground be FC, the reaction force acting between the ladder and the wall be RD and the friction force acting between the ladder and the wall be FD. 1 F D = RD 6 Resolving horizontally:

60g × 1.75 cos a + 70g × x cos a = RA × 3.5 sin a 105g + 70gx = 3.5RA tan a 70 105g + 70gx = R 13 A 13 RA = (105g + 70gx) 70 RA = 19.5g + 13gx FB ⩽ µRB 2 FB ⩽ RB 5 FB 2 5 ⩾ RB RA 2 5 ⩾ 130g 2 19.5g + 13gx 5⩾ 130g 2 × 130g 5 19.5g + 13gx ⩽ 52g

19.5g + 13gx ⩽ 13gx ⩽ 32.5g

FC = RD = 6FD Resolving vertically:

x ⩽ 2.5

FD + RC = 54g

The man can safely climb 2.5 m.

FC = 6(54g − RC) = 324g − 6RC Moments about D: 54g × a cos a + FC × 2a sin a = RC × 2a cos a 54g + 2FC tan a = 2RC 27g + FC tan a = RC Given that sin a =

4 4 , tan a = . 5 3

6 Let the reaction force acting between the ladder and the ground be RC, the friction force acting between the ladder and the ground be FC, the reaction force acting between the ladder and the wall be RD and the friction force acting between the ladder and the ground be FD. FD = µRD and FC = µRC

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WORKED SOLUTIONS

Resolving horizontally:

2 

c Resolving vertically:

FC = RD



RA + RC cos 25° = 12g

Resolving vertically:



RA = 12g − 81.57 cos 25°

RC + FD = 40g



RA = 46.07 N

RC + µRD = 40g

d Magnitude = 34.472 + 46.072 = 57.5 N

RC = 40g − µRD

Direction = tan−1 46.07 = 53.2° to the horizontal 34.47 2 Let the magnitude of the components of the reaction at A be X and Y and of the tension in the cable be T. a Moments about A:

RC = 40g − µ(µRC) RC = 40g − µ2RC (1 + µ2)RC = 40g 40g RC = 1 + µ2 Moments about D:



40g + 2FC tan a = 2RC 20g + FC tan a = RC Given that sin a =

4 4 , tan a = . 5 3

4 F = RC 3 C

Hence

(

T = 46.2 N b Resolving horizontally:

X = T cos 60° = 23.09 N



Resolving vertically:



T sin 60° + Y = 8g Y = 4g

)

20g 40g = 4 1 + µ2 1 − 3 µ

)

4 40g  1 − 3 µ = 20g (1 + µ2) 2 − 8 µ = 1 + µ2 3 6 − 8µ = 3 + 3µ2 (3µ − 1)(µ + 3) = 0 µ = 1 or −3 3 Since 0 ⩽ µ ⩽ 1, µ =



Magnitude = 23.092 + ( 4g ) = 46.2 N



4g  Direction = tan−1  = 60.0° to the  23.09  horizontal

2

3 Let the friction at the wall be F N, the reaction at the wall be R N and the tension be T N. 12 a From the right-angled triangle, cos a = , 13 from which sin a = 5 . 13 Moments about A:

3µ2 + 8µ − 3 = 0

1 3

Exercise 2.6B 1 Let the reaction at the ground be RA and the friction at the ground be FA. a Moments about A:

12g × 0.9 cos 25° = RC × 1.2



RC = 81.6 N

b Resolving horizontally:

8g × 0.8 = T sin 60° × 1.6

4g + Y = 8g

20g + 4 µRC = RC 3 4 20g = RC 1 − µ 3 20g RC = 1 − 43 µ

(

)

T sin 60° = 4g

40g × a cos a + FC × 2a sin a = RC × 2a cos a

20g +

(



RC = 40g − µFC

FA = RC sin 25° = 81.57 sin 25° = 34.5 N

Mg × 4a = T sin a × 12a 1 T sin a = Mg 3 5 1 T× = Mg 13 3 13 T = Mg 15 b Resolving vertically:



F + T sin a = Mg 1 F + Mg = Mg 3 2 F = Mg 3 Resolving horizontally: R = T cos a 13 12 4 R = Mg × = Mg 15 13 5 F ⩽ µR

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2 EQUILIBRIUM OF A RIGID BODY

F µ⩾ R 2 Mg 3 µ⩾ 4 Mg 5 5 µ⩾ 6 4 Let the magnitude of the components of the reaction at A be X and Y.

7 a Resolving horizontally:

Moments about A:



6g × 5a = 2F × 2a + F × 10a



14F = 30g F = 300  N 14 Resolving vertically: Y + F = 6g Y = 6g − 300 = 540  N 14 14 Resolving horizontally: X = 2F = 600  N 14 540 = arctan 0.9 to the horizontal Direction = arctan  600

( )

5 Let the friction at the ground be F N and the reaction at the ground be R N. a Moments about A:

7.5g × 2 cos a = 49 × 2.5 cos a = 0.8167 a = 35.2°

b Resolving horizontally:

F = 49 sin a = 28.28 N



Resolving vertically:



R + 49 cos a = 7.5g



R = 34.98 N 28.27 µ= F = = 0.808 R 34.98 6 Let the magnitude of the components of the reaction at A be X and Y and the thrust be T. Let AB be x m. a Moments about A: 3 0.6g × 1 x = T cos 75° × 4 x 2 T cos 75 = 0.4g 0.4g T= = 15.5 N cos75° b Resolving vertically:

Y + 0.4g = 0.6g



Y = 0.2g = 2.00 N



Resolving horizontally:



X = T sin 75° = 14.93 N



Magnitude = 22 + 14.922 = 15.1 N

18

TA cos 20° = TB cos 50° TB cos50° cos20° Resolving vertically: TA =

TA sin 20° + TB sin 50° = 2g TB cos50° sin 20° + TB sin 50° = 2g cos20° TB(cos 50° tan 20° + sin 50°) = 2g 2g TB = = 2g = 20.0 N cos50° tan 20° + sin 50°

TA = 20cos50° = 13.7 N cos20° b Moments about A:



2gx = TB sin 50° × 3



x = 3 sin 50° = 2.30 

Exercise 2.7A 1 a At the point of sliding, F = μR

R = 12g so F = 0.32 × 12g = 38.40 N



P = F = 38.4 N

b R = 12g − P sin 15°

So F = 0.32(12g − P sin 15°)



So P cos 15° = 0.32(12g − P sin 15°)



0.32 × 12g P = (cos15° + 0.32sin15°) = 36.6 N

2 a Let A be the bottom right corner, which is the pivot point.

Moments (A) = 2g × 5 − 20P



At the point of toppling, moments about A = 0



So 10g − 20P = 0 P = 5 N

b At the point of sliding, F = μR

R = 2g so F = 0.3 × 2g = 6 N



P = F = 6 N

c The cuboid would topple first. 3 a Sliding: At the point of sliding, F = μR

F = 0.2 × 2.5g = 5 N



P = F = 5 N



Toppling: Let A be the bottom right corner, which is the pivot point.



Moments (A) = 2.5g × 20 − 50P



At the point of toppling, moments about A = 0



2.5g × 20 − 50P = 0



P = 10 N



So cuboid will slide first.

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2 

WORKED SOLUTIONS

b Sliding: At the point of sliding, F = μR

2 Let the distance of the centre of mass from A be x m.



F = 0.2(2.5g + P sin 10°)

Moments about V:



6mgd = 8mg (4d − x)



P cos 10° = 0.2(2.5g + P sin 10°) 0.2 × 2.5g P= = 5.26 N cos10° − 0.2sin10° Toppling: Let A be the pivot point.



Moments (A) = 2.5g × 0.2 − 0.5P cos 10°



At the point of toppling moments about A = 0



2.5g × 0.2 − 0.5P cos 10° = 0

RX + RY = 7g + 5g + 9g



P = 10.2 N

RX + 2RX = 7g + 5g + 9g



So cuboid will slide first.

3RX = 21g



6d = 32d − 8x 8x = 26d x = 3.25d 3 Resolving vertically:

RX = 7g, RY = 14g

4 a Sliding:

Resolving horizontally: F = 4g sin a



Resolving vertically: R = 4g cos a



At the point of sliding, F = μR



4g sin a = 0.3 × 4g cos a



tan a  = 0.3



a   = 16.7°

Let the distance WY be x m. Moments about W: 5g × 7 + 9g × 14 = 7g × 2 + 14gx 161 = 14 + 14x 14x = 147 x = 10.5

b Toppling occurs when the centre of mass is directly above the lower corner of the cuboid. 2.5 This occurs when a  = arctan 5 = 26.6°

( )

WY = 10.5 m 4 a The frustum is a cone of base 10 cm and height h with a cone of base 5 cm and height h − 8 removed.

c The cuboid will slide first.



5 a Sliding:

Resolving horizontally: F = 12g sin 20°



Resolving vertically: R = 12g cos 20°



At the point of sliding, F = μR



12g sin 20° = μ × 12g cos 20°



μ = tan 20° = 0.364

b Toppling occurs when the centre of mass is directly above the pivot point. 5 This occurs when a = arctan  12 = 22.6°

( )

Exam-style questions



Shape

Volume

Large cone

1676 (4 s.f.)

4

Small cone

209.4

10

Frustum

1467

z

z-coordinate of centre of mass



1467z = 1676 × 4 − 209.4 × 10



z = 3.14 cm above the centre of the base.

b Shape

1 a Let the reaction force at C be RC.

Volume

z-coordinate of centre of mass

Frustum

1467

3.142

60g + 90g + 40g = RC + 95g

Hemisphere

261.8

9.875

RC = 95g = 950 N

New solid

1729

z



Resolving vertically:



b Moments about C:

By similarity, h = 16 cm.

90g(x − 1) + 40g × 9 = 95g × 6 + 60g × 1 90(x − 1) + 360 = 570 + 60 90x − 90 + 360 = 630 90x = 360



1729z = 1467 × 3.142 + 261.8 × 9.875



z = 4.161 cm above the base.



So the centre of mass of the new solid is 4.161 − 3.142 = 1.02 cm higher than the centre of mass of the old solid.

x=4

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2 EQUILIBRIUM OF A RIGID BODY

5 a The rope is light and inextensible.

Moments about D:



W × 5 + 28 × 11 = TC × 7 5 TC = W + 44 7 b Resolving vertically:

(

Using M y =

)

6 y = 1 × ×



( )



1800 × 8 = W(8 − x)



14 400 = 8W − Wx



Wx = 8W – 14 400



Moments about B:



1575 × 8 = W(x − 2)



12 600 = Wx − 2W



Wx = 2W + 12 600



8W – 14 400 = 2W + 12 600



6W = 27 000



W = 4500



3 +1×1+1× 1 +1×0 2 2

(

(

∑i =1mi yi you get n

)

3 +2 +1×

)

3 +1 ×

(

)

3 +1 +1×

3 +1× 2

(

3 +1

3 +1 2

)



5 2 3+ 6 3 So centre of mass = (1, 2.11) (to 3 s.f.) y=

b The distance from centre of mass to centre of AB = 3 + 2 – 2.11 = 1.62 1.62 So θ = tan −1 = 58.3º 1 So angle between AB and vertical is 58.3º

( )

Let the angle between the ladder and the floor be a, where tan a = 2. Moments about B: 50g × 1.75 cos a + 70gx cos a = RW × 3.5 sin a 87.5g + 70gx = 3.5RW tan a 87.5g + 70gx = 3.5RW × 2 = 7RW Resolving horizontally: RW = FF Resolving vertically:

8 a Let the bottom left of the diagram be the point (0,0).

n

9 Let the reaction force between the ladder and the wall be RW, the reaction force between the ladder and the floor be RF and the friction force between the ladder and the floor be FF.

b Wx = 2W + 12 600 x = 4.8

∑i =1mi xi you get

6 y = 5 3 + 4

6 Toppling will occur when the centre of mass is directly above bottom left corner. 2.5 This occurs when q  = arctan  = 32.0° 4 7 a Moments about C:



Assume the mass of each rod is 1

6 x = 6 x = 1 (this can also be found from the  symmetry of the shape)

W = 84

4500x = 9000 + 12 600



6x = 1 × 1 + 1 × 2 + 1 ×

5 26 W + 44 = W − 208 7 7 252 = 3W



)

Centre of mass of AE = 0, 3 + 1

Using Mx =

TC + TD = W + 28 5 7 W + 44 + TD = W + 28 2 TD = W − 16 7 5 2 W + 44 = 13 W − 16 7 7

(



( Centre of mass of BC = ( 2,

) 3 + 1)

Centre of mass of AB = 1, 3 + 2



  Centre of mass of CD =  3 , 3  2 2 



Centre of mass of CE = 1, 3



1 3 Centre of mass of DE =  ,  2 2 

(

)

RF = 120g FF ⩽ µRF 1 RW ⩽ 3 × 120g 87.5g + 70gx ⩽ 40g 7 87.5g + 70gx ⩽ 280g 70gx ⩽ 192.5g x ⩽ 2.75 The person can climb 2.75 m before the ladder slides.

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2 

WORKED SOLUTIONS

13 Let A = (0, 0)

10 Let A be the point (0, 0) Shape

Area

y-coordinate of centre of mass

Large circle

4πr2

0

Small circle

πr2

r

Lamina

3πr2

y

Shape

Semicircle

r 3

r 3 11 a Let the reaction at A be RA and the friction at A be FA The distance from A to the centre of mass is



Moments about B:



2.5g × 3.6 cos a + FA × 7.2 sin a = RA × 7.2 cos a



2.5g + 2FA tan a = 2RA 1 FA = RA 2 2.5g + RA tan a = 2RA 3 3 Given that sin a = , tan a = 5 4 3 2.5g + RA × = 2RA 4 5 2.5g = 4 RA RA = 2g = 20 N



b Resolving vertically:

RA + T sin b = 2.5g



T sin b = 2.5g − 2g = 0.5g



0.28T = 0.5g



T = 17.9 N

x-coordinate of centre of mass

pr 2 2

r+

y-coordinate of centre of mass

4r 3p

r2 2

2 r 3

1 r 3

Lamina

r2 (1 + π) 2

x

y

(

)

r2 4r r2 2 pr 2 a 2 (1 + p)x = 2 r + 3p + 2 × 3 r 2

3

3



r pr 2r r3 (1 + p)x = + + 2 2 3 3



p r2 (1 + p)x = r 3 + 1 2 2



(1 + π) x = r(π + 2)



x=

( )

r(2 + p) (1 + p)

r2 r2 1 b 2 (1 + p)y = 2 × 3 r

r2 r3 (1 + p)y = 2 6 r (1 + p)y = 3 r y= 3(1 + p)

c Let q  be the angle between the vertical and AO

c The rod is one-dimensional, straight and rigid, and the cable is a light inextensible string.



12 Sliding:

x=

(1 + p)(2 + p) = 2+p (1 + p)

y=

(1 + p) 1 = 3(1 + p) 3

 1  q  = tan −1  3  = 3.71°  2 + p  

14 a Let the weight of the rod be W N and the distance of the centre of mass from A be x m.



Resolving horizontally: F = 5g sin a



Resolving vertically: R = 5g cos a



At the point of sliding: F = μR





5g sin a = 0.6 × 5g cos a

20g × 4.8 + W(x − 1.2) = 100g × 1.2



a  = arctan 0.6 = 31.0°

W(x − 1.2) = 24g



Toppling: will occur when the centre of mass is directly above bottom right corner. 1 This occurs when a  = arctan 4 = 14.0

Wx − 1.2W = 24g



0

Triangle 3πr 2 y = 0 × 4πr 2 + πr 3 y=

Area

()

So the solid will topple when the angle a reaches 14.0°.

On the point of tilting about C, RD = 0.

Wx = 1.2W + 24g

On the point of tilting about D, RC = 0.

250g × 0.8 = 20g × 5.2 + W(5.2 − x) 96g = 5.2W − Wx Wx = 5.2W − 96g

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2 EQUILIBRIUM OF A RIGID BODY



Resolving vertically:

120g = 4W



FW + RG = Mg

W = 30g



RG = Mg − FW





RG = Mg − µWRW



RG = Mg − µWFG



RG = Mg − µWµGRG



RG(1 + µWµG) = Mg



RG =



Hence 1.2W + 24g = 5.2W − 96g

The mass of the rod is 30 kg.

b Wx = 1.2W + 24g

30gx = 1.2 × 30g + 24g

30x = 60 x = 2

The centre of mass is 2 m from A.

15 a Shape

Length

x-coordinate y-coordinate of centre of of centre of mass mass

Mg 1 + µ W µG

b Moments about the point of contact of the ladder with the wall:

RG × 8a cos q = FG × 8a sin q + Mg × 2a cos q



4RG cos q = 4µGRG sin q + Mg cos q

πr

4RG = 4µGRG tan q + Mg

0

2r (below) p



Semicircle



RG(4 − 4µG tan q) = Mg

Straight rod

2r

0

0



RG =

Mass at A

2r

–r

0

Mass at B

πr

r

0

2r(π + 2)

x

y

Lamina

2r(p + 2)x = pr2 − 2r2 2r(p + 2)x =

r2(p

− 2)

2



x=

r (p − 2) 2r(p + 2)



x=

r(p − 2) 2(p + 2)



x = r(p − 2) 2p + 4

b 2r(p + 2)y = pr ×

y=

2r 2 2r(p + 2)



y=

r (p + 2)

Mg Mg c Hence 1 + µ µ = 4 − 4µ tan θ W G G

1 + µWµG = 4 − 4µG tan q



µWµG = 3 − 4µG tan q



µW =

3 − 4µG tan θ µG

17 a Assuming origin is at centre of base of cross-section Shape

2r p

2r(p + 2) y = 2r2

Mg 4 − 4µG tan θ

c When r = 5, x = 5(p − 2) and y = 5 (p + 2) 2(p + 2) 2 −1  y  So q  = tan  x  = (p − 2) = 60.3° 16 a Let the friction force between the ladder and the wall be FW, the reaction force between the ladder and the wall be RW and the friction force between the ladder and the ground be RG.

Resolving horizontally:



RW = FG

Area

x-coordinate of centre of mass

y-coordinate of centre of mass

Rectangle

24

0

3

Triangle

6

0

7

Lamina

30

x

y

30 y = 24 × 3 + 6 × 7

x = 0



Height of shape = 6 + 3 = 9

y = 3.8

 o centre of mass is 9 – 3.8 = 5.2 below apex S of triangle. b Sliding:

Resolving horizontally: F = 8g sin q



Resolving vertically: R = 8g cos q



At the point of sliding: F = μR



8g sin q  = 0.25 × 8g cos q

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2 

WORKED SOLUTIONS



q  = arctan 0.25 = 14.0°



Toppling will occur when the centre of mass is directly above the pivot point. 2 This occurs when q  = arctan  3.8 = 27.8°



( )

So the solid will slide down the slope first, when the angle q  reaches 14.0°.



( )

2 This occurs when q  = arctan  6.37 = 17.4° So the solid will topple first, when the angle q reaches 17.4°.

20 Let A = (0, 0) Shape

Area

x-coordinate of centre of mass

Rectangle

60

5

3

Semicircle

39.27

5

8.122

Triangle

15.59

11.73

3

Lamina

114.86

x

y

18 a Let DC = x, angle OCD = 30º

r 3r In triangle DOC, tan 30º = r so x = 3 = x 3 3



Using Pythagoras’ theorem on triangle BCD



 6r   3r  BD 2 =  −  3   3 

2

=

2

36r 2 9r 2 + 3 3

a 114.86x = 60 × 5 + 39.27 × 5 + 15.59 × 11.73 114.86x = 679.22

27r 2 3 = 9r2

=





BD = 9r 2 = 3 r

b 114.86 y = 60 × 3 + 39.27 × 8.122 + 15.59 × 3 114.86 y = 545.72

y tan θ = = 3rr = 2 3 3 x



2 3



y = 4.751

Vertical distance between centre of mass and AB = 4.75 cm

So q = 49.1° 19 a Shape

x = 5.913

Horizontal distance between centre of mass and AE = 5.91 cm

b Let q be the angle between ED and the vertical

y-coordinate of centre of mass

Volume

z-coordinate of centre of mass

Cylinder

502.7

5

Hemisphere

134.0

11.5

Solid

636.7

z

c Let q  be the angle between the vertical and AB 4.751 q = tan −1 = 38.8° 5.913

(

Mathematics in life and work 1 Let h be the height of the full pyramid. By similarity, 2(h − 1.15) = h so h = 2.3 m.

636.7z = 502.7 × 5 + 134.0 × 11.5

z = 6.37



The centre of mass of the solid is 6.37 cm above the flat base.

b Sliding:

Resolving horizontally: F = 5g sin q



Resolving vertically: R = 5g cos q



At the point of sliding: F = μR



5g sin q = 0.32 × 5g cos q



q = 17.7°



Toppling will occur when the centre of mass is directly above the pivot point.

)

2

Shape

Volume

z-coordinate of centre of mass

Full pyramid

0.7667

0.575

Small pyramid

0.09583

1.4375

Model

0.6709

z

0.6709 z = 0.7667 × 0.575 − 0.09583 × 1.4375 z = 0.452 The centre of mass of the model is 45.2 cm above the base.

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2 EQUILIBRIUM OF A RIGID BODY

3 Sliding: At the point of sliding, F = μR. R = 125g so F = 0.4 × 125g = 500 N P = F = 125 N Toppling: Let A be the bottom right corner, which is the pivot point. Moments (A) = 125g × 0.50 − 1.15P At point of toppling moments about A = 0. So 125g × 0.50 − 1.15P = 0 P = 543.5 N So the solid will slide first.

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3

WORKED SOLUTIONS

3 Circular motion Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

Prerequisite knowledge 1 2 3 4 5 6

720° = 4π radians 5 × 2π = 10π radians π π 60° = 3 radians. Arc length = rq = 12 × 3 = 4π cm PE = mgh = 5 × 10 × 6 = 300 J 1 1 KE = mv2 = × 3000 × 82 = 96 000 J 2 2 Total energy = PE + KE 1 = 500 × 10 × 10 + × 500 × 152 = 106 250 J 2

Exercise 3.1A 1 2

3 4

5

6

7

8

60 r.p.m. = 60 × 2π = 120π radians per minute = 2π radians per second The second hand moves at a speed of 2π radians π per minute. This is the same as radians per 30 second. π π v = wr = × 0.1 = 300 m s−1 = 0.0105 m s−1 30 15 × 2π π a ω = = rad s−1 60 2 π b v = w r = × 1 = π m s−1 = 1.57 m s−1 2 2 5 × 2 π −1 = π rad s ω= 10 v = w r = π × 0.6 = 0.6π m s−1 = 1.88 m s−1 a t = 60π = 12π seconds = 37.7 seconds 5 b ω = v = 5 = 0.167 rad s−1 r 30 84π = 58.643 seconds a Priya: t = 4.5 88π Sanjay: t = = 61.436 seconds 4.5 Difference in time = 2.79 seconds v 4.5 b Priya: ω = = = 0.10714 rad s−1 r 42 Sanjay: ω = v = 4.5 = 0.10227 rad s−1 r 44 Difference = 0.00487 rad s−1 Distance travelled = v × t = 15 × 35 = 525 m = circumference of the circle C 525 r= = 2π 2π = 83.6 m

2π π 9 ω = 24 × 60 × 60 = 43 200 rad s−1, π v = w r = 43 200 × 6 371 000 = 463 m s−1

Exercise 3.2A 1 2 3

a=

v2 32 = = 36 m s−2 r 0.25

v 2 4.52 = = 6.75 m s−2 r 3 2 60 × 1000 2 60 × 60 v = 6.17 m s−2 (to 3 s.f.) a= = r 45 directed towards the centre of the circle a=

(

)

4

mv 2 0.03 × 4.8 2 = = 3.46 N directed towards r 0.2 the centre of the circle

5

a

F =

R rw

2

F

0.05 g b Friction force = mrw 2 = 0.05 × 0.25 × 32 = 0.113 N 6

Resolving vertically: R = 2000g = 20 000 N 2000 × 50 × 1000 2 60 × 60 mv Resultant force, F = = r 50 = 7716 N The car is in limiting equilibrium so F = µR:

(

)

2

µR = 7716 so µ = 0.386 7

Resolving vertically: R = 600g = 6000 N 2 2 Resultant force, F = mv = 600 × v r 42 The motorbike is in limiting equilibrium so F = µR: 600 × v 2 = 0.6 × 6000 42 So v = 15.9 m s−1

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3 Circular motion

8 Resolving vertically: R = 0.1g = 1 N mv 2 0.1 × v 2 = Resultant force, F = r 0.25 The toy is in limiting equilibrium so F = μR: 0.1 × v 2 = 0.6 × 1 0.25 v = 1.22 m s−1 The turntable must be rotating at a speed of 1.22 m s−1 for the toy to slide off. 9 Resolving vertically: R = 1500g = 15 000 N 1500 × 45 × 1000 2 60 × 60 mv Resultant force, F = = r r The car is in limiting equilibrium so F = μR:

(

(

1500 × 45 × 1000 60 × 60 r

)

By Newton’s second law (horizontally): 2       R cos 20° = mv r 0.35 × 52 10.233 cos 20° = r       r = 0.910 m

5 Let q be the angle between the normal reaction force and the horizontal. Resolving vertically: R sin q = 0.08g

)

2

2

= 0.2 × 15000 r = 78.1 m

10 Resolving vertically: R = mg = 10m Resultant force, F = mrw 2 = mxw 2 For the object not to slide, F  μR mxw 2  10mμ xw 2  10μ 10µ w2  x



1 a r = l sin q = 0.5 × sin 30° = 0.25 m l cos θ 0.5cos 30 = 2π = 1.31 s g 10 c Resolving vertically: T cos q = mg b t = 2π

T cos 30° = 2.5 × 10

T = 28.9 N 35 2 a sin q ° = so q ° = 39.5° 55 b Resolving vertically: T cos q = mg

Radius of circle is 0.3 cos q. So R cos q = 0.08 × 0.3 cos q × 102 R = 2.4 N

0.08g so q = 19.5°. 2.4 Perpendicular height of the ball below top of hemisphere, h = 0.3 sin 19.5° = 0.100 m

This gives sin θ =

So distance from base = 0.3 − 0.1 = 0.2 m 6 Let q be the angle between the normal reaction force and the horizontal. 0.4 So cos q ° = 0.6 , q = 48.1897° Resolving vertically: R sin q = 0.3g R = 4.0249 N

Exercise 3.3A



Resolving horizontally: R cos q = mrw2 where r is the radius of the circle

T cos 39.5212° = 0.4 × 10

T = 5.19 N g 10 = c ω = = 4.85 rad s−1 0.55cos 39.5212 l cos θ a = rw 2 = 0.35 × 4.85492 = 8.25 m s−2 d v = rw = 0.35 × 4.8549 = 1.70 m s−1 3 a Resolving vertically: T cos q = mg T × 0.6 = 0.6g 0.9 T = 9 N g 10 b ω = = 4.08 rad s−1 = l cos θ 0.9 × 0.6 0.9 4 Resolving vertically: R sin 20° = 0.35g R = 10.233 N

   Resolving horizontally: R cos q = mrw 2 ω 2 = R cosθ = 4.0249cos48.1897° = 22.4 mr 0.3 × 0.4         w = 4.73 rad s−1 7 Let q be the angle between the pendulum string and the vertical. Resolving vertically: T cos q = mg (where T is the tension) Resolving horizontally: T sin q = mrw 2 Equating to eliminate T: mg mrω 2 = cosθ sin θ g rω 2 = cosθ sin θ g sin θ ω2 = r cosθ Since r = sin q and h = cos q : gr ω2 = rh g ω= h 8 Resolving vertically: R sin q = mg Resolving horizontally (and using Newton’s second law): R cos q = mrw2

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3 

WORKED SOLUTIONS

Equating to eliminate R: mg mrω 2 = sinθ cos θ g r 2 = ω sin θ cos θ g cos θ ω2 = r sin θ g 2 ω = r tan θ

ω=

g r tan θ

Exercise 3.4A 1 a  Total energy = mgh + 1 mv2 2 = 2 × 10 × 1.5 + 0.5 × 2 × 32 = 39 J b At the lowest point the object will have no PE so: 1 mv2 = 39 2 1 × 2 × v2 = 39 2 v = 6.24 m s−1 2 At highest point of circle: Total energy = mgh + 1 mv 2 2 = 0.3 × 10 × 1.4 + 0.5 × 0.3 × 1.52 = 4.5375 J At the lowest point, the object will have no PE so: 1 mv2 = 4.5375 2 1 2 2 × 0.3 × v = 4.5375 v = 5.50 m s−1

3 a  Assuming the base of the swing has a height of 0 m:



Height of release point = 2.4 − x where x is the perpendicular distance of the swing to the point where the rope is attached. x=  2.4 cos 30° so height of swing = 2.4 − 2.4 cos 30° = 0.32154 m

b Total energy = mgh = 30 × 10 × 0.32154 = 96.5 J 1 c At lowest point: 2 mv 2 = 96.46 v = 2.54 The speed at the lowest point is 2.54 m s−1. 4 a  The maximum speed occurs at the bottom of the circle.

Assume h = 0 at base of circle.



Energy at start = mgh = 3 × 10 × 2 = 60 J 1 At lowest point: 2 mv2 = 60   v = 6.32 m s−1



b The maximum force occurs at the bottom of the circle. 2 F = mv = 3 × 40 = 60 N r 2

5 a  Height at point of release =  2.5 − 2.5 cos 40° = 0.58489 m

Total energy at point of release = mgh = 45 × 10 × 0.58489 = 263.2 N



Maximum speed occurs at the lowest point of the circle. At lowest point: 1 mv2 = 263.2 2 v = 3.42 m s−1



b When the angle made by the rope and the downward vertical is 25°, h =  2.5 − 2.5 cos 25° = 0.234 m 1 At this point mgh + mv2 = 263.2 2 1 45 × 10 × 0.23423 + × 45 × v2 = 263.2 2 v = 2.65 m s−1 6 a  When the angle with downward vertical is 110°, height = 0.75 + 0.75 sin 20° = 1.01 m b Total energy at the end of the swing = mgh = 0.5 × 10 × 1.0652 = 5.0326 J 1 Energy at start: mgh + 2 mv2 = 5.0326 1 0.5 × 10 × 0.75 + 2 × 0.5 × u2 = 5.0326  u = 2.27 m s−1 1 7 a At the bottom of the circle, total energy = mv2 2 1 = × m × (4v)2 = 8mv2 2 1 Total energy at top: mgh + mv2 2 1 = mg(2l) + 2 mv2

Energy is conserved so 2mgl + 1 mv2 = 8mv2 2 4gl + v2 = 16v2 4gl = 15v2 15v 2 l = 4g

b Maximum force is at the bottom of the circle. 2 2 64gmv 2 64gm Fmaximum = mv = m(4v2) = N = r 15 15v 15v 2 4g Minimum force is at the top of the circle. 2 2 2 Fminimum = mv = mv 2 = 4mv2 = 4m N r 15 15v 15v 4 8 At the bottom of the circle, total energy 2 1 1 3 = mv 2 = m 3gr = gmr J 2 2 2 Height at any point = r + r sin (90° − q ) = r(1 + cos q )

(

)

Total energy at any point: mgh + 1 mv2 = 3 gmr 2 2 1 3 mgr(1 + cos q ) + mv2 = gmr 2 2 2gr(1 + cos q ) + v2 = 3gr

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3 Circular motion

v2 = 3gr − 2gr(1 + cos q ) v2

= 3gr − 2gr − 2gr cos q

v2

= gr − 2gr cos q

v2

= gr(1 – 2 cos q )

v=

gr(1 − 2cos θ )

b It will complete a full circle if the speed at the top is greater than 0 and the reaction force does not decrease to zero before it reaches the top of the circle.

At top: PE = mgh = 0.1 × 10 × 1 = 1 J



In order for it to have KE at top, total energy must be greater than 1 J. So at bottom: 1 mv2 > 1 2 1 × 0.1 × u2 > 1 2 u > 4.47 m s−1

Exercise 3.5A 1 PE at top = 2 × 10 × 1.5 = 30 J  or complete motion, total energy must be F greater than 30 J when at the bottom to allow for the object to have some velocity at the top. 1 mv2 > 30 2 1 × 2 × v2 > 30 2 v > 5.48 m s−1 2 a  Total energy = mgh + 1 mv2 2 = 0.1 × 10 × 0.6 + 0.5 × 0.1 × 52 = 1.85 J 1 At bottom: mv2 = 1.85 2 v = 6.08 m s−1 b The speed at the top of the circle must be greater than 0 m s−1 and the reaction force must not decrease to zero before it reaches the top of the circle. c At top: total energy = 0.1 × 10 × 1.2 + 0.5 × 0.1v2 = 1.85

v = 3.61 m s−1

When v = 3.61 ms−1, 2 reaction force = 0.1 × 3.61 − 0.1 × 10 > 0. 0.6 This is greater than 0 so the marble will complete a full circle inside the pipe. 3 a Total energy = 1 mv2 = 0.5 × 0.25 × 42 = 2 J 2 When it comes to rest, it will have only PE.



0.25 × 10 × h = 2

h = 0.8 m 0.2 b q = cos = 78.5° 1 It will make an angle of 78.5° with the downward vertical when it comes to rest. −1

( )

4 a  It will oscillate around its starting point if it comes to rest before it reaches a height of 0.5 m. 1 Energy at start = 2 mv2 = 0.05u2 When h = 0.5, PE = mgh = 0.1 × 10 × 0.5 = 0.5

So 0.05u2 = 0.5, u = 3.16 m s−1



To oscillate around its starting point, u  3.16 m s−1.



2 Reaction force = 0.1 × 4.47 − 0.1 × 10 > 0 0.5 To complete a full circle, u > 4.47 m s−1. 5 Total energy at start = mgh + 1 mv2 2 = 0.04 × 10 × 0.6 + 0.5 × 0.04 × 2.32 = 0.3458 J



When v = 0, mgh = 0.3458 so h = 0.865 m.  he particle will come to rest when h = 0.865 m. T This is above the horizontal so the particle will leave the circle and follow projectile motion. 6 a Total energy at bottom = 1 mv2 2 = 0.5 × 0.4 × 62 = 7.2 J 1 At top: mgh + mv2 = 7.2 2 1 0.4 × 10 × 1 + × 0.4 × v2 = 7.2 2 v = 4 m s−1 2 2 b Total force = mg + mv =  0.4 × 10 + 0.4 × 4 0.5 r = 16.8 N 7 a  Energy at top of bowl = mgh + 1 mv 2 2 = 0.15 × 10 × 0.5 + 0.5 × 0.15 × 2.32 = 1.14675 J



The height at any point is h = 0.75 cos q. 1 So at any point mgh + mv2 = 1.14675 2 Total energy = 0.15 × 10 × 0.75 cos q + 0.5 × 0.15 × v2 = 1.125 cos q + 0.075v2 ( = 1.14675)

b Applying Newton’s second law towards the centre of the base: mv 2 mg cos q − R = r v2 When R = 0, 10 cos q = 0.75 1.14675 − 0.75cos θ 2 From total energy equation, v = 0.075 1.14675 − 1.125 cos θ So 7.5 cos q = 0.075 0.5625 cos q = 1.14675 − 1.125 cos q 1.6875 cos q = 1.14675 1.14675 q = cos−1 = 47.2° 1.6875

(

)

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3 

WORKED SOLUTIONS

8 At the top of the hemisphere, total energy = mgh = 3 × 10 × 2 = 60 J Let q be the angle between the line joining the ball and the centre of base of the sphere and the upwards vertical. At any point on the hemisphere total energy = mgh + 1 mv2 = 60 2 3 × 10 × 2 cos q + 0.5 × 3 × v2 = 60 60 cos q + 1.5v2 = 60 40 cos q + v2 = 40 Applying Newton’s second law towards the centre of the base, 2 mg cos q − R = mv r v2 When R = 0, 10 cos q = 2

0.6 so q = 27.5° 1.3 b Resolving vertically: T cos q = mg 1.5 × 10 T = cos 27.5 = 16.9 N 2 a sin θ =

3 a  Total energy at top = 20 × 10 × 2 + 0.5 × 20 × 22 = 440 J

0.5 × 20 × v2 = 440 v = 6.63 m s−1 mv 2 b T − mg = so T = 1080 N r g 10 = 4 a ω = = 1.5302 rad s−1 5cos 31.33 l cos θ since θ = sin −1

ω = 1.5302 = 0.589 m s−1 r 2.6 b Resolving vertically: T cos q = mg 50 × 10 T = cos 31.3° = 585 N

30 cos q = 20 q = 48.2° The ball leaves the hemisphere when q = 48.2°.

5 a Resolving vertically: R = 2500g = 25 000 N mv 2 2500 × v 2 = Resultant force, F = r 80

At this point v2 = 40 − 40 cos 48.2° v = 3.65 m s−1 The ball leaves the hemisphere at a speed of 3.65 m s−1. 1 1 9 a Total energy at bottom of circle = 2 mv2 = 2 mu2 1 1 At any point, total energy = mgh + mv2 = mu2 2 2 At any point, height = r − r cos q = r(1 − cos q ) 1 1 So mgr(1 − cos q ) + mv2 = mu2 2 2 2gr(1 − cos q ) + v2 = u2 v2 = u2 − 2gr(1 − cos q ) b Resolving towards the centre of the circle: mv 2 T − mg cos q = r m T = mg cos q + (u2 − 2gr(1 − cos q)) r





( )

0.5 = 19.5° 1.5

ω = 2.66 = 5.32 m s−1 r 0.5 l cos θ 1.5cos19.5° t = 2π = 2π b = 2.36 s g 10 v=

(

2500 × 45 × 1000 2 60 × 60 Resultant force, F = mv = r 80 = 4883 N The car is in limiting equilibrium so F = μR. µ = F = 4883 = 0.195 R 25 000

)

2

90 × 2π 6 a ω = 60 = 3π rad s−1 b Resolving vertically: R = 0.2g = 2 N

g 10 = = 2.66 rad s−1 1.5cos19.5 l cos θ

since q = sin −1

The car is in limiting equilibrium so F = μR. 2500 × v 2 = 0.7 × 25 000 80 Maximum speed, v = 23.7 m s−1 = 85.2 km h−1

b Resolving vertically: R = 2500g = 25 000 N

Exam style questions



( 2.65 ) = 31.33°

v=

v2 = 40 − 40 cos q 40 − 40cos θ So 10 cos q = 2 10 cos q = 20 − 20 cos q

ω= 1 a 

Maximum speed is at the bottom.





Resultant force, F = mrw 2 = 0.2 × d × (3π)2 = 17.765d N



When the object is in limiting equilibrium, F = μR.



17.765d = 0.4 × 2   d = 0.0450 m

 he maximum distance from centre that T the object can be without sliding off the turntable is 4.50 cm. r 7 a sin q = = 0.28 1.5 So r = 1.5 × 0.28 = 0.42 m

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3 Circular motion

(1.05 − 0.6 )

28 96 b sin θ = 100 or cos θ = 100



So sin q =

g 10 = = 2.64  rad s−1 lcos θ 1.5 × 96 100 c Resolving vertically,



So string becomes slack at an angle of 48.6º above the horizontal.

ω=



T cos q = 3g

T = 31.25 N 60 × 2π = 2π rad s−1 8 ω = 60 Resolving vertically: R = 0.3g = 3 N Resultant force, F = mrw 2 = 0.3 × 0.1 × (2π)2 = 1.184 N

0.6 So q = 48.6º

12 a Total energy = 0.5 × m × 30x = 15mx 1 At any point, total energy = mgh + 1 mv2 = mu2 2 2 At any point, height = x − x cos q = x(1 − cos q) 1 So mgx(1 − cos q ) + mv2 = 15mx 2 2gx(1 − cos q) + v2 = 30x

When the object is in limiting equilibrium, F = μR.

v2 = 30x − 20x(1 − cos q)

µ = F = 1.184 = 0.395. R 3 The least possible value of μ is 0.395.

v2 = 10x + 20x cos q

9 a  Resolving vertically: T = 2mg sin q (where q is angle between the horizontal and the string) 2mg 2mg 100m T= = 25m N = = sin θ 4 4 5 2 b Resolving horizontally: mv = 2mg cos q r v2 3 = 20 × 3x 5

v2 = 10x(1 + cos q)

Resolving towards the centre of the circle:



2 T − mg cos θ = mv r

m (20x(1 + cos q)) x T = 10m cos q + 20m(1 + cos q))

T = mg cos q +

()

v = 36x m s−1 10 a Height = 4 − 4 cos 50° = 1.4289 m Total energy = mgh + 1 mv2 2 = 0.05 × 10 × 1.4289 + 0.5 × 0.05 × 32 = 0.939 J At the bottom: 1 mv2 = 0.939 2 v = 6.13 m s−1 b When v = 0, mgh = 0.939 so h = 1.88 m.

It will reach a height of 1.88 m before changing direction and returning towards its starting point.

11 a Total energy = mgh + 1 mv2 2 = 0.03 × 10 × 0.6 + 0.5 × 0.03 × 32 = 0. 315 J

At q below the horizontal Total energy = 1 × 0.03 × v2 2 + 0.03 × 10 × (0.6 − 0.6 sin q ) = 0.315



0.015v2 + 0.18 − 0.18 sin q = 0.315



0.015v2



v2

= 0.135 + 0.18 sin q = 9 + 12 sin q

b When the string becomes slack, v = 0 and the reaction force < 0, so it only has PE.

When v = 0, mgh = 0.315 so h = 1.05 m.

T = 30m cos q + 20m T = 10m(3 cos q + 2) b When T = 0:

3 cos q + 2 = 0



cos  q = −2 3

( 32 )

v2 = 10x 1 −

10 x 3 13 a Let j be the angle between the horizontal and OB. 0.5 5 = but q = 90 − j, so Then sin j = 0.7 7

v2 =

5 = cos q 7 b  Applying Newton’s second law towards the centre of the base: mv 2 mg cos q − R = r When R = 0: v2  10 cos q = 2 v2 = 20 cos q

sin(90° − q ) =

v = 3.78 m s−1 c  At the top of the hemisphere: total energy = 0.4 × 10 × 0.7 + 0.5 × 0.4 × u2 = 0.2u2 + 2.8







At any point on hemisphere: total energy = mgh + 1 mv2 = 0.2u2 + 2.8 2



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3 

WORKED SOLUTIONS

0.4 × 10 × 0.7 cos q + 0.5 × 0.4 × v2 = 0.2u2 + 2.8 5 2.8 × 7 + 0.2v2 = 0.2u2 + 2.8 2 + 0.2v2 = 0.2u2 + 2.8



T + mg = 2m(u2 − g)



T + mg = 2mu2 − 2mg

0.2v2 = 0.2u2 + 0.8



v2 = u2 + 4 u = 3.21 m s−1 14 Let the radius of the cone be r and the height be 4 r . 3 The half-angle at the vertex of the cone q 3 satisfies tan θ = . 4 Resolving vertically: R sin q = mg R = 10m = 10m = 50m sin θ 3 3 5

T = 2mu2 − 3mg 3g . 2 c Let min speed (at top of circle) be v.

Therefore 2u2  3g and u 



This means the max speed (at bottom of circle) is 2v.



The speed of projection is u.



At top: loss in KE = gain in GPE 0.5m(u2 – v2) = 0.5mg u2 – v2 = g

()



0.5m (4v2 – u2) = 0.5mg

Resolving horizontally and using Newton’s second law:

4v2 – u2 = g

R cos q = mxw2 80xm mx 80 3r 3r 100xm R= = = 3r cosθ 4 Equating: 5 50m 100xm = 3 3r 150mr = 300xm x = r (i.e. the radius of the circle traced out 2 by the object is half the radius of the cone) Using trigonometry, d = r tan θ = 3r so the 2 4 perpendicular distance between the circle and the vertex of the cone is 3r where r is the 4 radius of the cone.

()

)





5g 3 16 a Tension in the string = 8g = 80 N

Tcos q = 4g cos q = 40 = 0.5 80 b Resolving vertically mv 2 r 2 4 v = 2v2 8g sin q = 2 1 3 If cos θ = , then sin θ = 2 2 8g 3 = 2v 2 so v 2 = 2 3 g 2 17 At lowest point, total energy = 0.5mv2 T sin θ =

The maximum speed occurs at the bottom of the circle where h = 0 At this points, KE = 1 mv2 = 5m 2 So v2 = 10 v = 3.16 m s−1.

So max angular speed ω = v = 3.16 = 6.32 rads−1. r 0.5 b At top: loss in KE = gain in GPE



0.5m(u2 – v2) = 0.5mg



u2 – v2 = g



so v2 = u2 – g 2 Also T + mg = mv but r = 0.5 so r T + mg = 2mv2



Substituting ➀ in ➁ gives

➁

➀ + ➁ gives 3v2 = 2g 2g v2 = 3 2g 2 So from  ➀  u = +g 3 5g u2 = 3   u =

15 a Initially, total energy = m × 10 × 0.5 = 5m

➀

At bottom: gain in KE = loss in GPE 0.5m ((2v)2 – u2) = 0.5mg

Let x be the radius of the circle followed by the object.

( ) (

For complete circles, T  0 so 2mu2 − 3mg  0

At highest point, total energy = 0.5mv + 1.5mg Energy is conserved so 0.5mv2 = 0.5mv + 1.5mg 0.5v2 = 0.5v + 15 v2 = v + 30

➀

v2 – v – 30 = 0  (v – 6)(v + 5) = 0

➁

v = 6 or v = –5 So v = 6 m s−1. So maximum speed is 6 m s−1.

31

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3 Circular motion

18 Let x = radius of circle and r = radius of bowl. Let q be the angle between OP and the circular face.



 esolving these horizontally, by Newton’s R second law:



T sin q = 2ma

a sin θ = 0.5r = 0.5. Therefore q = 30° r b Resolving vertically, 10sin q = mg 10 × 0.5 Therefore m = = 0.5 kg 10 2 c Resolving horizontally, 10cos q = mv x



T sin q = 2mrw2



2 T = 2mrω sin θ Equating the above two equations: 2mg 2mrω 2 = cosθ sin θ



3 and x = (r 2 − (0.5r )2) 2



cos θ =



3r 2 10 3 0.5v 2 = Therefore 2 3r 2 This gives v2 = 15r so v = 15r



x=



Total energy = 0.5 × 2 × 102 = 100 J



So at any point,



Energy = 0.5 × 2 × v2 + 2 × 10 × l (1 – cos q  ) = v2 + 20l (1 – cos q )



So v2 + 20l (1 – cos q  ) = 100 So v2 = 100 – 20l (1 – cos q  )



Resolving towards O



2 T – 20 cos q = 2v l



2 T = 20 cos q + 2v l 200 − 40l + 40lcos θ T = 20 cos q + l 200 T = 20 cos q + − 40 + 40 cos q l

T = 60 cos q + 200 − 40 l b The max l for complete circular motion would require v = 0 at top of circle and reaction force to be greater than 0.

So mgh = 100

2 × 10 × 2l = 100 40 l = 100



Resolving vertically:



T cos q = 2mg 2mg T= cosθ



As q increases towards 90°, w increases.



So range of w  values is w   1.83 rad s–1

Mathematics in life and work 1 Resolving vertically: R = mg = 10m N

mv 2 m × v 2 = r 25 The go-kart is in limiting equilibrium so F = μR.

Resultant force, F =

2 m × v = 0.8 × 10m 25

v2 =8 25 v = 14.1





The maximum speed at which the go-karts can travel without sliding is 14.1 m s−1 or 50.8 km h−1. 2 2 When wet m × v = 0.5 × 10m 25 2 v =5 25 v = 11.2

The maximum speed at which the go-karts can travel without sliding is 11.2 m s−1 or 40.2 km h−1.



So the speed would decrease by 2.9 m s−1 or 10.6 km h−1.

l = 2.5 m 20 a Let T be the tension in the string



10 3cos θ Min value = 1.83 rad s−1

b ω =

19 a The height of P at any point =  l – l cos q = l (1 – cos q )

Since sin q = r . 3 g rω 2 cos θ = r 3 g cos θ = 3ω 2

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WORKED SOLUTIONS

4

4 Hooke’s law Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

Prerequisite knowledge 1

RB T B

RA

T

m

4m

3

A

Loss in potential energy = mgh = 250 × 10 × 4 = 10 000 J Loss in kinetic energy = 1 mu2 – 1 mv2 2 2 (where u is speed at start and v final speed) 1 1 = 2 × 250 × 22 – × 250 × 12 = 375 J 2 Work done against resistances = 400 × 100 = 40 000 N

40° 30°

30°

mg

40° 4mg

a RA = 4mg cos 40° RB = mg cos 30° F = ma for

Work done by carriage is 40 000 – 10 000 – 375 = 29 625 J (as the energy lost by the motion provides some of the work required to overcome the resistances).

4mg sin 40° – T – 0.1mg cos 40° = 4ma 1 T – mg sin 30° – 0.15mg cos 30° = ma

2

Adding equations 1 and 2 together

Exercise 4.1A

4mg sin40° – 0.1mg cos 40 – mg sin 30° – 0.15mg cos 30° = 5ma

1

40sin 40 – cos 40° – 10sin 30° – 1.5cos 30° = 5a

a = 3.73 ms–2

2

b Using equation 2: T – mg sin 30° – 0.15mg cos 30° = 3.73m T = 3.73m + 10m sin 30° + 1.5m cos 30°

l = 86.6 cm

c Using parallelogram of forces and the cosine rule: F2 = (10.03m)2 + (10.03m)2 – 2 × 10.03m × 10.03m cos 70

3

F = 11.5m N a Resolve vertically at Z:

4

2T sin q = 4mg T sin q = 2mg 10mg T= 3 b Resolve vertically at X:

40 × 0.2 3 T = 2.67 N λx l= T 1.4 − l l = 60 × 37 37l = 84 − 60l T=

97l = 84

T = 10.03m N

2

3mgµ = T cos q 3mgµ = 8mg 3 8 µ= 9 Carriage has lost both kinetic energy and potential energy since it has both slowed and dropped

5

33 = l × 0.3 0.9 l = 33 × 3 = 99 N T = 32 × 0.75l l T = 24 N

Rx = 3mg

The percentage of the compression of the spring x = × 100 l 120 × (l − x) 1.4g = l 14l = 120l − 120x

Resolve horizontally at X:

120x = 106l

Rx = mg + T sin q

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4 HOOKE’S LAW

x 106 = l 120 Therefore percentage of the compression of the spring is 88.3%. 6

T = 2g T = 20 N

1.5 − l 20 20l = 45 − 30l l = 30 ×

50l = 45 7 8

l = 90 cm 350 × x 5.5 × 10 = 0.2 x = 3.14 cm The tension acts throughout the strings and is equal to 0.5g = 5 N. For S1: 5l1 = 17 × x1 5 × 1 = 17x1

2x = 10(0.7 − x) 12x = 7 x = 0.583 m y = 0.117 m 11 T = 130 × 0.1 0.5 T = 26 N Resolving parallel to the plane: 26 = 4g sin q 26 sin q = 40 q = 40.5° 12 Let x be the extension in the string attached to point A. Let y be the extension in the string attached to point B. 2+x+3+y=7 x+y=2 16x Tx = = 8x 2 14y Ty = 3 14y 8x = 3 24x = 14y

x1 = 0.294 m For S2: 5l2 = 20x2 5 × 0.6 = 20x2 x2 = 0.15 m Total length is 1 + 0.294 + 0.6 + 0.15 = 2.04 m 9

Resolving vertically: 2g + T2 = T1 Let x1 be the extension in SP and x2 be the extension in TP.

24x = 14(2 − x) 38x = 28

1.5 + x1 + x2 = 2.7 x1 + x2 = 1.2 45x1 T1 = = 60x1 0.75 45x 2 T2 = = 60x2 0.75 2g + 60x2 = 60x1

x = 0.737 m y = 1.26 m 13 Let x be the extension in string AP. Let y be the extension in string BP. l

x2 = 0.433 m

P

0.5 + x

20 + 60x2 = 60(1.2 − x2) 120x2 = 52

2l

A

1 kg

Tx

x1 = 0.767 m

x + y = 0.7 λx TX = 0.5 3λ y TY = 0.3 As the particle is in equilibrium TX = TY so: λ x = 3λ y 0.5 0.3 2x = 10y

Ty

B

1g

10 Let x be the extension in string attached to X. Let y be the extension in string attached to Y.

0.75 + y

1.5 m y = 0.15 m (given in the question) x = 1.5 − 0.5 − 0.75 − 0.15 = 0.1 m First calculate the tensions in the strings so you know in which direction friction will be acting. Tx = λ × 0.1 = 0.2l 0.5 Ty = 2λ × 0.15 = 0.4l 0.75 Given that Ty > Tx friction will act to oppose force Ty.

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4

WORKED SOLUTIONS

16

Tx + µR = Ty

l = 100 N

R

0.2l + 0.3 × 1 × 9.8 = 0.4l

0.3 m

0.2l = 2.94 l = 14.7 N 0.1 × 60 14 T = = 20 N 0.3 Resolve vertically: 20 cos q = 2g

40 N 4gcosq

cos q = 0.98 q = 11.478… = 11.5°

4g

4gsinq

q q

If sin q = 0.8 then cos q = 0.6.

l = 60 N

100 × 0.2 = 40 0.5 Resolving perpendicular to the slope: Thrust in compressed spring =

T

R = 40 sin q + 4g cos q = 40 × 0.8 + 39.2 × 0.6 R = 55.52 N Resolving parallel to the slope: F

2 kg

4g sin q + 40 = 40 cos q + µ × R 40 × 0.8 + 40 = 40 × 0.6 + µ × 55.52 72 = 24 + 55.52µ

2g

µ = 0.865

Resolve horizontally: 20 sin 11.48° = F

17 Let T be the force in the stretched string.

F = 3.98 N

F = ma

15

T − µR = 0.5 × 0.25 Tx P

2.6 m Ty

T − 0.5 × 4.9 = 0.125

1 + x S1 l = 25

T = 2.575 2.575 = 40 × x 0.2 x = 0.013

0.5 kg 1 + y S2 l = 25

Length of string = 0.2 + 0.013 = 0.213 m

Exercise 4.2A Let x be the extension in S1. Let y be extension in S2. x + y = 0.6 Tx = 25x Ty = 25y Resolving vertically: 25x = 25y + 0.5g 25x = 25(0.6 − x) + 5 50x = 20 x = 0.4 m

8 × 0.52 = 2J 2 × 0.5

1

W=

2

EPE =

3

T=

7 × 0.52 = 1.25 J 2 × 0.7

λx = 15x = 2g l 20 = 15x

x = 1.33 2 EPE = 15 × 1.33 = 13.3 J 2×1

y = 0.2 m

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4 HOOKE’S LAW

4

50 × 1.52 = 28.125 J 2×2 Using conservation of energy:

EPE =

KE gain = EPE loss 1 mv2 = 28.125 2 v2 = 56.25 5

w 2 = 16.96 w = 4.12 rad s−1

The stretched length of the string is 1.17 m. 48 × (3l)2 2×l 60 = 216l

60 =

8

EPE =

loss of EPE = PE gain 100 × 1.92 100 × 1.4 2 − = 41.25 J 2×2 2×2

2 1.3g(0.5 + x) = 20 × x 2 × 1.5 20x2 − 39x − 19.5 = 0

50 × 4 2 = 400 J 2×1 By conservation of energy:

x = 2.36 or −0.413

Loss of EPE = gain in KE + work done against friction 1 400 = 2 × 2.5 × v2 + 0.3 × 2.5 × 10 × 4 v2 = 296

Let the extension at equilibrium point be y. 20 × y 1.3g = 1.5 y = 0.975 m

v = 17.2 m s−1 9

10 loss of EPE = gain in KE 5 × 1.32 = 1 × 2 × v2 2 × 1.7 2 v = 1.58 m s−1 11 As at lowest point v = 0

l = 0.278 m Gain in EPE =

where

52.22 sin 40° = 4 × 0.4947 × w 2

v = 7.5 m s−1 2000 × x 2 30 = 2×1 x2 = 0.03

7

T sin q = mrw 2 T = 0.019581 × 2000 = 52.22 0.75 r = 0.7696 sin 40° = 0.4947

x = 0.173

6

F = ma

Resolving vertically: T cos 40° = 4g x × 2000 cos 40° = 40 0.75 x = 0.0196 The stretched length of the string is 0.770 m. A l = 2000 N 40° 0.75 m

So x = 2.36 and the lowest point is 3.86 m below X.

KE gained + EPE gained = PE lost 2 1 × 1.3v2 + 20 × 0.975 = 1.3 × 10 × 1.48 2 × 1.5 2 0.65v2 + 6.3375 = 19.24 v = 4.46 m s−1 12 WD =

1000 × 40 2 = 2670 J 2 × 300

13 Gain in EPE = loss in GPE λ × 1.52 2 × 0.5 = 0.7 × 10 × 2 l = 6.22 N 14 Model the doll as a particle and assume that motion is vertical (not side to side). At start total energy is EPE and PE: 80 × 0.22 EPE = = 5.33 J 2 × 0.3 PE = 0.6 × 10 × 0.1 = 0.6 J

x

4g

At maximum distance the spring will have extended by x so will still contain EPE, v = 0 so KE = 0 and it will have gained height so will also have gained PE. 80 × x 2 EPE = = 133.3x2 2 × 0.3 PE = 0.6 × 10 × (x + 0.2) = 6x + 1.2

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4

WORKED SOLUTIONS

v2 = 60 − 2 × 5 × 2

By conservation of energy: 133.3x2

v = 6.32 m s−1

+ 6x + 1.2 = 5.93

133.3x2 + 6x − 4.73 = 0 x = 0.167 or −0.212 (this is the start position) So the doll rises a maximum distance of 0.3 + 0.167 = 0.467 m.

The speed at which the car crashes into the wall is 6.32 m s−1. 17 Assume that man falls exactly 70 m and starts 70 m above ground. At start: GPE = 86 × 10 × 70 = 60 200

15 A l

P X

x

m

q

m

At bottom: gain in EPE = loss of GPE 5000 × x 2 = 60 200 2×l 120 400l = 5000x2 Given that the stretched length of rope must not exceed 70 m: x + l = 70 So the equation becomes:

mg q Relative to A: At point P: EPE = 0 GPE = −mgl sin q = − 5 mgl 13 λ × x2 At point X: EPE = 2×l 5 GPE = −mg(l + x) sin q = − mg(l + x) 13 At both points KE = 0 as at rest.

24.08l = x2 24.08l = (70 − l )2 l 2 − 164.08l + 4900 = 0 l = 125 or 39.3 Hence maximum value for the rope’s natural length is 39.3 m. 18 Length of elastic = 2 × (0.0752 + 0.302)0.5 = 0.61847 m Loss of EPE = gain in KE 50 × 0.418472 = 1 × 0.1 × v2 2 × 0.2 2 v2 = 437.8 v = 20.9 m s−1

By conservation of energy: −

5 5 λ × x2 5 mgl = − mgl − mgx + 13 13 13 2×l

2 5 mgx = λ × x 13 2×l 10lmg x= 13λ 8 × 22 16 At start: EPE = = 8J 2×2 After the car has travelled 2 m and the string becomes slack:

EPE lost = gain in KE + work done against resistance 8 = 1 × 0.2v2 + 2 × 1 2 v2 = 60 v = 7.75 m s−1 So the velocity at the point that the string stops pulling car is 7.75 m s−1. Calculate the deceleration: F = ma

19 The model plane starts at equilibrium point. 15x Resolve vertically: 0.4g = 0.4 x = 0.107 m When plane has been pulled down: 15 × 0.6 2 = 6.75 J 2 × 0.4 Find the speed when string becomes slack after release:

EPE =

Loss of EPE = gain in KE + gain in GPE 1 6.75 = 2 ×0.4 × v2 + 0.4 × 10 × 0.6 6.75 = 0.2v2 + 2.4 v = 4.66 m s−1 Using this as u for motion under gravity, calculate how far the plane would travel if the ceiling was not present. v2 = u2 + 2as

1 = 0.2a

0 = 21.75 + 2 (−10)s

a = 5 m s−2 (deceleration)

s = 1.09 m

v2 = u2 + 2as

So the plane would hit the ceiling.

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4 HOOKE’S LAW

20 At top: GPE = 5 × 10 × 3 = 150 J

1200 × 0.13852 = 1200 × 0.12 + 1 × 2 × v 2 2 × 0.4 2 × 0.4 2

After falling 2 m as the string is still slack:

v2 = 13.8

Loss in GPE = gain in KE 150 – 5 × 10 × 1 = 1 × 5 × v2 2 v = 6.32 m s−1

v = 3.71 m s–1 2 A

Loss in GPE = gain in EPE 2 150 − 5 × 10 × (1 − x) = 9400 × x 2 100 + 50x = 4700x2

5m

5m

q

4m

T

4700x2 − 50x − 100 = 0

B

T

x = 0.15 or −0.14 So the length of the string when the particle first comes to instantaneous rest is 1.15 m. 21 a KE = 1 × 300 × 102 = 15 000 J 2 b By conservation of energy: EPE transferred to elastic rope = 15 000 J

P mg a Length AP = 52 + 4 2 = 6.403 m T = 60 × 1.403 = 28.1 N 3 Resolve vertically: force up = force down as in equilibrium

2000 × x 2 = 15 000 2×5 x = 8.7 m Distance from O will be 13.7 m.

2T sin q = mg

c The calculation above assumes no resistance forces.

mg = 2 × 28.1 × m = 3.51 kg

2 b EPE = 60 × 2.806 = 39.4 J 2×6

Exam-style questions 1

a

3

λx l λ 300 = × 10 40 l = 1200 N T=

B

q T

a Loss in GPE = gain in KE + gain in EPE 2mg × x 2 1 × m × v2 + 2×l 2 2g × x 2 2 2g(l + x) = v + l 2gx 2 2 v = 2g(l + x) − l 4gx dv = 2g − b 2v l dx mg(l + x) =

25 cm

b A

10 cm

T (view from above)

P

4gx dv = 0,2g − =0 dx l 1 x = 2l 2g 1 l 2 1 v 2 = 2g l + l − 2 l

(

v2 = T = 1200 × 6.926 = 416 20 Resolving perpendicular to the line AB 2T sin θ = 2a 2 × 416 × 10 = 2a 26.926 a = 154 m s–2 c By conservation of energy EPE at start = EPE at O + KE at O

38

4 = 35.1 N 41

( )

)

2

5gl 2

5gl 2 c When v = 0: 2gx 2 2g(l + x) − =0 l x2 l+x− =0 l x2 – xl − l2 = 0 v=

x=

(1 + 5 )l 2

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4

WORKED SOLUTIONS

4

EPE lost = gain in GPE + gain in KE O

56 mg × 10.369l 2 ( ) 1 3 13 2× = 4mg × 7l + 2 mv2 2 × 2l

q

l = 0.3 m l = 60 N

56 g × 107.52l = 56gl + v2 3 13 v2 = 501gl 6

a GPE lost = EPE gained

45 N

P

l

X

6g Resolve vertically: T cos q = 60 Resolve horizontally: T sin q = 45 2mg

tan q = 0.75, so sin q = 0.6 T × 0.6 = 45

30°

T = 75 N

5mgx 2 2l 2l = 5x2

Compare the equation for EPE and tension in an extended string: EPE = xT 2 8 = 37.5x

2mg × sin 30° = x=

b Greatest speed is when a = 0.

x = 0.213 m

This is when force up the slope = force down slope.

So the extended length of the string is 0.3 + 0.213 = 0.513 m.

T = mg sin 30° 5mgx = mg sin 30° l

2

5

 13l  7l a AX = (3l )2 +  = 2  2  3l

A

0.4l

3l 13 2

T

l 10 Gain in KE + gain in EPE = loss in PE x=

B

q

5mg ( 10l ) 1 l 2+ mv = mg sin 30° 2l 2 10 lg lg − v2 = 10 20 2

T

X

l = 2l l = l mg

4mg 2T sin q = 4mg 13 2 = 2mg T 7 2 14mg T= 13 3l 14mg λmg 2 = 2l 13

λ = 56 3 13 b AX = (3l)2 + (12l)2 = 12.369l Extension in each string is 10.369l.

lg 10 a When at rest v = 0, hence kinetic energy = 0. v=

7

EPE gained = GPE lost 19.6x 2 = (2 + x) × 0.9g 4 19.6x2 − 36x − 72 = 0 x = 3.04 (or −1.21) OA = 5.04 m b Resolve vertically: F = ma 19.6 × 3.04 − 0.9 × 10 = 0.9a 2 a = 23.1 m s−2

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4 HOOKE’S LAW

8

a KE at start + EPE at start = EPE at end − loss in GPE

(

)

(

)

λ 2 13 − 5 13 mg = 5 3 65mg 5 13 mg = l= 78 − 15 13 3 2 13 − 5

3l 2

T=

2m l 2m

Using Hooke’s law:

10 a EPE released = KE gained

λ ( 2l ) m2gl = 2l 2 l = 8mg 2

60°

b EPE released = gain in PE + gain in KE 8mg ( 34l ) 3l 1 = mg + mv 2 4 2 2l 2

60° 1 × 2m × U 2 + 3mg ( 2 2l − 2mgl sin 60° U2

)

= 3mg ( 2l

l 2 2

9lg 3lg 2v 2 = + 4 4 4 2 v = 3lg

)

3l 2 2

v = 3lg

27gl 8 3 gl 3gl − − = 8 8 8

(

U2 = gl 3 − 3

11 a Resolve vertically: 0.45λ 0.9g = 1.3 l = 26 N

)

(

U = gl 3 − 3

)

b Original EPE = EPE at instantaneous rest − loss in GPE 26 × 0.4 2 26 × x 2 − 0.9g(x + 0.4) = 2 × 1.3 2 × 1.3 2 4.16 = 26x − 23.4x − 9.36

b Gain in KE + gain in PE = loss of EPE 1 × 2m × v2 + 2mg × 1.5l sin 60 = 3mg ( 32l )2 2 2l v2 = v=

27gl 12 3 gl − 8 8

26x2 − 23.4x − 13.52 = 0

27gl 12 3 gl − 8 8

x = 1.3 (or −0.4) AB = 2.6 m

9

12 a Resolving vertically:

4l X

2l

2l

Y

q T

3l

P

XP = (2l)2 + (3l)2 = 13l 2T sin q = 2mg 3 T× = mg 13 T=

13 mg 3

T

200x = 20g 0.5 x = 0.5 m Initial extension = 1 m 200 × 12 EPE = = 200 J at a distance of 2 × 0.5 0.5 m below equilibrium point. b EPE at start = gain in GPE + gain in KE + EPE when particle first comes to rest 200 = 20 × 10 × (1.5 − x) + 1 × 20 × v2 2 2 200 × (1.5 − x ) + 2 × 0.5 200 = 300 − 200x + 10v2 + 200x2−600x + 450 v2 = 20x2 − 80x + 55 v = 0 as particle is at rest. x = 0.882 m

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0.8 + x) =

4 

WORKED SOLUTIONS

c  Assumptions: no air resistance, and that the string remained taut and therefore had some EPE.

3.24 − 7.2y + 4y 2 17.64 − 25.2y + 9y 2 29 + = 3.6 3.6 45

13 a In equilibrium

13y2 – 32.4y + 18.56 = 0

λ × 0.5 = 1.5 × g 2 l = 60 N

y = 1.6 or 0.89



y = 1.6 is start position so AC = 2 – 0.89 = 1.11 m

b From conservation of energy (taking point A as h = 0)

PE when raised = EPE at 2 m + KE at 2 m + GPE at 2 m



At 2 m, extension of spring = 0, therefore EPE = 0 1.5 × g × (0.8 + x) =

0.75v2 = 12

v2 = 16

From equations of motion

v2 = u2 + 2as

At A, v = 0, a = –10



0 = 16 + (2 × –10 × x)



20x = 16

 x = 0.8 m

So OA = 0.8 + 2 = 2.8 m

14 a Let x be the extension in spring one and, let y be the extension in spring 2.

(0.6 + x) + (0.9 + y) = 2



x + y  = 0.5



y  = 0.5 – x



In equilibrium so:



2y 3x = 0.6 0.9



3x 1 − 2x = 0.6 0.9



x = 0.15



AP = 0.75 cm

b EPE at start = EPE in AP + EPE in PB



EPE =

50 × 0.1442 2 × 0.3 = 1.73 J

b Mass comes to rest when v = 0. 60 × (0)2 1 + × 1.5 × v 2 + 1.5 × g × x 2×2 2 EPE before = EPE when next at rest + gain in PE

60 × (0)2 1 2 + × 1.5 × v + 1.5 × g × x 2×2 2 12 + 15x = 0.75v2 + 15x

15 a Resolving vertically: 50x = 2.4g 0.3 x = 0.144 m

3 × 0.22 2 × 0.7 2 29 + = 2 × 0.6 2 × 0.9 45 When particle is next at instantaneous rest v = 0 and let distance of particle from B = y From conservation of energy 2 × (0.9 − y)2 3 × (1.4 − y)2 29 + = 2 × 0.9 2 × 0.6 45



2 1.728 = 50 × x + 1.2 × 10 × (0.144 + x) 2 × 0.3 50x2 + 7.2x = 0

x = 0, i.e. the spring has returned to its natural length. 16 a EPE at X = work done against friction + EPE at Y

36 × 3.12 = 6F + 36 × 1.12 2 × 0.9 2 × 0.9 6F = 168

F = 28 N 1.1 × 36 b At Y tension = = 44 N which is greater 0.9 than 28 N so the particle starts to move again.

When particle is next at rest the work done against friction has used up all of the EPE from Y.



EPE at Y = work done against friction



36 × 1.12 = 28s 2 × 0.9 s = 0.86 m Total distance = 6 + 0.86 = 6.86 m

17 a l = 0.8, l = 30 N m = 0.3 kg

Let x be extension at equilibrium



At equilibrium 30x 0.3g = 0.8 0.3 × 10 × 0.8 x= = 0.08 m 30 b Particle is extended by a further distance y

30 × (0.08 + y)2 2 × 0.8 When particle is released and has returned to original length, by conservation of energy EPE at extension =

EPE loss = gain in KE + gain in GPE

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4 HOOKE’S LAW

30 × (0.08 + y)2 1 = × 0.3 × v2 + 0.3 × 10 (0.08 + y) 2 × 0.8 2 18.75(0.0064 + 0.16y + y2) = 0.15v2 + 0.24 + 3y 0.15v2 = 18.75y2 – 0.12 v2 = 125y2 – 0.8 Then motion under gravity u2 = 125y2 – 0.8, v = 0.1, s = 0.8, a = –10 v2 = u2 + 2as 0.01 = 125y2 – 0.8 + 2 × (–10) × 0.8 125y2 = 16.81 y = 0.367 cm   So AB = 0.8 + 0.08 + 0.367 = 1.25 m 18 l =1, modulus of elasticity = l 2

EPE =  λ × 2  = 2l 2×1 When string is no longer stretched by conservation of energy EPE lost = gain in KE + gain in GPE 2l = 1 × 0.5 × v2 + 0.5 × 10 × 2 2 2l = 0.25v2 + 10 v2 = 8l – 40 Motion under gravity Assuming that it reaches v = 0 when s > 1, u2 = 8l – 40, a = –10 v2 = u2 + 2as 0 < 8l – 40 + 2 × (–10) × 1 8l > 60 l > 7.5

λmg × 0.3L 19 a T = = 0.3lmg L b Resolving vertically: mg = T cos 60° mg = 0.3lmg × 0.5 20 l= 3 c T = 2mg

F = ma



2mg sin 60° = mrw 2



3 g = (1.3L sin 60°) w 2

ω=

2g 1.3L

20 a Loss in GPE = Gain in KE + gain in EPE 1 30mx 2 10m(l + x) = 2 mv2 + 2×l 2 20(l + x) = v2 + 30x l v2 = 20(l + x) – 30x l

42



 aximum velocity occurs when particle passes M though equilibrium point. 3mgx mg = l which is when l = 3x

2 v2 = 20(l + x) – 30x l 30x 2 2 v = 20(3x + x) – 3x = 80x – 10x

= 70x v = 70x 70l 3 b At A v = 0 so KE = 0 =

so by conservation of energy 3mgx 2 = mg(l + x) 2×l 3x 2 = (l + x) (Note this equation can also be 2l found by setting v2 = 0 in equation from part a) 3x2 = 2l2 + 2lx 3x2 – 2lx – 2l2 = 0 x= x=

2l ±

((−2l )

2l ±

(52l )

2

− 4 × 3 × ( −2l ) 6

2

)

2

6

x = l ± 2l 13 3

So distance AB = l + l + 2l 13 3 =

(

l 4 + 13 3

)

Mathematics in life and work 1 When speed is at a maximum, a = 0 and jumper is in equilibrium. Let x be the distance from platform. Resolving vertically: 3645(x − 36) = 88g 36 x = 44.69 m Extension of cord at maximum speed is 8.69 m. Gain in EPE + KE = loss in GPE 3645 (x − 36)2 88v 2 + = 88gx 2 × 36 2 v 2 = 2gx − v 2 = 806.9

405(x − 36)2 352

v = 28.4 m s−1

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WORKED SOLUTIONS

4 

2 From equation above: 3645 (x − 36)2 88v 2 + = 88gx 2 × 36 2 At maximum extension, v = 0 50.625(x − 36)2 = 880x 50.625(x2 − 72x + 1296) = 880x 50.625x2 − 3645x + 65 610 = 880x 50.625x2 − 4525x + 65 610 = 0 x = 71.2 m (or 18.2 m) x = 71 m to nearest metre So assuming the man is less than 3 m tall, the minimum value of y is 74 m. 3 No air resistance, always assumed extension is greater than 36 m, assume man is less than 3 m tall. Modelled man as particle until final stage.

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5 Linear motion under a variable force

5 Linear motion under a variable force Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

Prerequisite knowledge 1 2

So −y1 = ln x − 1 4 1 = 1 − ln x    y 4

x 1 =1− x +1 x +1 dy = xy dx 1 ∫ y d y = ∫ x dx ln y = 1 x2 + c 2  2 ln y = x2 + d

8 x

∫e

dy =

1

∫ x dx

−e−y = ln x + c y = 0 when x = 4 −e−0 = ln 4 + c

∫ y d y = ∫ sin x dx

−e−y

c = − ln 4

1 y2 = −cos x + c 2 y2 = −2 cos x + d

−e

−x

)

2

− 25 dx

dy = 4x2 − 25 dx

4x 2 − 25 dx x 25 ∫ 1 d y = ⌠⌡ 4x − x dx

∫ 1 d y = ⌠⌡

7 x

 y =

dy = y2 dx

(

2x2

)

− 25 ln x + c

1 d y = ⌠ dx ⌡x −1 = ln x + c y  When y = 4, x = 1 = ln 1 + c −1 4  c = − 1 4

∫y

−2

v = 1 t3 − t2 + c 3 When t = 0, v = 14 So c = 14 v = 1 t3 − t2 + 14 3

4 y = x3 − 25x + c 3



x 4 4 = ln x = ln

1 a v = ∫ t 2 − 2t dt

dy = 4x2 − 25 dx

6 x

= ln x − ln 4

Exercise 5.1A

dx

y = −5e−x + c

∫ 1 d y = ∫ ( 4x

−y

e −y  

dy 5 = dx e x

∫ 1 d y = ∫ 5e 5

−y

dy sin x = y dx

3

4

dy = ey dx

b s = ∫ 1 t 3 − t 2 + 14 dt 3 s = 1 t 4 − 1 t 3 + 14t + c 12 3 When t = 0, s = 0 So c = 0 s = 1 t 4 − 1 t 3 + 14t 12 3 d v 2 a a = dt

v = 5(5t + 2)−1 dv = −5 × 5(5t + 2)−2 dt −25 a= (5t + 2)2

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5 

WORKED SOLUTIONS

v2 = 100 − 10x2

b s = ⌠ 5 dt ⌡ 5t + 2

Changes direction when v = 0.

s = ln(5t + 2) + c When t = 0, s = 0 So c = −ln 2 s = ln(5t + 2) − ln 2 5t + 2 s = ln 2

(

)

3 a a = 5e−2t

v = ∫ 5e −2t dt



v = −2.5e−2t + c

When t = 0, v = 50 So c = 50 + 2.5 = 52.5 So v = 52.5 − 2.5e−2t b When t = 0, v = 50 As t → ∞, v → 52.5 – 0 = 52.5 So 50 ⩽ v < 52.5



4 a x = 5 sin 2πt v = dx = 10π cos 2πt dt d 2x a = 2 = −20π2 sin 2πt dt When t = 5, a = 0 m s−2 Maximum magnitude of acceleration = 20π2 m s−2

b

5 a a = 3x − 2 v dv = 3x − 2 dx

∫ v dv = ∫ (3x − 2) dx



1 2 3 2 v = x − 2x + c 2 2 When x = 0, v = 0, c = 0

0 = 100 − 10x2 x = 10 = 3.16 m 7 a = 10e0.5x dv v = 10e 0.5x dx

∫ v dv = ∫ 10e

0.5x

dx

1 v 2 = 20e0.5x + c 2 When x = 0, v = 2. 2 = 20 + c So c = −18 1 v 2 = 20e0.5x − 18 2 v = 40e 0.5x − 36 When x = 5, v = 21.2 m s−1. 8 a a = x3(3 − x) dv v = x 3(3 − x) dx

∫ v dv = ∫ 3x



3

− x 4 dx

1 2 3 4 1 5 v = x − x +c 2 4 5 When x = 0, v = 0 so c = 0 v2 = 3 x4 − 2 x5 2 5 v=

3 4 2 5 x − x 2 5

b When v = 0 0 = 3 x4 − 2 x5 2 5 x4 3 − 2 x = 0 2 5

(

)

v2 = 3x2 − 4x

x = 0 or 3 − 2 x = 0 2 5

v = 3x 2 − 4x



So the object is at rest when x = 0 m and when x = 3.75 m.

b When x = 3

v = 3(3)2 − 4(3)



v = 15 = 3.87 m s−1

6 a = 10x dv v = −10x dx v dv = ∫ −10x dx  ∫

1 2 v = −5x2 + c 2 When x = 0, v = 10, c = 50 1 2 v = 50 − 5x2 2

9 a 10a = 2x − 3x2  a = 0.2x − 0.3x2 v dv = 0.2x − 0.3x2 dx



2 ∫ v dv = ∫ (0.2x − 0.3x ) dx



1 2 v = 0.1x2 − 0.1x3 + c 2 When x = 0, v = 0 so c = 0

v 2 = 0.2x2 − 0.2x3



v = 0.2x 2 − 0.2x 3

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5 Linear motion under a variable force

b When v = 0 0.2x2 − 0.2x3 = 0

x2 − x3 = 0 x2(1

− x) = 0

So x = 0 or x = 1

The object will travel 1m before instantaneously coming to rest.

Exercise 5.2A

100  t = 1 ln  16  500 − 4v 2  When t = 5 5 = 1 ln  100 2  16  500 − 4v  100 = e 80 500 − 4v 2 500 − 4v2 = 100e−80 v = 11.2 m s−1 4 a 70 km h−1 = 70 × 1000 = 19.4 m s−1 60 × 60 b 800a = 1600 − 4v v

1 3a = −21v a = −7v dv = −7v dt



1 dv = −7 dt ∫  ∫v

800

dv 1600 − 4v 2 = v dt

ln v = −7t + c



When t = 0, v = 30 so c = ln 30 t = 1 ln 30 7 v

⌠ v 800 dv = ∫ 1 dt 2 ⌡ 1600 − 4v



⌠ −8v −100 dv = ∫ 1 dt 2 ⌡ 1600 − 4v



−100 ln (1600 − 4v2) = t + c

( )

When v = 15 t = 1 ln 30 = 0.0990 s 7 15

( )

2 a = −15v v dv = −15v dx

(

)

When t = 0, v = 15

c = −100 ln (1600 − 4 × 152)



c = −100 ln (700)

When v = 19.444

v = −15x + c When x = 0, v = 20 so c = 20 v = −15x + 20 When v = 5 5 = −15x + 20





 700  2  1600 − 4 × 19.444 

t = 100 ln 



t = 208 s 5 1000a = 4900 − 0.4v 2 v 1000 v dv = 4900 − 0.4v 2 v dx

x = 1 m 3 0.5a = 500 − 4v v dv 500 − 4v = v dt

)

700   So t = 100 ln   1600 − 4v 2 

∫ 1 dv = ∫ −15 dx

0.5

(

2

v 1⌠ dv = ∫ 1 dt  2 ⌡ 500 − 4v 2

⌠ −8v − 1  dv = ∫ 1 dt 16 ⌡ 500 − 4v 2 − 1 ln (500 − 4v2) = t + c 16 When t = 0, v = 10 c = − 1 ln (500 − 4 × 102) 16 c = − 1 ln (100) 16

1000 v 1000 −

dv 4900 − 0.4v 3 = v dx

v2 dv =1 4900 − 0.04v 3 dx

25000 ⌠ −0.12v 2 dv = ∫ 1 dx  3 ⌡ 4900 − 0.04v 3

− 25 000 ln(4900 − 0.04v3) = x + c 3 When x = 0, v = 5 c = − 25 000 ln(4900 − 5) 3 25 000 ln(4895) c=− 3

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WORKED SOLUTIONS

4895  So x = 25 000 ln   4900 − 0.04v 3  3



∫ 450(v + 3) dv = ∫ 45 dt + c

When v = 12



450v 2 + 1350v = 45t + c 2

4895  x = 25000 ln  3  4900 − 0.04 × 123 

When t = 0, v = 0 so c = 0 1 t = 45 (225v2 + 1350v)

  = 110 m (3sf )

When v = 4, t = 200 s

6 a F = μR

Exercise 5.3A

= μmg

= 0.6 × 0.02 × 10



= 0.12

b 0.02a =

−(5v2

0.3a = 1 − 0.5v2 + 0.12)

a = −(250v2 + 6)

v dv = −(250v2 + 6) dx

v ⌠  dv = ∫ −1 dx ⌡ 250v 2 + 6

1 ⌠  500v dv = ∫ −1 dx 500 ⌡ 250v 2 + 6 1 ln(250v2 + 6) = −x + c 500 When x = 0, v = 10 c = 1 ln (250 × 102 + 6) 500 c = 1 ln (25 006) 500 25 006  So x = 1 ln  500  250v 2 + 6 

When v = 5    x = 1 ln  25 006 500  250 × 52 + 6 

x = 0.002 77 m 7 a 450a = 45 v+3 dv 45 = dx v + 3



450v



∫ 450v (v + 3) dv = ∫ 45 dx + c 2 ∫ ( 450v + 1350v ) dv = ∫ 45 dx + c



450v 3 1350v 2 + = 45x + c 3 2

150v3 + 675v2 = 45x + c When x = 0, v = 0 so c = 0 x = 1 (150v3 + 675v2) 45 When v = 4, x = 453 m b 450a = 45 v+3

450

1 0.3a = 0.3 × 10 − (2 + 0.5v2)

dv 45 = dt v + 3

For terminal velocity: 1 − 0.5v2 = 0 v = 1.41 m s−1 2 0.5a = −0.5 × 10 − 0.6v 0.5a = −(5 + 0.6v) 0.5v dv = −(5 + 0.6v) dx 0.5v

∫ 5 + 0.6v dv = ∫ −1 dx

(

)

25 ⌠  5 − 6(5 + 0.6v) dv = ∫ −1 dx ⌡ 6

( ) 0.6 5⌠  1 − 25 5 + 0.6v ) dv = ∫ −1 dx 3 6 ⌡( 5 5⌠ 1− dv = ∫ −1 dx 6 5 + 0.6v ⌡

5 v − 125 ln(5 + 0.6v) = −x + c 6 18 When x = 0, v = 20 So c = 50 − 125 ln (17) 3 18 50 125 So x = ln(17) − 5 v + 125 ln(5 + 0.6v) − 6 18 3 18 Maximum height is when v = 0. So maximum height = 8.17 m. 3 0.8a = −0.8 × 10 − 0.5v 0.8 dv = −(8 + 0.5v) dt ⌠ 0.8 dv = −1 dt ∫ ⌡ 8 + 0.5v 8 0.5 dv = ∫ −1 dt 5 ∫ 8 + 0.5v 8 5 ln(8 + 0.5v) = −t + c When t = 0, v = 15 So c =

8 ln (15.5) 5

(

15.5 So t = 8 ln 5 8 + 0.5v

)

When v = 5 t = 0.623 seconds

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5 Linear motion under a variable force

4 0.3a = −(0.3 × 10 + 0.5v2) 0.3v dv = −(3 + 0.5v2) dx 0.3v ⌠ dv = ∫ −1dx  ⌡ 3 + 0.5v 2 5 ln (3 + 0.5v2) = −x + c 3 When x = 0, v = 10 So c = 5 ln(53) 3 So x =

So c = −3 ln 30 So t = 3 ln 30 30 − v

(

So maximum height = 4.79 m. 5 a 7a = 7 × 10 − 20v 7 dv = 70 − 20v dt 7 ∫ 70 − 20v dv = 1 dt − 27 ln(70 − 20v) = t + c 0 When t = 0, v = 0 7 ln(70) So c = − 20 60 7 =e7 70 − 20v 7 So t = 7 ln 70 − 20v 20

(

)

When t = 3 7 60 ln = 70 − 20v 7



)

( ) 4 30 = ln ( 3 30 − v )



4 30 = e3 30 − v



30 − v = 30e 3 v = 22.1 m s−1

75a = 75 × 10 − 20v − 40v2 0 = 750 − 20v − 40v2



v = −4.59 or v = 4.09 So her terminal velocity is 4.09 m s−1.

Exam-style questions

70 − 20v = 7e−

1 a a =

So 70 − 20v = 0

∫ v dv = ∫ 16 ( x + 2)



3.5 m s−1

−2

dx

  1 v2 = −16(x + 2)−1 + c 2

2v2)

6 a = −(10 + dv = −(10 + 2v2) dt



1 ⌠  dv = ∫ −1 dt ⌡ 10 + 2v 2 1 tan −1  v  = −t + c   2 5 5 When t = 0, v = 25   1 tan −1  25  So c =  5 2 5 1 tan −1  25   5 2 5

16

( x + 2 )2

 v dv = 16(x + 2)−2 dx

b For terminal velocity, a = 0

So t =

−4

b Terminal velocity is when a = 0.



60 7 =e7 70 − 20v

So v =

)

When t = 4 30 4 = 3 ln 30 − v

v = 3.50 m s−1

−3 ln(30 − v) = t + c

When t = 0, v = 0

Maximum height is when v = 0.



75a = 750 − 25v 3 dv = 30 − v dt 3 ∫ 30 − v dv = ∫ 1 dt

5  53  ln 3  3 + 0.5v 2 

(

7 a 75a = 75 × 10 − 25v

When x = 2, v = 0.5 m s−1 1 8 = −4 + c So c = 33 8 1 2 33 −16 + So 2 v = ( x + 2) 8

−

1 tan −1  v  2 5  5

Maximum height is when v = 0. So time to reach maximum height is 0.331 seconds.

1 2 −16 v = +c 2 ( x + 2)

v=

33 − 32 4 x+2

33 = 2.87 m s−1 4 2 a 0.4a = − 0.4 × 10 − 0.8v 0.4v dv = −(4 + 0.8v) dx b As x → ∞, v →

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WORKED SOLUTIONS

0.4v

∫ 4 + 0.8v dv = ∫ −1 dx



v





12 000 − v 3 = 11 936 e

∫ 10 + 2v dv = ∫ −1 dx

v 3 = 12 000 − 11 936 e

5   1− ∫ 2 10 + 2v dv = ∫ −1 dx 1 5 v − ln (10 + 2v ) = −x + c 2 2

v = 12 000 − 11 936 e

When x = 0, v = 25 So c = 25 − 5 ln ( 60 ) 2 2 So x =

25 2

−

(

1 5 10 + 2v v + ln 2 2 60

3

)



Maximum height is when v = 0.



So maximum height = 8.02 m.

b 0.4 dv = −(4 + 0.8v) dt 0.4

∫ 4 + 0.8v dv = ∫ −1dt



1 ln(4 + 0.8v) = −t + c 2 When t = 0, v = 25 So c = 1 ln(24) 2

(

24 So t = 1 ln 2 4 + 0.8v

)



Maximum height is when v = 0.



So time to reach maximum height = 0.896 seconds.

3 a 2000a = 60 000 − 5v 2 v 2000v dv = 60 000 − 5v 2 v dx d v 12 000 400v = − v2 v dx

− 3 x 400

400v

dv 12 000 − v 3 = v dx

v2 400⌠ dv = ∫ 1 dx  ⌡ 12 000 − v 3 − 400 ln(12 000 − v3) = x + c 3 When x = 0, v = 4 400 So c = − 3 ln(11 936)



So x =

400  11 936  ln 3  12 000 − v 3 

 11 936  3 x = ln  400  12 000 − v 3  3 x

e 400 =

11 936 12 000 − v 3

− 3 x 400 − 3 x 400

b When v = 15 x = 400 ln  11 936  3  12 000 − 153  x = 43.3 m 4 a x = ∫ v dt

x = ∫ e −5t + 5t 2 dt

1 5 x = − e −5t + t 3 + c 5 3 When t = 0, x = 3 1 t 5 c3 = 16 x = − e −5 So + t + 5 3 5 1 5 16 So x = − e −5t + t 3 + 5 3 5



When t = 5 625 16 1 x = e −25 + + 5 3 5

   = 211.533… = 212 (3 s.f.) b a = dv dt v = e−5t + 5t2 dv = −5e−5t + 10t dt When t = 3 a = 30.0 m s−2 1 5 a a = e −5t 3 1 v = ∫ 3 e −5t dt 1 v = − e −5t+ c 15 When t = 0, v = 5 5 = − 1 e −5× 0 + c 15 So c = 5 + 1 = 76 15 15 76 1 −5t So v = − e 15 15 b When t = 3 v = 5.07 m s−1 c 5  v  76 − 1 e −5t 15 15 6 a a = −(1 × 10 + 3v2) v dv = −(10 + 3v2) dx

v ⌠  dv = ∫ −1dx ⌡ 10 + 3v 2

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5 Linear motion under a variable force

1 2 6 ln(10 + 3v ) = −x + c



When x = 0, v = 25 So c = 1 ln(1885) 6 1  1885  ln 6  10 + 3v 2 



So x =



Maximum height is when v = 0.

So x = 0.873 m

⌠ 30 dv = ∫ 1 dt ⌡ 120 − 7v −30 ln(120 − 7v) = t + c 7 When t = 0, v = 0 So c = − 30 ln (120 ) 7 30 120 So t = ln 7 120 − 7v



So maximum height of object =  1.2 + 0.873 = 2.07 m.

b dv = −(10 + 3v2) dt

1 ⌠ dv = ∫ −1 dt  ⌡ 10 + 3v 2



⌠  ⌡

(

10

1

) +( 2

3v

)

2

dv = ∫ −1 dt

1 tan −1  3 v  = −t + c  10  30 When t = 0, v = 25   So c = 1 tan −1  1875   10  30  1875   3  1 1 tan −1  tan −1  v −  10   10  30 30

So t =

( ) 120 7 ln ( = t 120 − 7v ) 30

7t 120 = e 30 120 − 7v



120 − 7v = 120e

− 7 t 120  1 − e 30  7   120 b As t → ∞, v → = 17.1 m s−1. 7 9 50a = 50 × 10 − 0.08v2 50v dv = 500 − 0.08v2 dx



v=

50v ⌠ dv = ∫ 1 dx  ⌡ 500 − 0.08v 2 50 ln(500 − 0.08v2) = x + c 0.16 625 ln(500 − 0.08v2) = x + c − 2 When x = 0, v = 0 −



Maximum height is when v = 0.

So c = − 625 ln ( 500 ) 2



So time to reach maximum height = 0.273 seconds.

So x =

−1 x

7 a a = 1 e 2 3 dv 1 − 12 x v = e dx 3 1 −1 x ∫ v dv = ⌠⌡ 3 e 2 dx 1 2 2 −1 x v =− e 2 +c 2 3 When x = 0, v = 2 1 2 8 So c = 2 × 4 + 3 = 3



1 2 8 2 − 12 x v = − e 3 3 2

16 4 − 12 x v= − e 3 3

b As x → ∞, v → 2.31 m s−1. 8 a 1500a = 9000 − 3000 − 350v 30a = 120 − 7v 30 dv = 120 − 7v dt

−7t 30

625  500  ln 2  500 − 0.08v 2 

When x = 50 625  500  50 = ln 2  500 − 0.08v 2  4 500  = ln  25  500 − 0.08v 2  4 500 = e 25 500 − 0.08v 2

− 4

  500 − 0.08v2 = 500e 25 v = 30.4 m s−1 10 a 0.2a = ex − (1 + 2x) 0.2a = ex − 1 − 2x  a = 5ex − 10x − 5 v dv = 5ex − 10x − 5 dx b ∫ v dv =

∫ (5e

x

)

− 10x − 5 dx

1 v2 = 5e x − 5x2 − 5x + c 2 When x = 0, v = 4

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WORKED SOLUTIONS



8=5+c

So c = 3 1 So 2 v 2 = 5e x − 5x2 − 5x + 3 When x = 10 1 2 v 2 = 5e10 − 500 − 50 + 3 v = 468 m s−1 11 a a = 0.2e0.2x v dv = 0.2e0.2x dx

∫ v dv = ∫ 0.2e



1 2 0.2x +c 2 v  = e

0.2x

dx

When x = 0, v = 0 So c = −1 So 1 v 2 = e0.2x – 1 2 v = 2e 0.2x − 2 b When x = 10 v = 3.57 m s−1 c a = 0 12 a F = 6a = 1202 kv a = 202 kv dv 20 = dt kv 2 20 2 ∫ v dv = ⌠⌡ k dt



v 3 20t = +c 3 k v 3 = 60t + d k



When t = 0, v = 1 and when t = 1, v =2 b

1 = 0 + d so d = 1 60t v3 = +1 k 60 8= +1 k 60 7= k k = 60 7 60t 3 v = +1 k 60t v3 = +1 60 7 v3 = 7t + 1

c a = 7 2 3v v dv = 7 2 dx 3v v 3 dv = 7 dx 3 7 3 ∫ v dv = ∫ 3 dx v4 7 = x+c 3 4 28 4 v = x+d 3 When x = 0, v = 1 14 = 0 + d so d = 1 v4 = 28 x + 1 3 13 a F = 60a = 10 cosec v cosv a = cosec v 6cos v d v cosec v = 6cos v dt 6cos v ∫ cosec v dv = ∫ 1 dt

∫ 6cosv sin v dv = ∫ 1 dt     ∫ 3sin 2v dv = ∫ 1 dt

3   − cos 2v = t + c 2 When t = 0, v = π 6 3 π So c = − cos 3 2 3 c= − 4 So t = 3 − 3 cos 2v 4 2 4t = 3 − 6 cos 2v



6 cos 2v = 3 − 4t cos 2v = 3 − 4t 6

b When t = 1 1 cos 2v = − 6 2v = 1.738 or 2v = 4.454 Since 0 < v < π 2 2v = 1.738

v = 0.869

So v = 0.869 m s–1 when t = 1 14 a F = ma = 0.6a = 0.5v2 +1 – 0.25v2 – 0.5 0.6a = 0.25v 2 + 0.5

0.6v

dv = 0.25v2 + 0.5  dx

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5 Linear motion under a variable force



v 5 ⌠ dv = ∫ dx  3 ⌡ 0.25v 2 + 0.5



2 ln (0.25v 2 + 0.5) =

5 x+c 3

2 ln (1) = c = 0



2 ln (0.25v 2 + 0.5) =



ln (0.25v 2 + 0.5) =

8 80 So t = 3 ln 80 − 3v 5 x 3

5 x 6

5x

When t = 0, v = 0 So c = − 8 ln(80) 3

(

When x = 0, v = 2 .

8 − ln(80 − 3v) = t + c 3



When t = 5 80 15 ln = 80 − 3v 8

(





 5x  = 4  e 6 − 0.5   5x

v = 4(e 6 − 0.5)

b When x = 1.5 5x 4(e 6



v=

− 0.5)



v = 3.46 m s–1.

15 a 4a = v3 e5x − 4 4v dv = v3 e5x − 4 dx

4 ⌠  2 dv = ∫ e 5x − 4 dx ⌡v

−

4 1 5x − 4 = e +c v 5

When x =0.8, v = 2 −2 = 1 e0 + c 5 c = − 11 5 4 − = 1 e 5x − 4 − 11 v 5 5 − 20 = e 5x − 4 − 11 v

v=−



v=

20 e 5x − 4 − 11

20 11 − e 5x − 4

)



15 80 =e8 80 − 3v



80 − 3v = 80e



v=



v = 22.6 m s−1

0.25v 2 + 0.5 = e 6 v 2

)

− 15 8

− 15  80  1−e 8  3  

b For terminal velocity, a = 0 80a = 800 − 30v − 45v2

0 = 800 − 30v − 45v2



0 = 9v2 + 6v − 160



v = −4.56 or 3.90

So terminal velocity = 3.90 m s−1



2  17 a ma = −  mg + mv2   4λ 



2   a = − 10 + v 2   4λ 



2 2  a = − 40λ +2 v   4λ 

v

 40λ 2 + v 2  dv = − dx  4λ 2 

⌠ 4λ 2v dv = −1 dx  ∫ ⌡ 40λ 2 + v 2

2l2 ln(40l2 + v 2) = −x + c When x = 0, v = 40l

So c = 2l2 ln(40l2 + (40l2 ))



 40λ 2 + ( 40λ )2  x = 2λ 2 ln   2 2  40λ + v 

v = 1.86 m s−1



Maximum height is when v = 0.

16 a 80a = 80 × 10 − 30v



 40λ 2 + ( 40λ )2  x = 2λ 2 ln   40λ 2  



x = 2l2 ln(1 + 40)



x = 2l2 ln(41)

b When x = 0.5 20 v= 11 − e 5(0.5)− 4

80a = 800 − 30v 8a = 80 − 3v 8 dv = 80 − 3v dt

8

∫ 80 − 3v dv = ∫ 1 dt

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WORKED SOLUTIONS

2   b a = − 10 + v 2   4λ  2 2  a = − 40λ +2 v   4λ 



 40λ 2 + v 2  dv = − dt  4λ 2 



4λ 2 ⌠ dv = ∫ −1 dt  ⌡ 40λ 2 + v 2



 v  4λ tan −1  = −t + c  λ 40  λ 40  v  4λ tan −1  = −t + c  λ 40  40

(

t = 4λ tan −1 40

(

40 −

When v = 0 t = 4λ tan −1 40

(

40

)

)

 v  4λ tan −1   λ 40  40

)

18 a ma = −(mg + mkv 2) kv 2)

a = −(g + v dv = −(g + kv 2) dx   v ⌠  dv = ∫ −1 dx ⌡ g + kv 2

1 ln(g + kv 2) = −x + c 2k  

When x = 0, v = u So c = 1 ln(g + ku2) 2k

x=

 g + ku 2  1 ln 2k  g + kv 2 

Maximum height is when v = 0.  g + ku 2  1 ln  Maximum height =  g 2k 

b ma = −(mg + mkv2) a = −(g + kv2) dv = −(g + kv2) dt



Maximum height is when v = 0.



So time to maximum height  k 1 tan −1  u  g  gk

=

2

When t = 0, v = 40l So c = 4λ tan −1 40 40

 k  k 1 1 tan −1  u tan −1  v −  g   g  gk gk

t =

1 ⌠ dv = ∫ −1 dt  ⌡ g + kv 2  k 1 tan −1  v = −t + c  g  gk

When t = 0, v = u  k 1 tan −1  u So c =  g  gk

19 a 0.5 a =

x sec x 2

( )

0.5  a = xcos(x2) 0.5 v dv = xcos(x2)   dx 2 ∫ 0.5v dv = ∫ x cos x dx

( )

( ) = ∫ 4x cos ( x ) dx



1 2 v = ∫ x cos x 2 dx 4



v2

2

b Let u = x2 du = 2x dx 1 2 So v = ∫ 4x cos(u) 2x du v 2 = 2cos (u) du ∫

v2 = 2 sin (u) + c



v2 = 2 sin (x2) + c

When x = 0, v = 4 So c = 16

v = 2sin(x 2) + 16

c Max of 2 sin (x2) + 16 = 18

Min of 2 sin (x2) + 16 = 14



So max of v is 18 = 3 2



Min of v is 14

20 a 10a = x2 e−v 10v dv = x2 e−v dx

∫ 10ve



x3 = ∫ 10ve v dv 3



x 3 = 30∫ ve v dv

b

∫ ve

v

v

dv = ∫ x 2 dx

dv = ve v − ∫ e v dv = vev − ev + c = ev(v − 1) + c

So

x3

= 30ev(v − 1) + c

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5 Linear motion under a variable force

When x = 0, v = 0 so c = 30 So

x3

=

30ev(v

− 1) + 30

c When v = 2

x3 = 30e2(2 − 1) + 30



x3 = 251.7



x = 6.31 m

3 Terminal velocity, a = 0 dv 180 = 1800 − (20v + 28v2) dt 28v 2 + 20v − 1800 = 0 v = −8.38 or 7.67

So her maximum possible speed is 7.67 m s−1.

Mathematics in life and work 1 180a = 180 × 10 − 25v 180 dv = 1800 − 25v dt d v 36 = 360 − 5v dt

36 ⌠  dv = ∫ 1 dt ⌡ 360 − 5v 36 − ln(360 − 5v) = t + c 5

When t = 0, v = 0 So c = − 36 ln(360) 5

(

360 So t = 36 ln 5 360 − 5v

)

When t = 45 360  225 ln  = 36  360 − 5v  225 360 = e 36 360 − 5v

− 225

360 − 5v = 360e 36 v = 71.861… = 71.9 m s−1 dv = 360 − 5v dx 36v ⌠  dv = ∫ 1 dx ⌡ 360 − 5v 36 v dv = ∫ 1 dx 5 ∫ 72 − v 36 v dv = ∫ 1 dx 5 ∫ 72 − v 36 72 −1 + dv = ∫ 1 dx 5 ∫ 72 − v 36 5 (−v −72 ln(72 − v)) = x + c

2 36v

When x = 0, v = 0 2592 ln(72) So c = − 5 2592 72 36 ln − v x= 5 72 − v 5

(

)

When v = 71.861, x = 2723 m When she opens her parachute she will be 3200 − 2723 = 477 m above the ground.

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6 

WORKED SOLUTIONS

6 Momentum Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

Prerequisite knowledge 1 Momentum = mv = 2.6 × 3 = 7.8 kg m s−1 2 Momentum = mv = 0.56 × 2 = 1.12 kg m s−1 3 2 × 5 + 3 × 2 = 2 × v1 + 3 × 4 v1 = 2 m s−1 in the same direction as before 4 4 × 2 + 4.5 × −3 = 4 × −1 + 4.5 × v2 v2 = − 1  m s−1 i.e. 0.333 m s−1 in opposite direction 3 5 a) Add both equations together 2x = 4 x=2 Substituting x = 2 into the first equation gives: 2 + y = 5 so y = 3 b) Add both equations together 5x = 2.5 x = 0.5 Substituting x = 0.5 into the second equation gives: 1.5 – y = 1.2 so y = 0.3

Exercise 6.1A v 2.6 = = 0.65 u 4

2 e =

v 10 = = 0.833 u 12



= 02 − 2 × −10 × 2.1



= 42 u = 6.48 m s−1

Speed immediately after impact is 6.48 m s−1 Speed immediately before impact is v 6.48 u= = = 8.53 m s−1 e 0.76 s=

v 2 − u 2 0 2 − 8.532 = = 3.64 m 2a 2 × −10

7 Speed before first impact: v2 = u2 + 2as = 60 v = 7.75 m s−1 Speed immediately after first impact = 7.75 × 0.7 = 5.42 m s−1

v 3.2 = = 5.81 m s−1 e 0.55 4 a v2 = u2 + 2as 3 u =

Speed immediately after second impact = 5.42 × 0.7 = 3.80 m s−1 Speed immediately after third impact = 3.80 × 0.7 = 2.66 m s−1

= 02 + 2 × 10 × 1.5 = 30 v = 5.48 m s−1 b v = eu = 0.4 × 5.48 = 2.19 m s−1 2

2

2

v −u 0 − 2.19 = = 0.24 m 2a 2 × −10 2 2 5 v = u + 2as c s =

Speed immediately after impact is 3.87 m s−1 3.87 So e = = 0.726 5.33 6 u2 = v2 − 2as

= 02 + 2 × 10 × 3

1 e =

2

u2 = v2 − 2as = 02 − 2 × −10 × 0.75 = 15 u = 3.87 m s−1

= 02 + 2 × 10 × 1.42 = 28.4 m s−1

v = 5.33 Speed immediately before impact is 5.33 m s−1

Height reached after third impact: v 2 − u 2 0 2 − 2.66 2 = = 0.353 m 2 × −10 2a So the maximum height reached after the third bounce is 35.3 cm. s=

8 a Speed after impact with wall = 15 × 0.6 = 9 m s−1 b U = 9 m s−1, q = 0°, ay = 10 m s−2, y = 0 m, h = 1.2 m 1 Substitute into y = Ut sin q − gt2 + h 2 1 0 = 9 sin 0° × t − × 10 × t2 + 1.2 2

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6 Momentum



Simplify:





5t2

vB − vA = 0.8(0.5 − −0.9)

− 1.2 = 0 t2

vB − vA = 1.12

= 0.24

t = 0.490 or −0.490

The time of flight for the ball is 0.490 seconds.



= 4.41

c  mA = 5 kg, uA = 6 m s−1, mB = 3 kg, uB = −5 m s−1, e = 0.2 By conservation of momentum:

The ball will hit the ground 4.41 m from the base of the wall.

9 U = ? m s−1, q = 0°, ay = 10 m s−2, y = 0 m, h = 2 m, t = ? 1 Substitute into y = Ut sin q − gt2 + h 2 0 = Ut sin 0° − 0.5 × 10 × t2 + 2 t2 = 0.4 t = 0.63246 U = ? m s−1, q = 0°, t = 0.63246

0.36 − vA = 1.12 so vA = −0.76 m s−1 i.e. 0.76 m s−1 in the opposite direction.

Substitute into x = Ut cos q x = 9 × 0.490 × cos 0

Substitute into x = Ut cos q



5 = U × 0.63246 × cos 0° 0.63246U = 5



U = 7.906

Solving these simultaneous equations:

2vB = 0.72 so vB = 0.36 m s−1

c U = 9 m s−1, q = 0°, t = 0.4899 s

By Newton’s experimental law:

5 × 6 + 3 × −5 = 5 × vA + 3 × vB 5vA + 3vB = 15

By Newton’s experimental law:

vB − vA = 0.2(6 − −5) vB − vA = 2.2

Solving these simultaneous equations:

8vB = 26 so vB = 3.25 m s−1

3.25 − vA = 2.2 so vA = 1.05 m s−1

d mA = 4 kg, uA = 2 m s−1, mB = 3 kg, uB = −6 m s−1, e = 1

By conservation of momentum:



4 × 2 + 3 × −6 = 4 × vA + 3 × vB

The ball leaves the wall with a speed of 7.906 m s−1.

4vA + 3vB = −10

So the speed before impact is 7.906 = 9.88 m s−1. 0.8

vB − vA = 1(2 − −6)

By Newton’s experimental law:

vB − vA = 8

Exercise 6.2A 1 a  mA = 3 kg, uA = 5 m s−1, mB = 4 kg, uB = 2 m s−1, e = 0.4

By conservation of momentum:



3 × 5 + 4 × 2 = 3 × vA + 4 × vB 3vA + 2vB = 23





By Newton’s experimental law:

vB − vA = 0.4(5 − 2) vB − vA = 1.2

Solving these simultaneous equations:



7vB = 26.6 so vB = 3.8 m s−1



3.8 − vA = 1.2 so vA = 2.6 m s−1

b  mA = 1 kg, uA = 0.5 m s−1, mB = 1 kg, uB = −0.9 m s−1, e = 0.8

By conservation of momentum:



1 × 0.5 + 1 × −0.9 = 1 × vA + 1 × vB vA + vB = −0.4



Solving these simultaneous equations:

7vB = 22 so vB = 3.14 m s−1 3.14 − vA = 8 so vA = −4.86 m s−1 i.e. 4.86 m s−1 in the opposite direction 2 mA = 2 kg, uA = 3 m s−1, mB = 3 kg, uB = 0 m s−1, e = 0.7 By conservation of momentum: 2 × 3 + 3 × 0 = 2 × vA + 3 × vB 2vA + 3vB = 6 By Newton’s experimental law: vB − vA = 0.7(3 − 0) vB − vA = 2.1 Solving these simultaneous equations: 5vB = 10.2 so vB = 2.04 m s−1 in the same direction that A was originally moving in. 3 mA = 4m kg, uA = 5 m s−1, mB = 5m kg, uB = −5 m s−1, e = 0.55 By conservation of momentum: 4m × 5 + 5m × −5 = 4m × vA + 5m × vB 4vA + 5vB = −5

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6 

WORKED SOLUTIONS

By Newton’s experimental law:

By conservation of momentum:

vB − vA = 0.55(5 − −5)

m × 4 + m × 0 = m × vA + m × vB

vB − vA = 5.5

vA + vB = 4

Solving these simultaneous equations:

By Newton’s experimental law:

9vB = 17 so vB = 1.89 m s−1

vB − vA = 0.6(4 − 0)

1.89 − vA = 5.5 so vA =

−3.61 m s−1

vB − vA = 2.4

So both balls are moving in the opposite direction compared to before the collision with vA = −3.61 m s−1 and vB = 1.89 m s−1. 4 mA = 4 kg, uA =

5 m s−1,

mB = 3 kg, uB =

0 m s−1,

e = 0.6

By conservation of momentum:

Solving these simultaneous equations: 2vB = 6.4 vB = 3.2 m s−1 Second impact:

4 × 5 + 3 × 0 = 4 × vA + 3 × vB

mB = m kg, uB = 3.2 m s−1, mC = m kg, uC = 0 m s−1, e = 0.6

4vA + 3vB = 20

By conservation of momentum:

By Newton’s experimental law:

m × 3.2 + m × 0 = m × vB + m × vC

vB − vA = 0.6(5 − 0)

vB + vC = 3.2

vB − vA = 3

By Newton’s experimental law:

Solving these simultaneous equations: 7vB = 32 so vB =

4.57 m s−1

4.57 − vA = 3 so vA =

1.57 m s−1

So both balls are moving in the same direction with vA = 1.57 m s−1 and vB = 4.57 m s−1. 5 a  mA = 2m kg, uA = u m s−1, mB = 3m kg, uB = −3u m s−1, e = e

By conservation of momentum:



2m × u + 3m × −3u = 2m × vA + 3m × vB 2mvA + 3mvB = −7mu 2vA + 3vB = −7u



By Newton’s experimental law:

vB − vA = e(u − −3u) vB − vA = 4eu

Solving these simultaneous equations:



2vA + 3vB = −7u



2vB − 2vA = 8eu

5vB = 8eu − 7u u vB = (8e − 7) 5 b vB − vA = 4eu u (8e − 7) − vA = 4eu 5 8eu − 7u − 5vA = 20eu 5vA = −12eu − 7u −u vA = (12e + 7) 5 6 First impact: mA = m kg, uA = 4 m s−1, mB = m kg, uB = 0 m s−1, e = 0.6

vC − vB = 0.6(3.2 − 0) vC − vB = 1.92 Solving these simultaneous equations: 2vC = 5.12 vC = 2.56 m s−1

Exercise 6.3A 1 Parallel to the surface: Speed before = 4.5 cos 30°, speed after = 4.5 cos 30° Perpendicular to the surface: Speed before = 4.5 sin 30°, speed after = 0.45 × 4.5 sin 30° = 2.025 sin 30° m s−1 Velocity after = (4.5cos30°)2 + 1.01252 = 4.03 m s−1 2.025 sin 30°  Direction θ = tan −1  = 14.6° above  4.5cos 30°  the horizontal 2 Parallel to the surface: Speed before = 12 cos 50°, speed after = 12 cos 50° Perpendicular to the surface: Speed before = 12 sin 50°, speed after = 0.4 × 12 sin 50° = 4.8 sin 50° Velocity after =  (12cos 50°)2 + (4.8sin 50°)2 = 8.55 m s−1 4.8sin 50°  Direction θ = tan −1  = 25.5° above  12cos 50°  the horizontal 3 Parallel to the surface: Speed before = 2.5 cos 80°, speed after = 2.5 cos 80°

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6 Momentum

Perpendicular to the surface:



Speed before = 2.5 sin 80°, speed after = 0.7 × 2.5 sin 50° = 1.75 sin 50°

speed after = 0.6 × 3.2 sin 40° = 1.92 sin 40°



Velocity after =



= 2.74 m s−1 1.92sin 40°  Direction θ = tan −1  = 26.7°  3.2cos 40°  above the horizontal

Velocity after =

(2.5cos 80°)2 + (1.75sin 80°)2

= 1.78 m s−1 1.75sin 80° Direction θ = tan −1 = 75.9° 2.5cos80° to the wall.

(

)

b Maximum height =

4 Parallel to the cushion: Speed before = 4 cos 72°, speed after = 4 cos 72° Speed before = 4 sin 72°, speed after = 0.78 × 4 sin 72° = 3.12 sin 72° Velocity after =  (4cos 72°)2 + (3.12sin 72°)2 = 3.21 m s−1 3.12sin 72°  Direction θ = tan −1  = 67.4° between  4cos 72°  the direction of the ball and the cushion 5 a  Let a  be the angle at which the particle hit the surface. tan 30° = 0.3 tan a so a = 62.5°

b Parallel to the surface:

Speed after = 3.2 cos 30°, speed before = 3.2 cos 30°



Perpendicular to the surface:



Speed after = 3.2 sin 30°, speed before = 3.2 sin 30° ÷ 0.3 = 10.67 sin 30°



Velocity before =

(3.2cos 30°)2 + (10.67sin 30°)2

= 6.01

m s−1

6 a Parallel to the surface:

Speed before = 3.2 cos 36°, speed after = 3.2 cos 36°



Also speed after = 2.8 cos a  so 3.2cos 36°  = 22.4° α = cos−1  2.8   b tan 22.393° = e tan 36°

e = 0.567

7 Let a be the angle between the wall and the ball before the collision. Then 0.88 tan a = tan 50° So α = tan −1 tan 50° 0.88 a = 53.6°

(

)

8 a Parallel to the surface:

Speed before = 3.2 cos 40°, speed after = 3.2 cos 40°



Perpendicular to the surface:



Speed before = 3.2 sin 40°,

U 2 sin 2 θ 2g

2.74452 sin 226.7234° 20 = 0.0762 m or 7.62 cm

=

Perpendicular to the cushion:



(3.2cos 40°)2 + (1.92sin 40°)2

c Range =

U 2 sin 2θ g

2.74452 sin 53.45° 10 = 0.605 m or 60.5 cm =

9 a Time of flight: y = Ut sin q − 1 gt2 + h 2 0 = 5.1t sin 30° − 0.5 × 10 × t2 + 2

5t2 − 2.55t − 2 = 0



t=



2.55 ± (−2.55)2 − 4 × 5 × −2 2×5 t = −0.427 or t = 0.937



So the ball is in air for 0.937 seconds before it hits the ground.



So the distance to point of impact is x = Ut cos q = 5.1 × 0.937 × cos 30° = 4.14 m.



At point of impact:



Horizontal velocity = 5.1 cos 30° = 4.42 m s−1



Vertical velocity = U sin q − gt



= 5.1 sin 30° − 10 × 0.937



= −6.82 m s−1



i.e. 6.82 m s−1 downwards



Horizontal velocity before bounce = 4.42 m s−1 so horizontal velocity after bounce = 4.42 m s−1



Vertical velocity before bounce = 6.82 m s−1 so vertical velocity after bounce = 0.7 × 6.82 = 4.77 m s−1





Resultant velocity = 4.422 + 4.772 = 6.50 m s−1 4.77 Angle after bounce = tan −1 = 47.2° 4.42 Distance to highest point U 2 sin 2θ = 0.5 × range = 2g 2 6.50 sin 94.4° = 20 = 2.11 m



So total horizontal distance = 4.14 + 2.11







( )

= 6.25 m

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6 

WORKED SOLUTIONS

b Time of flight to bounce = 0.937 s

Time from bounce to catch



= 0.5 × 2U sin θ = 13sin 47.2° g 20



= 0.477 seconds



So total time of flight = 0.937 + 0.477 = 1.41 s

2 2 2 2 = U sin θ = 10.37 sin 15.355° = 0.377 m, 2g 20 which is less

than 0.5 m

c Motion occurs only in two dimensions.

Air resistance is negligible.



Gravity remains constant.



There is no spin applied to the ball.

10 After each bounce, the horizontal velocity will remain 10 m s−1. Find time of first flight by substituting y = 0 into 1 y = Ut sin q − 2 gt2 + h 0 = 10t sin 0° − 0.5 × 10 × t2 + 5 5t2 = 5 t = 1 s Vertical velocity = U sin q − gt = 10 sin 0° − 10 × 1 = 10 m s−1 downwards After first bounce: vx = 10 m s−1, vy = 0.65 × 10 = 6.5 m s−1 2 2 Resultant velocity = 10 + 6.5 = 11.927 m s−1 6.5 Angle after bounce = tan −1 = 33.024° 10 Maximum height after bounce

So the maximum height of the ball is less than 50 cm for the first time after the third bounce.

Exercise 6.4A 1 a  Perpendicular to the line joining the centres of the spheres:

Sphere A: speed after collision =  3 sin 52° = 2.364 m s−1



Sphere B: speed after collision = 0 m s−1



Parallel to the line joining the centres of the spheres:



By the conservation of linear momentum:



3 cos 52° = vA + vB



By Newton’s experimental law:

vB − vA = 0.6 × 3 cos 52° vB − vA = 1.8 cos 52°

2vB = 4.8 cos 52°

( )

2

2

2

vB = 1.478 m s−1 vA + 1.478 = 3 cos 52° vA = 0.369 m s−1 to 3 s.f.

2

= U sin θ = 11.927 sin 33.024° = 2.11 m, which is 2g 20 greater than 0.5 m Second stage of journey: Immediately before second bounce: vx = 10 m s−1, vy = 6.5 m s−1 After second bounce: vx = 10 m s−1, vy = 0.65 × 6.5 = 4.225 m s−1 2 2 Resultant velocity = 10 + 4.225 = 10.856 m s−1 −1 4.23 Angle after bounce = tan 10 = 22.904° Maximum height after bounce U 2 sin 2 θ 10.8562 sin 2 22.904° = = = 0.893 m, which is 2g 20 greater than 0.5 m

( )

Solving simultaneously:



Sphere A:





Velocity after impact =  2.3642 + 0.3692 = 2.39 m s−1 2.364 a = tan−1 = 81.1° above the line 0.369 joining the centres of the two spheres.



Sphere B:



Sphere B is travelling at a velocity of 1.48 m s−1 parallel to the line joining the centres of the two spheres.



(

)

b  Perpendicular to the line joining the centres of the spheres:

Sphere A: speed after collision =  10 sin 45° = 7.0711 m s−1

Immediately before third bounce: vx = 10 m s−1, vy = 4.225 m s−1



Sphere B: speed after collision =  8 sin 30° = 4 m s−1

After third bounce: vx = 10 m s−1, vy = 0.65 × 4.225 = 2.746 m s−1 2 2 Resultant velocity = 10 + 2.746 = 10.37 m s−1 −1 2.746 Angle after bounce = tan 10 = 15.355°



Parallel to the line joining the centres of the spheres:



By the conservation of linear momentum:



10 cos 45° − 8 cos 30° = vA + vB

Third stage of journey:

(

Maximum height after bounce

)

vA + vB = 0.1429

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6 Momentum





By Newton’s experimental law:

vB − vA = 0.5 × (10 cos 45° − (−8 cos 30°))



vB − vA = 6.9996





Solving simultaneously: 2vB = 7.1425

vA + 3.5713 = 0.1429 Sphere A: 2

Sphere A: speed after collision = 5 sin 60° = 4.3301 m s−1 (to the left)



Sphere B: speed after collision = 4 sin 40° = 2.5712 m s−1 (to the right)



Parallel to the line joining the centres of the spheres:



By the conservation of linear momentum:



5 cos 60° − 4 cos 40° = vA + vB

2



Velocity after impact =  7.0711 + (−3.4284) = 7.86 m s−1 7.0711 a = tan−1 = 64.1° below the line joining 3.4284 the centres of the two spheres.



Sphere B:



Velocity after impact = 4 2 + 3.57132 = 5.36 m s−1 3.5713 a = tan−1 = 41.8° below the line joining 4 the centres of the two spheres in the opposite direction to B.





( (

centres of the two spheres in the opposite direction to B (upwards).



vA = −3.4284 m s−1

( )

d Perpendicular to the line joining the centres of the spheres:

vB = 3.5713 m s−1



Velocity after impact = 12 + 2.49642 = 2.69 m s−1 1 a = tan−1 2.5 = 21.8° to the line joining the

)

)

c  Perpendicular to the line joining the centres of the spheres:

vA + vB = −0.5642

By Newton’s experimental law:

vB − vA = 0.3 × (5 cos 60° − (−4 cos 40°)) vB − vA = 1.6693

Solving simultaneously:



2vB = 1.1051 vB = 0.5526 m s−1



Sphere A: speed after collision = −3 sin 25° = −1.2679 m s−1 (to the left)

vA + 0.5526 = −0.5642



Sphere B: speed after collision = 2 sin 30° = 1 m s−1 (to the right)



Sphere A:



Parallel to the line joining the centres of the spheres:



Velocity after impact = (4.3301)2 + (−1.1168)2 = 4.47 m s−1



By the conservation of linear momentum:



a = tan−1



3 cos 25° − 2 cos 30° = vA + vB



centres of the two spheres



Sphere B:



Velocity after impact =  2.57122 + 0.55262 = 2.63 m s−1 a = tan−1 2.57 = 77.9° to the line joining the 0.56 centres of the two spheres in the opposite direction to B.

vA = −1.1168 m s−1

vA + vB = 0.9869

By Newton’s experimental law:

vB − vA = 0.9 × (3 cos 25° − (−2 cos 30°)) vB − vA = 4.0059



Solving simultaneously:



2vB = 4.9928 vB = 2.4964 m s−1

( )

2 Perpendicular to the line joining the centres of the spheres:

vA + 2.4964 = 0.9869 vA = −1.5095 m s−1

4.3301 ( 1.1168 ) = 75.5° to the line joining the

Red ball: speed after collision = 0.5 sin 15° = 0.129 m s−1

Sphere A:

Blue ball: speed after collision = 0 m s−1 2

2

Parallel to the line joining the centres of the spheres:



Velocity after impact =  (−1.2679) + (−1.5095) = 1.97 m s−1 a = tan−1 1.2679 = 40.0° to the line joining the 1.5095 centres of the two spheres (downwards)



Sphere B:

By Newton’s experimental law:



(

)

By the conservation of linear momentum: 0.5 cos 15° = vR + vB vR + vB = 0.4830

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WORKED SOLUTIONS

Before the collision there was an angle of 157° between their directions of motion.

vB − vR = 0.3 × (0.5 cos 15° − 0) vB − vR = 0.1449

4 Perpendicular to the line joining the centres of the spheres:

Solving simultaneously: 2vB = 0.6279

Sphere A: speed before collision = 0 m s−1

vB = 0.314 m s−1

Sphere B: speed after collision = u m s−1

vR + 0.314 = 0.483

Parallel to the line joining the centres of the spheres:

vR = 0.169 m s−1

By the conservation of linear momentum:

Red ball: Velocity after impact =  (0.129) + (0.169) = 0.213 m s−1 0.129 a = tan−1 = 37.4° to the line joining the 0.169 centres of the two spheres

u = vA + vB

Blue ball moves with a velocity of 0.314 m s−1 parallel

2vB = 0.5u

2

(

2

)

to the line joining the centres. 3 Perpendicular to the line joining the centres of the spheres:

By Newton’s experimental law: vB − vA = −0.5u Solving simultaneously: vB = 0.25u m s−1 vA + 0.25u = u vA = 0.75u m s−1

Marble A: speed after collision =  2 sin 25° = 0.8452 m s−1

Sphere A will travel at 0.75u m s−1 parallel to the line joining the centres of the two spheres.

Marble B: speed after collision =  1.5 sin 30° = 0.75 m s−1

Sphere B:

Parallel to the line joining the centres of the spheres: By the conservation of linear momentum: uA + uB = −2 cos 25° + 1.5 cos 30° uA + uB = −0.5136 By Newton’s experimental law: 1.5 cos 30° − (−2 cos 25°) = 0.4 × (uA − uB) 0.4uA − 0.4uB = 3.1117 Solving simultaneously: 2uA = 7.2655

Parallel to the line joining the centres of the spheres: By the conservation of linear momentum:

By Newton’s experimental law:

uB = −4.147 m s−1

vB − vA = 0.22 × (16 cos q − (−10 cos q))

Marble A: Velocity before impact = (0.845)2 + (3.633)2 = 3.73 m s−1

0.845 a = tan−1 3.633 = 13.1° to the line joining the centres of the two marbles Marble B: Velocity before impact =  0.752 + (−4.147)2 = 4.21 m s−1 a = tan−1 0.75 = 10.3° to the line joining the 4.147 centres of the two marbles.

)

Sphere A: speed after collision = −16 sin q = −12.4939 m s−1

4vA + vB = 33.7335

uB + 3.633 = −0.5136

(

)

4 × 16 cos q − 10 cos q = 4vA + vB

uA = 3.633 m s−1

)

(

Sphere B: speed after collision = 10 sin q = 7.8087 m s−1

uA − uB = 7.7791

(

Velocity after impact = u 2 + (0.25u)2 = 1.03u m s−1 u a = tan−1 = 76.0 (3 s.f.) to the line joining 0.25u the centres of the two spheres. 5 4 5 5 If tan q = then cos θ = . and sin q = 4 41 41 Perpendicular to the line joining the centres of the spheres:

vB − vA = 3.5733 Solving simultaneously: 5vB = 48.0267 vB = 9.605 m s−1 4vA + 9.605 = 33.7335 vA = 6.032 m s−1 Sphere A: Velocity after impact = (−12.494)2 + (6.032)2 = 13.9 m s−1

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6 Momentum

( )

a = tan−1 12.5 = 64.2° to the line joining the 6.03 centres of the two spheres. Sphere B: Velocity after impact =  7.8092 + 9.6052 = 12.4 m s−1 7.81 a = tan−1 = 39.1° to the line joining the 9.60 centres of the two spheres in the opposite direction to B.

( )

After the collision there will be an angle of 76.7° between their directions of motion.

4 mA = 3m kg, uA = u m s−1, mB = 5m kg, uB = 0, e = e By conservation of momentum: 3m × u + 5m × 0 = 3m × vA + 5m × vB 3vA + 5vB = 3u By Newton’s experimental law: vB − vA = e(u − 0) vB − vA = eu

Exam-style questions 1 mA = 5m kg, uA = 2u m s−1, mB = 3m kg, uB = –4u, e = e By conservation of linear momentum 10mu – 12mu = 5mvA + 3mvB –2u = 5vA + 3vB By Newton’s experimental law: vB – vA = e (2u + 4u) vB – vA = 6eu Solving these simultaneous equations: 8vB = 30eu – 2u u vB = (15e − 1) 4 For B to travel in the opposite direction we need vB to be greater than 0. 1 This will be when e > 15 2 Parallel to the surface: Speed before = 5 cos 45°, speed after = 5 cos 45° Perpendicular to the surface: Speed before = 5 sin 45°, speed after = 5e sin 45° Velocity after =

256 – 163.84 sin2 q = 144 163.84 sin2 q = 112 sin q = 0.8268 q = 55.8º

(5cos45°)2 + (5e sin 45°)2 = 4

(5 cos 45°)2 + (5e sin 45°)2 = 16 12.5 + 12.5e2 = 16 12.5(1 + e2) = 16

Solving these simultaneous equations: 8vB = 3u(1 + e) so vB = 3 u(1 + e) 8 vB − vA = eu 3 u(1 + e) − v = eu A 8 3 3 vA = u + eu − eu 8 8 3 3 vA = u + eu − eu 8 8 vA = 3 u − 5 eu 8 8 1 vA = u(3 − 5e) 8 5 a Parallel to the surface:

Speed before = 5.2 cos 42°, speed after = 5.2 cos 42°



Perpendicular to the surface:



Speed before = 5.2 sin 42°, speed after = 5.2e sin 42°



Velocity after = (5.2cos42°)2 + (5.2e sin 42°)2 = 3.9



14.933 + 12.107e2 = 15.21

e = 0.151 b tan  q = 0.1513 × tan 42°

6 a Speed before first impact: v2 = u2 + 2as = 02 + 2 × 10 × 2

e = 0.529

= 40

3 Parallel to the wall:

Before = 16 cos q, speed after = 16 cos q



Perpendicular to the wall:



Speed before = 16 sin q, speed after = 9.6sin q



After the collision v = 12 m s−1



So (16cos θ )2 + (9.6sin θ )2 = 12



256 cos2 q + 92.16 sin2 q = 144



256 (1 – sin2 q) + 92.16 sin2 q = 144

q = 7.76°

v = 6.32 m s−1 b Speed immediately after first impact : v2 = u2 + 2as

02 = u2 + 2 × −10 × 1.2



u2 = 24

u = 4.90 m s−1 4.90 c e = 6.32 = 0.775

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6 

WORKED SOLUTIONS

7 a Speed before first impact: v2

=

u2

+ 2as

=

02

+ 2 × 10 × 6





Speed of B after impact with wall = 0.5 × 3.4 = 1.7 m s−1

10 a This is motion in a straight line.

= 120 v = 10.9545 m s−1

b Speed of B before impact with wall = 3.4 m s−1



Speed immediately after first impact = 10.9545 × 0.8 = 8.764 m s−1

mA = 0.05 kg, uA = 4 m s−1, mB = 0.04 kg, uB = 0 m s−1, e = ?, vA = 1 m s−1



By the conservation of linear momentum:

Height reached after first impact:



s=

2

2

2

2

v −u 0 − 8.764 = = 3.84 m 2 × −10 2a

v −u a 0 − 8.764 =2× = 1.75 s −10 c Speed immediately before second impact is 8.76 m s−1 b Time between bounces =  2×

8 a Speed after impact = 0.7 × 12 = 8.4 m s−1 b Lands when y = 0 y = Ut sin q − 1 gt2 + h 2 0 = 8.4 × t × sin 0 − 1 ×10 × t2 + 2 2 5t2 = 2 t = 0.632 s

4 × 0.05 + 0 × 0.04 = 0.05 × 1 + 0.04vB 0.15 vB = = 3.75  m s−1 0.04 b By Newton’s experimental law:

3.75 − 1 = e( 4 − 0)  

 e = 0.688

11 a tan 30° = e tan 60° 1  e = 3 b Parallel to the surface:

Speed before = 5u cos 60°, speed after = 5u cos 60°



Perpendicular to the surface:



5 Speed before = 5u sin 60°, speed after = u sin 60° 3



Speed after = (5u cos 60°)2 + 5 u sin 60° 3

(

c At time of landing:

vx = U cos q = 8.4 cos 0 = 8.4 m s−1



vy = U sin q − gt = 8.4 sin 0° − 10 × 0.63246 = 6.3246 m s−1 downwards



After impact:



vx = 8.4 m s−1

vy = 0.5 × 6.3246 = 3.1623 m s−1

So speed = 8.4 2 + 3.16232 = 8.976 m s−1 θ = tan −1 3.1623 = 20.63° 8.4

(

)

2 × 8.976 × sin 20.63 = 0.633 s 10 9 a  mA = m kg, uA = 4 m s−1, mB = m kg, uB = 0 m s−1, e = 0.7

T=

)

2

= lu

25 u2 + 25 u2 = l2u2 12 4 25 + 25 = l2 12 4 l = 2.89 12 a Time of flight: 1 y = Ut sin q − gt2 + h 2 0 = 4t sin 40° − 0.5 × 10 × t2 + 3

5t2 − 2.57115t − 3 = 0



t=



2.57115 ± (−2.57115)2 − 4 × 5 × −3 2×5 t = − 0.559 s or t = 1.07327 s



So the ball is in air for 1.07327 seconds before it hits the ground.



At point of impact:



Horizontal velocity = 4 cos 40° = 3.0642 m s−1



Vertical velocity = U sin q − gt



= 4 sin 40° − 10 × 1.07327 = −8.1615 m s−1



i.e. 8.1615 m s−1 downwards



Horizontal velocity before bounce = 3.0642 m s−1 so horizontal velocity after bounce = 3.0642 m s−1



Vertical velocity before bounce = 8.1615 m s−1 so vertical velocity after = 0.6 × 8.1615 = 4.8969 m s−1

vB = 3.4 m s−1



Resultant velocity after bounce



3.4 − vA = 2.8



vA = 0.6 m s−1



= 3.06422 + 4.89692 = 5.78 m s−1 4.88 Angle after bounce = tan −1 = 58.0° 3.06



By conservation of momentum:



m × 4 + m × 0 = m × vA + m × vB



vA + vB = 4



By Newton’s experimental law:

vB − vA = 0.7(4 − 0) vB − vA = 2.8

Solving these simultaneous equations:



2vB = 6.8

( )

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6 Momentum

b Time of flight to first bounce = 1.07327 s

Distance travelled to first bounce = Ut cos q = 4 × 1.07327 × cos 40° = 3.289 m 2U sin θ Time of flight to second bounce = g 2 × 5.7766 sin 57.964 ° = = 0.9794 s 10 Distance travelled to first bounce = Ut cos q = 5.7766 × 0.9794 × cos 57.964° = 3.001 m

v2 = u2 + 2as =

02



+ 2 × 10 × 10 10 2 v=



Time to reach ground after dropping:



t=



v − u 10 2 = = 2 10 a Speed after first bounce = e 2 Speed after second bounce = e 2 2 , etc. 2u Time of flight = g So total time of flight 20e 2 20e 2 2 20e 3 2 + + +… = 2+ 10 10 10

(

)



= 2 + 2 2e + 2e 2 + 2e 3 + …



= 2 + 2 × 2e 1−e = 2 1 + 2e 1−e = 2 (1 − e) + 2e (1 − e) (1 − e)



=

( ) ( ) 1+e 2( 1− e)

14 a Perpendicular to the line joining the centres of the spheres:

Solving simultaneously: 5vB = 1.1049 vB = 0.22098 m s−1



0.22098 − vA = 1.01771 vA = −0.79673 m s−1



Sphere A:



Velocity after impact = (2.0075)2 + (−0.7967)2



= 2.16 m s−1 2.0075 a= = 68.4° to the line joining 0.7967 the centres of the two spheres



Sphere B:



Velocity after impact = (1.8641)2 + (0.221)2   = 1.88 m s−1 1.8641 a = tan−1 = 83.2° to the line joining 0.221 the centres of the two spheres

Total horizontal distance = 6.29 m

13 Time to first bounce:







tan−1

( (

)

)

b Sphere B will hit the wall at an angle of 83.24° between the wall and the direction of travel with a speed of 1.877 m s−1.

Parallel to the surface:



Speed before = 1.877 cos 83.24° = 0.2209 m s−1, speed after = 0.2209 m s−1



Perpendicular to the surface:



Speed before = 1.877 sin 83.24° = 1.864 m s−1, speed after = 0.6 × 1.864 = 1.118 m s−1



Velocity after = 0.22092 + 1.1182 = 1.14 m s−1 Direction = tan −1 1.118 = 78.8° between the 0.2209 wall and the direction of travel



(

)

15 a mA = 3 kg, uA = 4 m s−1, mB = 4 kg, uB = −2 m s−1, e = 0.2

By conservation of momentum:



3 × 4 + 4 × −2 = 3 × vA + 4 × vB

Sphere B: speed after collision = 2.9 sin 40° = 1.8641 m s−1



By Newton’s experimental law:



Parallel to the line joining the centres of the spheres:

vB − vA = 1.2



By the conservation of linear momentum:





2m × 3.5 cos 35° − 3m × 2.9 cos 40° = 2mvA + 3mvB



2vA + 3vB = −0.93052



By Newton’s experimental law:



Sphere A: speed after collision = 3.5 sin 35° = 2.0075 m s−1



vB − vA = 0.2 × (3.5 cos 35° − (−2.9 cos 40°))

3vA + 4vB = 4 vB − vA = 0.2(4 − −2) Solving these simultaneous equations: 7vB = 7.6 vB = 1.09 m s−1

1.086 − vA = 1.2 vA = −0.114 m s−1

vB − vA = 1.01771

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6 

WORKED SOLUTIONS

b Time ball B is in the air: 1 s = ut + at 2 2 1 0.8 = 0 × t + × 10 × t2 2 So t2 = 0.16

So t = 0.4 s



Horizontal range:



x = 1.086 × 0.4 × cos 0°



Collision between B and C By conservation of linear momentum 3 × 0.4375 + 1 × 0 = 3vB + vC 3vB + vC = 1.3125 By Newton’s experimental law: vC – vB = 0.8 (0.4375) vC – vB = 0.35 3vC – 3vB = 1.05

= 0.434 m

Solving simultaneously,

So the ball will land 0.434 m away from the base of the table.

16 a Perpendicular to the line joining the centres of the spheres: Sphere A: speed after collision = u sin 50°

4vC = 2.3625 vC = 0.591 m s−1 The speed of C after the collision with B is 0.591 m s−1. 18 a By conservation of linear momentum

Sphere B: speed after collision = 0 m s−1



5 × 4 + 3 × 0 = 5vA + 4vB

Parallel to the line joining the centres of the spheres:



5vA + 4vB = 20



By Newton’s experimental law:

By the conservation of linear momentum:



vB – vA = 0.8 (4)

u cos 50° = vA + vB



vB – vA = 3.2

By Newton’s experimental law:



5vB – 5vA = 16

vB − vA = e × (u cos 50°)



Solving simultaneously,

vB − vA = eu cos 50°



9vB = 36

Solving simultaneously:



vB = 4 m s−1.

2vB = u cos 50° (1 + e) u vB = cos 50°(1 + e) 2 vA = u cos 50° − u cos 50°(1 + e) 2 u vA = cos 50°(1 − e) 2 Velocity of A after impact



4 – vA = 3.2



vA = 0.8 m s−1 (i.e. in the same direction as before the collision)

=

u 2sin 2 50° +

2 u2 cos2 50° (1 − e ) 4

b Velocity of B after impact =

u cos 50°(1 + e) 2

17 mA = 5 kg, mB = 3 kg, mC = 1 kg

b Parallel to the surface:

Speed before = 4 cos 45°, speed after = 4 cos 45°



Perpendicular to the surface:



Speed before = 4 sin 45°, speed after = 4e sin 45°



Velocity after = (4cos 45°)2 + (4e sin 45°)2  = 2.9

Collision between A and B By conservation of linear momentum



8e2 = 0.41

5 × 0.5 + 3 × 0 = 5vA + 3vB

e = 0.05125

5vA + 3vB = 2.5

  = 0.226

By Newton’s experimental law: vB – vA = 0.4 (0.5) vB – vA = 0.2 5vB – 5vA = 1 Solving simultaneously,

8 + 8e 2 = 2.9

c q  = angle between direction of B and wall after collision

0.226 tan 45° = tan q So q  = 12.7º

8vB = 3.5 vB = 0.4375 m s−1.

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6 Momentum

19 Collision between A and B

By the conservation of linear momentum



1 × 2 + 1 × 0 = (1 + 1) vAB

Vertical velocity before bounce = 6.32456 m s−1 so vertical velocity after = 0.8 × 6.32456 = 5.0596 m s−1

2 = 2vAB vAB = 1 m s−1.

Resultant velocity after bounce =  10 2 + 5.05962 = 11.21 m s−1 5.0596 Angle after bounce = tan −1 = 26.84° 10 To reach the assistant’s mouth, y = 1.1 1 1.1 = 11.21t sin 26.84° − × 10 × t2 2 5t2 − 5.0613t + 1.1 = 0

(

Collision between AB and C KE of AB before = 1 × 2 × 12 = 1 J 2 1 1 KE of AB after = × 1 = J 3 3 So velocity of AB after collision = = 0.5774 m s−1.

Horizontal velocity before bounce = 10 m s−1 so horizontal velocity after bounce = 10 m s−1

1 2

1 3

×2

Collision between AB and C By the conservation of linear momentum 2 × 1 + 1 × 0 = 2vAB + vC 2vAB + vC = 2 By Newton’s experimental law vC – vAB = e (1) 2vC – 2vAB = 2e Solving simultaneously,

)

t = 0.316 or t = 0.696 Total time taken to reach the assistant’s mouth is either 0.948 s or 1.33 s. 2 At these times, the distance since the bounce is: distance = 11.21 × 0.316 × cos 26.84° = 3.161 m and distance = 11.21 × 0.696 × cos 26.84° = 6.962 m  istance between release and first bounce D = 10 × 0.6325 = 6.325 m  o distance between the two entertainers is S 9.49 m or 13.3 m.

3vC = 2 + 2e vC = 2 (1 + e) 3 vAB = vC – e = 2 (1 + e) − e 3 2 1 = − e 3 3 From above, vAB = 0.5774 2 1 − e = 0.5774 3 3 e = 0.268

Mathematics in life and work 1 Time of flight until first bounce: 1 y = Ut sin q − gt2 + h 2 0 = 10t sin 0° − 0.5 × 10 × t2 + 2 5t2 = 2 t = 0.632456 s So the ball is in air for 0.632 seconds before it hits the ground. At point of impact: Horizontal velocity = 10 m s−1 Vertical velocity = U sin q − gt = 10 sin 0° − 10 × 0.632456 = −6.32456 m s−1 i.e. 6.32456 m s−1 downwards

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WORKED SOLUTIONS

Summary Review Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

Warm-up questions

Since F = 0.42R

1 tan a = 8 so cos a = 15 and sin a = 8 17 17 15 Resolving vertically: 15 T× =3 17 T = 3.4 N

F 2 + P × = 5g  0.42 7 Substituting 1 in 2 and making P the subject: P=

Resolving horizontally: 8 F = 3.4 × = 1.6 17

3

So F = 1.25W As W < F, the minimum vertical force to move the block is less than the minimum horizontal force to move the block.



P = 99 

2 i F = ma 1 0.4v 2 = 0.6 v dv dx

3 i Work done = Fd

= (30 cos a + 40 cos β) × 20



= (30 × 0.6 + 40 × 0.8) × 20 = 1000 J

ii  Constant speed so friction is equal to driving force and F = μR.

Resolving vertically: R = W



F = μR. 5 50 = W 8





2 = 3 v 2 dv dx



1

2 3 2x = 3 × v 2 + c 3 3

2x = 2v 2 + c When x = 1, v = 1 2=2+c So c = 0 3 x =v2 2 v = x3

1 sin a = 2 so cos a = 45 7 7 Sliding: Resolving horizontally:

2

P cos a = F

R + P sin a = 5g

Dividing by 0.2v 2 :

1

A Level questions

45 P 7 Resolving vertically:

1



2∫dx = 3∫v 2dv ii

W = 80 N

F=

4096 = 16 cm = 0.16m

Moments about A: 5g × 0.08 = P sin a × 0.16 + P cos a × 0.16 8P 0.16 45P 4= + 175 7 P = 20.10 N P for sliding is less than P for toppling, so the cube will slide first.

ii Using Newton’s second law: P − 1.25 × 60 = 6 × 4

50 = 19.47 N  745 2  +  0.42 7   

Toppling:

2 i Using F = μR and resolving horizontally, R = W



2

1

dx iii = x 3 dt −2 ∫ x 3 dx = ∫dt 1 3x 3 = t + c When t = 0, x = 1

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SUMMARY REVIEW



So c = 3 1 3x 3 = t + 3 At B, x = 8 3×2=t+3 t=3 3 Let vA and vB be the speeds of A and B after the collision. Conservation of momentum: 2mu = 2mvA + mvB 2u = 2vA + vB Newton’s experimental law: eu = vB – vA Solving simultaneously: 2u – eu = 3vA u(2 − e) vA = 3 2u + 2eu = vB + 2vB

2t t + 1 = 2 2 2 −1 t =1  t = 2 + 1 = 2.414 213 5 = 2.414 seconds to 3 decimal places, as required.

(

1 2

2u(1 + e) = 3vB 2u(1 + e) vB = 3 Newton’s experimental law for B bouncing off wall: 2u(1 + e) u(2 − e) 0.4 = 3 3 0.8(1 + e) = 2 – e 2 e= 3 Speed = distance ÷ time For B (from first collision to wall): 10u d = 9 t 9d t= 10u

(

V cos 45° t = V cos 60° (t + 1) cos 45° t = cos 60° (t + 1) t t +1 = 2 2

)

So the distance travelled by A during that time is given by dA where: 4u d A = 9 9d 10u 4u 10ud A = 9 9d 2d dA = 5 When B reaches the wall, A has travelled 2d metres. 5 From the wall to the second collision, the particles are travelling towards each other at the same constant speed, so (by symmetry), they will meet in the middle. The distance between A and B is 3d , so B will be 5 3d metres away from the wall when the particles 10 collide again. 4 i  At point of collision P and Q must have travelled the same distance horizontally.

)

Or using your calculator from line 2 in the above working: 0.707 11t = 0.5(t + 1) Make sure you write sufficient decimal places to show the answer is correct to 3 d.p. As before, t = 2.414 to 3 decimal places as required. ii At point of collision P and Q must have travelled the same distance vertically. 1 Use s = ut + at 2 vertically. 2 Vertical distance = gt 2 g(t + 1)2 V sin45° t – = V sin60° (t + 1) − 2 2 gt 2 g(t + 1)2 V sin45° t − V sin60° (t + 1) = − 2 2 Substitute in for t: 10(2.414)2 10(3.414)2 V 1.707 – V 2.957 = − 2 2 −1.250 V = 29.13698 – 58.27698 V = 23.3 iii Greatest height, U 2(sinq )2 23.32sin2 60° = = 20.35 2g 20 At point of collision, use S = Ut + 1 at 2 2  h = 23.3 × sin 60° × 3.414 − 5 × (3.414)2 = 68.89 – 58.28 = 10.61 H =

Vertical distance below greatest height = 9.74 m NB: You may have slightly different figures depending on the number of decimal places you work with. 5 i tan q = 5 so cos q = 12 and sin q = 5 12 13 13 Conservation of energy (with bottom of circle as zero potential energy): mg × 2 = mg ( 2 − 2cos q ) + 1 mv 2 2    20 = 20 − 20 × 12 + 1 v 2 13 2 1 240 2   v = 2 13    v = 480 ms−1 13

As acceleration = 0 horizontally, use S = Ut

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WORKED SOLUTIONS

ii  At B, the particle is h m above the ground, where h = 3 − 2cos q = 15 . 13 Let vG be the speed at impact with the ground. Conservation of energy (with ground as zero potential energy): Energy at B = energy at impact with the ground 10m × 15 + 1 m × 480 = 1 mv 2 G 13 2 13 2 150 + 240 = 1 v 2 13 13 2 G 780 2 v = G 13 v = 780 ms−1 G 13

iii Speed = distance ÷ time, so the horizontal displacement of the particle at time t is: x =  480 × 12  t  13 13   13 13  t = x  12 480  Using s = ut + 1 at 2 , the vertical displacement of 2 the particle at time t is: y =  480 × 5  t + 1 (−10)t 2 13  2  13



7 i Let m be the mass of the particle.

T = 3mg



Resolving vertically:



T cos q = mg

3mg cos q  = mg 8 cos q = 1 so sin q = 3 3

By Newton’s second law:



T sin q  = mrw 2 8 3mg × = mr × 52 3



r = 1.13 m



Using trigonometry on the triangle: sin q = r l 8 1.13… = 3 l l = 1.2 m



ii v = rw 8 i

 5 480  y = t − 5t 2  13 13  Substituting for t: y = −5

13 (1312 ) ( 480 )x 2

2

+

5 480 13 13 × x 13 13 12 480

2197 2 5 x + x 1152 12 480 2197 2 y= x− x 1152 1152 x y= ( 480 − 2197x ) 1152 y=−

6 Let x be the extension when the particle is initially at rest.

v = 1.13… × 5 = 5.66 m s–1 Shape

Arc

Rod

Total

Mass (kg)

1.8π

3.6

1.8π + 3.6

Distance of centre of mass from O (m)

1.8sin π 2 π 2

0

y

()

Centre of mass: 1.8π × 1.8 × 2 = (1.8π + 3.6 ) y π 6.48 = y 1.8π + 3.6      y = 0.700m ii

Shape

Frame Lamina

Mass (kg) M Distance of centre of mass from O (m)

By Hooke’s law: 28x T= = 17.5x 1.6 Resolving vertically: 17.5x = 0.35g

0.70…

Total

2.75

M + 2.75

()

y 2 × 1.8sin π 2 = 2.4 π 3π 2

x = 0.2 m



y = 1.8tan 22° = 0.7272

Conservation of energy: 28 × 0.22 1 1 + × 0.35 × 1.8 2 = 0.35 × 10 × 0.2 + × 0.35v 2 2 × 1.6 2 2



Centre of mass: 2.75 × 2.4 0.7002 M + = ( 2.75 + M ) × 0.7272 π 6.6 0.7002 M + = 2.000 + 0.7272 M π 6.6 − 2.000 π M= = 3.74 kg 0.7272 − 0.7002



0.35 + 0.567 = 0.7 + 0.175v2 v2 = 1.24 v=



1.11 m s–1

So the weight of the frame is 37.4 N.

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SUMMARY REVIEW

ii

9

14 m s–1

R2 1m q 3m

2g

1m

u m s–1

4m 74.5°

q 8g

Newton’s experimental law: q

F sin q = 3 and cos q = 7 4 4 Resolving vertically: R1 = 10g = 100 N F = μR1 = 100μ Resolving horizontally: F = R2 = 100μ Moments about the base of the ladder: 2 × 8g cos q + 3 × 2g cos q = 4R2 sin q 220 ×

q

e = 0.4

R1

7 3 = 4 × × 100µ 4 4

0.4 ×

× sin 74.5° u sin 14 cos74.5 ° = u=cos q q

u sin q = 1.4422 Motion parallel to the ground: 14 cos74.5° = u cosq u cos q = 1 u2 = 1.44222 + 12 = 3.080 u = 1.75 m s–1 (3 sf) Substituting in (1): sin q = 1.4422 = 0.82178 1.7550 q = 55.3° (3 sf) For vertical motion after the bounce: using v2 = u2 + 2as

µ = 11 7 60

u = 1.755 sin 55.26° m s–1, a = –10 m s–2, s = ?, v = 0 m s–1

v2 = 3 + 2 × (–10) × (–0.5) = 13 v = ± 13 As the ball is moving towards the ground, v = − 13 m s−1. Horizontal motion is a constant speed of 2 cos 60° = 1 m s−1. Let vG be the speed on impact with the ground. By Pythagoras: vG2 = 12 + 13 vG = 14 m s−1 By trigonometry: tan α = 13 a = 74.5° The ball hits the ground at a speed of 14 m s−1 at an angle of 74.5° to the horizontal ground.

2

Square and add 1 and 2 :

220 7 = 1200µ

10 i Using v2 = u2 + 2as u = 2sin 60° = 3 m s−1, a = –10 m s–2, s = – 0.5, v=?

1

0 = 1.4422 + 2 ×(–10) × s s = 1.4422 = 0.0721m = 7.21cm (3 sf ) 20 11 For complete circular motion, tension ⩾ 0 at the top of the circle. By Newton’s second law: mu 2 T + mg = a mu 2 T= − mg a But T ⩾ 0, so mu 2 − mg  0 a u2  g a u  ag By conservation of energy: 1 1 mg ( 2a ) + mu 2 = mv 2 2 2 1 1 2 2ag + ag = v 2 2 5 1 2 ag = v 2 2

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WORKED SOLUTIONS

v2 = 5ag v = 5ag (where v is the speed of P before the collision) Let vC be the speed of the combined particle after the collision. By conservation of momentum: 5m m 5ag = × vC 4 4 vC = 5ag 5 By conservation of energy: 5mg 1 5m 16 1 5m 2 × × 5ag = a + a cos q ) + v 2 4 25 4 ( 2 4

( )

( )

5mga 5mga cos q 5mv 2 + + 4 4 8 2 6mga = 10mga cos q + 5mv  2mga =

T +

2 in 1 : 6mga = 10mga cos q + 4aT + 5mga cos q 4aT = 6mga − 15mga cos q 3mg 15mg cos q − 2 4 The string becomes slack when T = 0: 3 15 − cosq = 0 2 4 cos q = 6 = 2 15 5

T =

12 i Horizontally acceleration = 0 so velocity is constant and = 14cos 60° = 7 Vertically use v = u + at v = 14 sin 60° – 1.8g = –5.8756 q

7 m s–1

v

5.8756 m s–1

v2 = 72 + 5.87562 = 83.52 v = 9.139 m s–1 tan  q  = 5.8756/7 q  = 40.0° Speed = 9.14 m s–1 in direction 40.0˚ below the horizontal

ii Use s = ut + 1 at 2 in vertical direction 2 −2 = (14sin60 )t − 1 gt 2 2 5t2 – 12.124t – 2 = 0



2

   3sin q − cos q = 0.8sin q





(





tanq =



q  = 47.0°

1

By Newton’s second law: 5mg cos q 5m v 2 = 4 4 a 4aT + 5mga cos q = 5mv 2 

Use the quadratic formula, taking only the positive root (the negative root is before the motion started) t = 2.58 s 13 i Take moments about A 10 × 1.2 sin (q – 30°) = 6 × 0.8 sinq Use: 3 1 sin (q − 30° ) = sin q cos30° − cos q sin 30° = sin q − cos q 2 2  3  1 12 ×  sin q − cos q  = 6 × 0.8sin q 2  2 

)

3 − 0.8 sin q = cos q

(

1 = 1.0729 3 − 0.8

)

ii Resolve horizontally Reaction at A = 10 sin 30° (as the rod is in equilibrium) Resolve vertically Friction force at A = 10cos 30° – 6 Use F ⩽ μR, Limiting equilibrium, so F = μR 10cos 30° – 6 = μ × 10 sin 30° (10cos30° − 6) µ= = 0.532 10sin30° 1 1 3 3 14 i At A: KE = 2 mv 2 = 2 m 10 ga = 2 ma , PE = mgh = 10ma

1 At bottom: KE = 2 mv 2





Conservation of energy: 3 1 ma + 10ma = mv 2 2 2 3a + 20a = v2 v 2 = 23a





Acceleration towards centre =



F = ma = 23m



R = 23m + mg = 33m

v 2 23a = = 23 r a

ii At B: PE = mgh = 10m ( a − asin q ) = 30 ma, 4 1 2 KE = mv 2 Conservation of energy: 1 m × 23a = 30 + 1 mv 2 2 4 2 46ma = 30ma + 2mv2 v 2 = 8a v = 8a = 2.83 a

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SUMMARY REVIEW

iii Vertical component of speed at B = 8a cos q sin q = 1 , ∴ cos q = 15 (Note: as q  is acute  4 4 sin2 q + cos2 q  = 1) Use v2 = u2 + 2aS for vertical motion of projectile. At greatest height above point of projection v = 0 0 = 8acos2 q  – 20S 20S = 15a 2 S = 15a = 3a 40 8 At moment of projection let h be the distance of the bead below the level of O. sin q = h a Greatest height of bead above the level of O 3a 1 1 = − a = a as required. 8 4 8 15 i At point of release PE = mgh = 6h, KE = 0, 2 EE = λ x = 45 × 1 = 15 2 × 1.5 2l When P comes to rest PE = 0, KE = 0, 2 2 EE = λ x = 45 × h = 15h 2 2 × 1.5 2l

iii At top position, resultant force λ x = 6 + 45 × 1 = 36 = mg + 1.5 l F = ma, 0.6a = 36, a = 60 m s–2 At bottom position, resultant force λ x − mg = 45 × 1.22 − 6 = 30.6 = 1.5 l

F = ma, 0.6a = 30.6, a = 51 m s–2

Greatest magnitude of the acceleration of P is 60 m s–2 16 i Vertically, resultant force = 0.6g – 3v Use F = ma with acceleration as dv to find t dt 0.6g – 3v = 0.6 dv dt 0.6 dv = 0.6g − 3v dt Divide both sides by 0.6       dv = g − 5v dt

  





1

∫ 10 − 5v dv = ∫dt

−

1 −5 dv = dt ∫ 5 ∫ 10 − 5v

     − 1 ln (10 − 5v ) = t + c Conservation of energy, 5 15h2 = 15 + 6h Substitute initial conditions, i.e. V = 0 when t = 0 5h2 – 2h – 5 = 0 1 − ln (10 ) = c Solve for h, taking the positive solution 5 h = 1.22 m 1 1 t = ln (10 ) − ln (10 − 5 × 1.95) 5 5 ii Greatest speed occurs at equilibrium position. 1 1 t = ln (10 ) − ln ( 0.25) Let x be the distance between the point of 5 5 1 release and the equilibrium position. t = ln ( 40 ) 5 Resolving vertically using Hooke’s Law   t = 0.738 seconds λ x λ (1 − x ) = + mg l l ii Use F = ma with acceleration as v dv to find x dx 45x = 45(1 − x) + 6   −3v = 0.6v dv 1.5 1.5 dx 45x = 45 – 45x + 9 dv = −5 dx 90x = 54 x = 0.6 ∫dv = −∫5dx Distance below A = 0.6 + 1.5 = 2.1m    v = –5x + c Energy at equilibrium position S  ubstitute initial conditions, i.e. v = 1.95 and 2 2 1 λ x 2 + λ (1 − x ) = 1 0.6v 2 + 45 × 0.6 2 + 45 (1 − 0.6 ) = mv 2 + x = 00.3v 2 + 5.4 + 2.4 2 2 × 1.5 2 × 1.5 2l 2l 2 c = 1.95 2 2 2 2 λ x + λ (1 − x ) = 1 0.6v 2 + 45 × 0.6 + 45 (1 − 0.6 ) = 0.3v 2 + 5.4 + 2.4 mv 2 + v = –5x + 1.95 2 2 × 1.5 2 × 1.5 2l 2l 2 P comes to rest when v = 0 × 0.6 2 45 (1 − 0.6 ) 2 + = 0.3v + 5.4 + 2.4 × 1.5 2 × 1.5 5x = 1.95 = 7.8 + 0.3v2 x = 0.39 m Conservation of energy: 17 Let vA and vB be velocities of spheres A and B after 7.8 + 0.3v 2 = 0.6g × 0.6 + 45 × 1 2 × 1.5 impact, respectively. Momentum before = momentum after 7.8 + 0.3v2 = 3.6 + 15 4mu = 4mvA + 2mvB 0.3v2 = 10.8 1 2u = 2vA + vB v2 = 36 v = 6 m s–1

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WORKED SOLUTIONS



e × (speed of approach) = speed of separation eu = vB – vA vB = eu + vA

2

Substitute 2 into Eqn 1

18 i Horizontally, s = ut, x = (15cos41°)t Vertically, s = ut + 1 at 2 , y = (15sin41°)t – 5t2 2 x Substitute t = 15cos41°

2u = 2vA + eu + vA = eu + 3 vA



2u = eu + 3 vA (2 – e) u = 3 vA  Consider the change in KE of sphere A 1 1 1 (4m)v 2A = × (4m)u 2 2 4 2 1      v 2A = u 2  4

3

4

y = (15sin41°) ×

(

x x −5 15cos41° 15cos41°

)

2

y = 0.869x – 0.0390x2 ii When x = 1.5 y = 0.869 × 1.5 – 0.0390 × 1.52 = 1.216 Remember to add the height of O above the horizontal. Height of fence, h = 1.6 + 1.216 = 2.82 m

Substitute 4 into the square of 3



Ball lands when y = –1.6

(2 – e)2 u2 = 9 (vA )2 1 (2 – e)2 u2 = 9 u 2 4



–1.6 = 0.869x – 0.0390x2



0 = 1.6 + 0.869x – 0.0390x2



Solve using the quadratic formula



x = 24.0



Distance from fence to A = 24.0 – 1.5 = 22.5 m



4 (2 – e)2 = 9

 4e2 – 16e + 7 = 0   e 2 − 4e + 7 = 0 4       e = 1 or e = 7 2 2 1 As 0  e  1, e = 2 2 From above v A = 1 u 2, so v A = 1 u 4 2 eu = vB – vA 1 1 u = v B − u , so vB = u 2 2 Consider the collision between spheres B and C Let the speeds of spheres B and C after collision be v'B and vC, respectively e × (speed of approach) = speed of separation 1 e × u = vC − v' B 2 1 v' B − vC = − u  5 4 Conservation of momentum 1 2mu + mu = 2mv 'B + mvC 2 5 u = 2v' + v  6 B C 2 Solve 5 and 6 simultaneously 5 2v' B + vC = u 2    v' − v = − 1 u B C 4      3v' = 9 u B 4      v' = 3 u B 4 3 As v' B = u and v A = 1 u and 3 u > 1 u, spheres 4 2 4 2 A and B will not collide again, so no further collisions between the spheres.

19 i For triangle, centre of mass lies on the line of symmetry. Height of triangle = 0.36 × 2 ÷ 0.6 = 1.2 Distance of centre of mass from base of triangle = 1.2 ÷ 3 = 0.4 For semicircle, centre of mass lies on line of symmetry at a distance of 4r from the straight 3π edge: 4r = 0.8 π 3π Let mass per sq metre be m Take moments about the axis OB If the centre of mass of the lamina lies on OB then these moments will be equal. For triangle: mg × 0.36 × 0.4 = 0.144mg



2 For semicircle: mg × π0.6 × 0.8 = 0.144mg 2 π



Centre of mass lies on OB.

ii Let x be the distance of the centre of mass of the lamina from the point O. Take moments about the axis OC: For the parts: 0.36mg × 0.3. (Note: centre of mass of semicircle is on this axis.)



2 For the whole: (0.36mg + π × 0.6 mg ) × x 2



2 0.36mg × 0.3 = (0.36mg + π × 0.6 mg ) × x 2



2 0.36 × 0.3 = (0.36 + π × 0.6 ) × x 2    x = 0.117 m

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SUMMARY REVIEW

20

O a

q

R

v m s–1 q mg

h mg

Initially energy = KE = 1 m × 7 ga = 7 mga 2 2 4 After motion has started, P also has PE due to the vertical height gained. Conservation of energy: 7 1 mga = mgh + mv 2 4 2 7 mga = mg(a − acos q ) + 1 mv 2 4 2 v 2 = 7 ga − 2ga + 2gacos q 2     3 2 v = ga + 2gacos q 2     2 Acceleration towards centre = v r mv 2 mv 2 F = ma = = r a 2 Resolving forces towards centre: R = mv + mgcos q a Substituting for v2 R = m × 3 ga + 2gacos q + mgcos q a 2 3mg R= + 2mgcos q + mgcos q 2 3mg R= + 3mgcos q 2 3mg R= 1 + 2cos q ) as required 2 ( i It loses contact with the sphere when R = 0 1 + 2cos q  = 0     cos q = − 1 2       v 2 = 3 ga + 2gacos q = 3 ga − ga = 1 ga 2 2 2       1 v= ga 2 ii Consider energy when P passes through the horizontal plane level with O. 1 mv 2 + mga 2 Conservation of energy: 7 mga = 1 mv 2 + mga 4 2 7 1   ga = v 2 + ga 4 2 2    v = 2 × 7 ga − ga = 3 ga 4 2

(

)

(

  

v=

3 ga 2

)

21 i Consider forces vertically. Resultant force, F = mg + T T = λ x = 19.2 × 1.5 1.2 l F = 0.4 × 10 + 19.2 × 1.5 = 28 1.2 Use F = ma 28 = 0.4a a = 28 = 70 m s–2 0.4 ii Initial energy = PE + EE 2 = mgh + λ x 2l 19.2 × 1.52 = 0.4 × 10 × 2.7 + 2 × 1.2 = 10.8 + 18 = 28.8 At A, energy = 1 mv 2 2 = 0.2v 2 Conservation of energy: 0.2v 2 = 28.8 28.8    v2 = = 144 0.2   v = 12 m s−1 22 i Use F = ma 4e −x − 2.4x 2 = 0.8 × v dv dx 4e −x 2.4x 2 dv − =v 0.8 0.8 dx v dv = 5e −x − 3x 2 dx dv v = 5e −x − 3x 2 ii dx



Separate and integrate.

∫v dv = ∫(5e

−x

− 3x 2)dx

2

 v = −5e −x − x 3 + c 2 Substitute initial conditions, i.e. v = 6, x = 0  18 = –5 + c, c = 23 v2 = −5e −x − x 3 + 23 2 Now substitute x = 2 to find the required value of v. v2 = −5e −2 − 23 + 23 2 v2 = 14.32 2 2    v = 28.64    v = 5.35 m s–1

23 First collision: Let vA and vB be the speeds of A and B, respectively, after the collision. Momentum before = momentum after 2mu = 2mvA + mvB    2u = 2vA + vB e × (speed of approach) = speed of separation −2 u = v A − vB 3  2u = 2vA + vB

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WORKED SOLUTIONS

Adding gives: 4 4 u = 3v A vA = u 3 9 2u = 2vA + vB 2u = 2 × 4 u + vB 9 10 10 u = vB vB = u 9 9

20 t 24 i Horizontally, use s = ut  x = ucos 45°t = 2 x = 10 2 t Vertically, use s = ut + 1 at 2   y = 20sin 45°t – 5t2 2 y = 10 2t − 5t 2

Second collision (sphere B and the barrier): Let the speed of B be v'B after the collision. e × (speed of approach) = speed of separation 10 e u = v' B 9





x 10 2 Substitute this into y = 10 2t − 5t 2





y = 10 2 ×

ii x = 10 2 t ∴ t =

2

 x  x 5x 2 − 5  = x − 200  10 2 10 2





After A and B collide again, they move either in the same direction or in opposite directions.

y=x−

x2 40 iii Ball strikes the ground when y = 0

If they move in the same direction:





0=x−

Conservation of Momentum:

( ) 1 4u 10eu + =w 9 ) 6( 9 

1

2 in 1

(

8u 10eu 7 4u 10eu − = + 9 9 6 9 9

 x = 40, so ball first strikes the ground 40 m from O. 25 Centre of mass: distance from O = 2rsinα 2α on line/plane of symmetry 2r sin π 3 2r sinα 3r = = 3α π π 3 3

() ()

2 4u 10eu + = 5w A − w A = 4w A 3 9 9 2

A



)



48u − 60eu = 28u + 70eu 130eu = 20u

( ) 1 4u 10eu + = 3w 9 ) 3( 9  2 in 1

(

8u 10eu 1 4u 10eu − = + 9 9 3 9 9 24u − 30eu = 4u + 10eu 40eu = 20u e=

1 2

a

P mg

T q

A

P 2

)

O

1

2 4u 10eu + = 5wA + wA = 6wA 3 9 9 A

( )

q  = 0.224 radians

( ) ( )

26

8mu 10meu − = −2mwA + 5mwA 9 9

( )



e=

Conservation of Momentum:

()





2 13 If they move in opposite directions:

Take moments about O: π = 7.5r For 15N force  15rcos 3 π 3 For centre of mass  20 cos  3 − q × π r

π 3 At limit of equilibrium: 7.5r = 20cos 3 − q × π r cos π − q = 7.5 × π = 0.6802 3 20 3 π − q = 0.8228 3



8u 10eu − = 3wA 9 9  Newton’s Experimental Law:

)



8mu 10meu − = 2mwA + 5mwA 9 9 8u 10eu − = 7wA 9 9  Newton’s Experimental Law:

(

x2 x = x 1− 40 40

mg Initial energy = PE = mga At base of motion energy = KE = 1 mv 2 2 Conservation of energy : 1 mv 2 = mga 2    v  2 = 2ga Circular motion about A, when angle is q Energy = 1 mv 2 + mg ( a − x ) (1 − cos q ) 2

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SUMMARY REVIEW

Conservation of energy: 1 mv 2 + mg a − x 1 − cosq = mga ( )( ) 2 2 v = 2gx + 2gacos q  – 2gxcos q   = 2gx + 2g(a – x)cos q 2 Acceleration towards centre = v r mv 2 v2 F = ma = m = r (a − x) Resolving forces towards centre: 2 T = mgcosq + mv (a − x) m(2gx + 2g ( a − x ) cosq ) T = mgcosq + (a − x) 2mgx T = mgcosq + + 2mgcosq a−x 2mgx T = 3mgcos q + a−x T = mg(3cos q + 2x ) a−x P completes a circle if T ⩾ 0 at the top of the motion; for minimum value, T = 0 at the top of the circle, which is when q = π radians. 0 = mg(3cos q + 2x ) a−x    2x 3=     a − x 3a – 3x = 2x   3a = 5x x 3 =    a 5 27 i Tension = 0.4g = 4 N

Use Hooke’s law T = λ x l 4 = 16 × x 0.8   x = 0.2 m ii When P comes to rest all energy is elastic, 2 2 16 × ( 0.6 ) EPE = λ x = 2 × 0.8 2l When P is projected, PE = mgh =  0.4 × 10 × 0.4 = 1.6

1 1 2 2 KE = 2 mv = 2 × 0.4u 2 PE = λ x = 16 × ( 0.2) = 0.4 2 × 0.8 2l Conservation of energy 3.6 = 1.6 + 0.2u2 + 0.4 1.6 = 0.2u2  u2 = 8 2

−1    u = 8 = 2.828427… = 2.83m s iii When string becomes slack, PE = mgh = 0.4 × 10 × (1.4 – 0.8) = 2.4 1 1 KE = mv 2 = × 0.4 × v 2 = 0.2v 2 2 2 EPE = 0 (as there is no extension x = 0)







  v = 2.45 m s–1

Conservation of energy 3.6 = 2.4 + 0.2v2 1.2 v2 = =6 0.2

28 i Resultant force = mg – 0.8v = 2 – 0.8v Use F = ma 2 – 0.8v = 0.2a a = 2 − 0.8v = (10 − 4v) m s–2 0.2 0.2 ii Use a = dv dt dv = (10 − 4v) dt −4 dv 1 = ∫dt −4 ∫ 10 − 4v −1 ln (10 − 4v ) = t + c 4 v = 0 when t = 0    −1 ln 10 = c ( ) 4 −1 ln (10 − 4v ) = t − 1 ln (10 ) 4 4 When t = 0.6 −1 ln (10 − 4v ) = 0.6 − 1 ln (10 ) 4 4    10 – 4v = 0.9072         v = 2.27 m s–1 29 Conservation of momentum Momentum before = momentum after 8mu – 3mu = 2mvA + vB 5mu = 2mvA + vB

1

Coefficient of restitution e × (speed of approach) = speed of separation –e(4u + 3u) = vA – vB –7eu = vA – vB

2

1 + 2 5mu – 7eu = 3vA 1 v A = (5u − 7ue) 3 vA < 0 if 5u – 7ue < 0 5u < 7ue 5 e> 7

Extension questions 1 Let h be the ‘height’ of the cone. 3 2 2π Volume of the hemisphere = π ( 0.1) = 3 3000 Volume of the cone = 1 π ( 0.1)2 h = π(h) 3 300 Mass = density × volume

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WORKED SOLUTIONS

Moments about the common circle: D × 2π × 3 × 0.1 = D × πh × h 3000 8 300 4



For particle A



1 After collision KE = 2 m (1.1 2gl )2

0.6 h2 = 24 000 1200

At point at which it comes to rest PE = mgh where h = height above point of collision

h2 = 0.03



h = 0.1732 m



( )(

)

( )()

For the hemisphere end: r = 0.1 so v = 0.1 × 4 = 0.4 m s−1 For the cone end: 0.693 m s−1

2 i Before collision for A initially: PE = mgl











Vertically below O, before collision KE = 1 mv 2 2 Conservation of energy: 1 mv 2 = mgl 2 v2 = 2gl v = 2gl



Similarly for B  PE = 3mgl



KE = 1 (3m)v 2 2 1 3m v 2 = 3mgl and similarly v = 2gl 2( )



ii e(speed of approach) = speed of separation 0.4 2 2gl = v A + v B 0.8 2gl = v + v

(

)

A





q  = 102°

3 For the initial fall: using v2 = u2 + 2as u = 0 m s–1, a = 10 m s–2, s = 20 m, v = ? v2 = 2 × 10 × 20 = 400 v = 20 m s–1 The speed at which the ball leaves the ground is equal to the speed at which the ball hits the ground again when it comes back down. For a height of ⩽ 5 m: u = ?, a = –10 m s–2, s ⩽ 5 m, v = 0 m s–1 02 ⩽ u2 + 2 × (–10) × 5 u2 ⩾ 100 u ⩾ 10 m s–1 By Newton’s experimental law: v = 0.85u

Conservation of momentum



(left as positive)

So after n bounces, the speed of the ball immediately after the bounce is given by:



Momentum before = momentum after

vn = 20 × 0.85n

−m 2gl + 3m 2gl = mv A − 3mv B



2 2gl = v A − 3v B

For the height of the bounce is to remain below 5 m, vn ⩾ 10.



Solving the two equations leads to v B = −0.3 2gl v A = 1.1 2gl

iii For particle B

B

cosq  = –0.21







Conservation of energy: 1 × m(1.1 2gl )2 = mgh 2 1.21l = h 1.21l = l(1 – cosq )

r' = 0.1732 v = 0.1732 × 4 =

h = l(1 – cosq )

1 After collision KE = 2 3m (0.3 2gl )2 At point at which it comes to rest PE = 3mgh where h = height above point of collision h = l(1 – cos q ) Conservation of energy: 1 × 3m (0.3 2gl )2 = 3mgh 2 0.09l = h 0.09l = l(1 – cos q ) cosq = 0.91 q = 24.5°

20 × 0.85n ⩾ 10 0.85n ⩾ 0.5 ln 0.85n ⩾ ln 0.5 n ln 0.85 ⩾ ln 0.5 n  ln 0.5 ln 0.85 n ⩾ 4.265 So the ball will remain below 5 m after 5 bounces. 4 Use F = ma dv x2 = 2v dx v sin 2 Separate and integrate v 2 ∫ x dx = 2∫vsin 2 dv

()

()

Integrate the RHS using integration by parts

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SUMMARY REVIEW

() () x = 2  −2vcos v + 4sin v  + c ( 2 ) ( 2 )   3 x = −4vcos v + 8sin v + c ( 2) (2) 3

x 3 = 2  −2vcos v − − 2cos v dv    3 2 ∫ 2

x

3

T

3

Substitute in the given initial conditions, i.e. when t = 0, v = π, x = 0 0 = −4π cos π + 8sin π + c 2 2 c = –8 x3 v v = −4vcos + 8sin −8 3 2 2

()

()

() () v v x = 3  −4vcos ( ) + 8sin ( ) − 8  2 2   v v x = −24 + 24sin ( ) − 12vcos ( ) 2 2 3

3

5 Initially, elastic energy =



0.25

λ x 2 = 10 × 0.25 × 0.25 = 0.625 1 2l At point of projection all energy is converted to KE 1 mv 2 = 0.625 2 v 2 = 12.5 v = 12.5 For motion from point of projection Horizontally x = 12.5 × cos 45° × t = 2.5t x = 5 t , or t = 2 x 2 5 Vertically using s = ut + 1 at 2 2 1 y = 12.5 sin 45° t − 10 t 2 2 y = 2.5t – 5t2 4 Substituting for t, y = x − x 2 5 This is the equation of motion for the particle from the point of launch.

T – mgcosq = 0

λ x = mgcosq l 10x 1 = 0.1 × 10 × 0.5 2 x=

1 2 = v 20 2 40

mg

0.5 sin 45 = 1 2 2

45° 0.5 cos 45 = 1 2 2 Relative to O, point of launch has coordinates  1  − 1 , 1 − 1  2 2 40 2 2 40  Therefore equation of trajectory relative to O is 2

 1  4 1 1  1 1 1 y = x − + + −  − x −  +  2 2 40  5  2 2 40  2 2 40 6 i If P and Q collide, l < 4.2 m ii For particle P: 2 Energy at P = λ x = 1 6 2l Energy at A = 1 mv 2 = 1 due to conservation of 2 6 energy

3 6 For particle Q:

v=

2 Energy at Q = λ x = 5 6 2l Energy at B = 1 mv 2 = 5 2 6 5 v= 3 e(speed of approach) = (speed of separation)  3 5 0.7  + = v A + vB 3   6 Momentum before = momentum after



For maximum speed, acceleration = 0 F = ma

1 40

2 40 1 40



3 5 ×4− × 3 = 3v B − 4v A 6 3  3 5 4 × 0.7  + = 4v A + 4v B 3   6

solve simultaneously for vB leads to vB = 0.2591 For Q: energy after collision = KE = 1 mv 2 = 1 × 3 × (0.2591)2 = 0.1007 2 2 When Q stops moving all energy is elastic,

λ x 2 = 2x 2 = 0.1007 2.4 2l x = 0.3476 Distance from B = 1.2 + 0.34766… = 1.55 m (3 s.f.)

78

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WORKED SOLUTIONS

7 Use F = ma dv e sin 2x = 4v dx 2x

Separate and integrate

9 Use s = ut + 1 at 2 2 At collision height, for P: s = uT – 5T 2 At collision height, for Q: s = 2u(T – 2) – 5(T – 2)2

∫ 4vdv = ∫ e2x sin 2xdx For the RHS use integration by parts twice. e 2xsin 2x e 2xcos2x 2v 2 = − +c 4 4 Substitute initial conditions, x = 0, t = 0, v = 2 1 8 = 0 − + c , c = 8.25 4 e 2xsin 2x e 2xcos2x 2 2v = − + 8.25 4 4 e 2xsin 2x e 2xcos2x 33 − + 8 8 8

v=

At collision height uT – 5T 2 = 2u(T – 2) – 5(T – 2)2



uT – 5T 2 = 2uT – 4u – 5T2 + 20T – 20

4u + 20 = uT + 20T = T(u + 20) 4u + 20 T = u + 20 10 5 5 sin 60 =5 3 2

8 Initial energy = 1 mv 2 + mgh = 1 × 100 × u 2 + 100 × 10 × (5sin 30° + 5) 2 2 = 50u2 + 7500 At point of collision energy = 1 mv 2 = 0.5 × 100 × v 2 = 50v 2 2





60° 5 cos 60 = 5 2 8 8 sin 30 = 4

Conservation of energy 50v2 = 50u2 + 7500 v2 = u2 + 150 e(speed of approach) = speed of separation 2

e u + 150 = vQ + v P e u 2 + 150 = 5u + v P Conservation of momentum Before = after 2

100 u + 150 = 5vQ − 100v P 2

100 u + 150 = 25u − 100v P



30° 8 cos 30 =4 3

Use s = ut horizontally and s = ut + 1 at 2 vertically 2 5 For P horizontally x = t 2 5 3 vertically y = t − 5t 2 2 For Q horizontally x = 4 3t

vertically y = 4t – 5t2

When P and Q collide the time of flight for Q is t and time of flight for P is t + T

100 u 2 + 150 = 25u − 100 e u 2 + 150 − 5u   

  5 (t T ) +  4 3t    2   =  2  4t − 5t   5 3 (t + T ) − 5(t + T )2    2

100 u 2 + 150 = 25u + 500u − 100e u 2 + 150

Using the x component (t + T ) =

Eliminate vP using the equation above for e.

100 u 2 + 150 + 100e u 2 + 150 = 525u

(1 + e )100 (1 + e ) = e=

u 2 + 150 = 525u

525u 100 u 2 + 150

525u −1 100 u 2 + 150

2 4 3t 5 Substitute this into the y component 2 4t − 5t 2 = 5 3 2 4 3t − 5 2 4 3t 2 5 5 4 4t − 5t 2 = 12t − × 48t 2 5 0 = 8t – 33.4t2

(

) (

)

0 = t(8 – 33.4t) Solving, t = 0.2395

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SUMMARY REVIEW

Substitute into (t + T ) = T =

( 52 × 4 3 − 1)t

2 4 3t 5

Substitute v = 4 when x = 3 65

(

)

2 = ln 3 65 + 24 + c c = 2 − ln(24 + 3 65 )

T = 0.424 s

2 11 i For a projectile the range of flight u sin 2α g 2 For particle A, range of flight = u 10 u2 For particle B, range of flight = 10 u2 Given that the particles collide, 0 < l < 5 ii Horizontal speed before collision =  10cos 45° = 10 2

(

)

v 2 = ln x + x 2 − 9 + 2 − ln(24 + 3 65 ) 8 When x = 6

(

)

v 2 = ln 6 + 6 2 − 9 + 2 − ln(24 + 3 65 ) 8 v2 = 4.324 v = ±2.08 m s–1

Horizontally:

e(speed of approach) = speed of separation



 10  0.5  2 ×  = v P + vQ  2

v P = vQ =



Collision takes place 7 m from O



Consider horizontal motion

5 2

s = ut  7 = ucos 45°t, 7 = 10 t  i.e. t = 7 2 10 2 Collision takes place 7 2 seconds after 10 particles are projected.



Total time of flight – consider vertical motion s = ut + 1 at 2 2



  10 0 = 10sin 45°t − 5t 2 = t  − 5t   2 

t = 0, or 5t = 10 ie t = 2 seconds 2 Time of flight left after collision takes place = 2 −7 2 10





Horizontal distance covered in this time



  5 35 =5− = 1.5 =  2 − 7 2 × 10  10  2

Both particles cover this distance, ∴ distance between A and B is 3 m 4 x2 − 9 d 4 v v = 2 dx x −9 Separate and integrate 1 v dv = 1 ∫ x 2 − 9 dx 4∫

12 a =

(

)

v 2 = ln x + x 2 − 9 + c 8

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