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Cambridge International AS & A Level Mathematics

Mechanics STUDENT’S BOOK: Worked solutions

Tom Andrews, Michael Kent Series Editor: Dr Adam Boddison

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

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1

WORKED SOLUTIONS

Worked solutions 1 Forces and equilibrium 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

c

1 5x + 4y = 46

1

2

2x = 3y Rearrange equation 2 . x= 3y 2 Substitute for x in equation (1).

( )

5 3 y + 4y = 46 2

15y + 8y = 92

23y = 92

2x = 3 × 4 = 12

R P W

2 A = C, B = D 3

R

X

T

y=4 4N

x=6

4 a

58 N

2 14 ÷ sin 58 = 16.5 cm 3 cos–1(15 ÷ 19) = 37.9°

12 N

4 a d2 = 192 + 132 − 2 × 19 × 13 × cos 122° = 791.78

T

d = 791.78 = 28.1 cm

b sin BAD = sin122° 28.1 13

T = 12 N c W − 58 = 0

Exercise 1.1A 1 a

W

b T − 12 = 0

W = 58 N

Exercise 1.2A

R

1 a q

mg

R

b

F

R

b

T W

F

mg R

F

T 60° mg

1

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1 Forces and equilibrium

c

R

R − X sin θ c W = X cosθ − F = cosθ sin θ

T

X cos2 θ – F cos θ = R sin θ – X sin2 θ

30º mg

X(sin2 θ + cos2 θ) = R sin θ + F cos θ d

R F

q

R = X sin 12° + 250 cos 12°

mg

F=

e

R F 26°

f

4X cos 12° = X sin 12° + 250 cos 12° + 1000 sin 12°

F mg

50°

250 cos12° + 1000 sin 12° = 122 N 4 cos12° − sin 12°

X=

P

9 T cos 30 = F T sin 30 + R = W F W = cos 30° sin 30° + R

20° T2

W= mg W 2sin θ

2 a 2T sin θ − W = 0

b T =

3 a T cos 19° − F = 0

b T sin 19 + R − W = 0

4 a TB cos 65° − TA cos 25° = 0 b TA sin 25° + TB sin 65° − W = 0

F 1 × +R 2 3 2

W=

3F +R 3

W=

3R + 3F . 3

10 T cos 15° = F + W sin 27°

5 a Mary has not included X.

1 (X sin 12° + 250 cos 12°) + 250 sin 12° 4

X(4 cos 12° – sin 12°) = 250 cos 12° + 1000 sin 12°

R

T1

1 1 R = (X sin 12° + 250 cos 12°) 4 4

X cos 12° =

P

mg

26° g

X = R sin θ + F cos θ 8 X cos 12° = F + 250 sin 12°

D

R + T sin 15° = W cos 27°

Correct answer: R = W cos 24 + X sin 24

c Mary has written W cos 24° rather than W sin 24°.

1 R 2 1 T cos 15° – W sin 27° = (W cos 27° – T sin 15°) 2 2T cos 15° – 2W sin 27° = W cos 27° – T sin 15°

2T cos 15° + T sin 15° = W cos 27° + 2W sin 27°

F=

b Mary has written X cos 24° rather than X.

Correct answer: F + X = W sin 24° Correct answer: X cos 24° = F + W sin 24°

d Mary has written X cos 24° rather than X sin 24°.

T(2 cos 15° + sin 15°) = W(cos 27° + 2 sin 27°)

cos 27° + 2sin 27° T = W 2 cos15° + sin 15°

Correct answer: R = X sin 24° + W cos 24°

e Mary has written X cos 64° rather than X cos 40°.

Correct answer: X cos 40° = F + W sin 24° 1

R = W cos θ

2

1 a R( )

6 F = W sin θ

Exercise 1.2B

F sin θ 1 ÷ 2 : R = cosθ = tan θ

b R = X sin θ + W cos θ

R − 50 cos 28° = 0

R = 44.1 N

b R( )

F = R tan θ. 7 a X cos θ = F + W sin θ

F − 50 sin 28° = 0

F = 23.5 N 2 R(↑)

2T sin θ − W = 0

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1

WORKED SOLUTIONS

W = 2 × 32 × 12 35 = 21.9 N

3 Both solutions are correct. Victrix’s solution requires resolving twice whereas Colin’s solution requires resolving once. Victrix’s solution is possibly simpler since it just requires horizontal and vertical resolving. By resolving perpendicular to the tension, Colin’s solution does not require the tension to be found.

Hence friction is acting up the slope with a magnitude of 6.34 N.

sin θ =

7 Let the tension in the string at 55° be T1 and the tension in the string at 35° be T2. R(→)

T2 cos 35° − T1 cos 55° = 0

R(↑)

T1 sin 55° + T2 sin 35° − 15 = 0

b R(→)

Make T1 the subject of equation 1 .

A cos 30° − Q = 0

Q = A cos 30°

T1 =

R(↑)

P − A sin 30° = 0

Substitute for T1 in equation 2 . T2 cos 35° sin 55° + T2 sin 35° − 15 = 0 cos 55°

P = A sin 30°

Since Q is 4 N larger than P

A cos 30° = 4 + A sin 30°

A cos 30° − A sin 30° = 4

A(cos 30° − sin 30°)  = 4 4 = 10.9 N cos30° − sin 30°

A=

5 Let the angle between the cord and the floor be  θ.

R(→) T cos θ − 6 = 0

6 cosθ sin θ = 1 5

T=

52 − 12 = 5

cos θ =

T=

R(↑)

R + T sin θ − 30 = 0

R = 30 − T sin θ

= 30 − 2.5 × 0.2

= 29.5 N

6 24 5

T2

24 5

= 2.5 N

6 Let the angle between the plane and the horizontal be θ and assume the friction is acting down the slope R( )

20 − 120 sin θ − F = 0 cos θ = 40 41

1 2

T2 cos 35° cos 55°

35° sin 55° + sin 35° ( cos ) = 15 cos 55°

15 = 8.60 N T2 = cos35° sin 55° + sin 35° cos55° T1 = T2 cos35° = 8.60 × cos35° = 12.3 N cos55° cos55° The tensions are 8.60 N and 12.3 N.

8 a R (→) U cos 2a = T cos a T =

4 a  All three forces are in the horizontal plane, whereas weight acts vertically and perpendicular to all the three forces.

412 − 402 = 9 41 41 9 F = 20 − 120 × = −6.34 41

U cos2α cosα

R (↑) 20 = U sin 2a + T sin a

20 = U sin 2 a +

20 =

U sin 2α cosα + U cos2α sin α cosα

20 =

U(sin 2α cosα + cos2α sin α ) cos α

U cos 2α sin a cosα

U =

20cosα cos 2α sin α + sin 2α cos α

b U =

20 cos α 20cos 30 = sin(2α + α ) sin(90)

U =

20 3 = 10 3 2

T =

1 10 3 cos 60° 10 3 × 2 = = 10 cos 30° 3 2

3

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1 Forces and equilibrium

9 Q = P cos(180° – b ) = –P cos b

5 a  Hita has drawn out the diagram with the forces tip to tail.

R = P cos(b – 90°) = P sin b

Q + R = P sin b  – P cos b Q + R = P(sin b  – cos b )

12 35 , sin a = 37 37

10 Given that cos a = T1 cos a = T2 cos b

Using alternate angles she has calculated the angle opposite the overall resultant force as (90 + 38.2) = 128.2°.

She has then applied the cosine rule again to the 4 N and 7 N forces to obtain the overall resultant force of 9.98 N.

She has then applied the sine rule again to obtain the angle opposite the 4 N force as 18.4°.

T1 = 2T2 T cos b = 1 cosα = 2 cos a T2     12 24 = 37 37

cos b = 2 ×

( )

24 sin2 b = 1 – 37

2

793 = 1369

She has then added 38.2° and 18.4° to obtain the angle 56.6°. The direction of the resultant force is 56.6° to the horizontal. b R(→) (3 + 5 cos 60°) N

793 sin b = 37   W = T1 sin a + T2 sin b W = 2T2 × W=

35 793 + T2 × 37 37

(

T2 70 + 793 37

)

( )

b tan−1  8 = 28.1° 15 2 a Magnitude = 38 2 + 232 = 44.4 N

( )

b tan−1  23 = 31.2° 38 Bearing = 270° − 31.2° = 239°

3 a R(→) (9 + 7 cos 50°) N

R(↑) (7 sin 50°) N

Magnitude =

(9 + 7cos50°)2 + (7sin 50°)2

= 14.5 N

(

)

7sin 50° = 21.7° 9 + 7cos50°

4 R(→) (11 − 7 cos 70°) N

R(↑) (7 sin 70°) N

Magnitude =

(11 − 7cos70°)2 + (7sin 70°)2

= 10.8 N 7sin 70° −1 = 37.4° to Ox Angle = tan   11 − 7cos70°

(

R(↑) (4 + 5 sin 60°) N

Magnitude =

1 a Magnitude = 152 + 8 2 = 17 N

b Angle = tan−1

Exercise 1.3A

She has applied the cosine rule to the 3 N and 5 N forces to obtain the resultant of these two as 7 N. She has then applied the sine rule to obtain the angle opposite the 5 N force as 38.2°.

)

(3 + 5cos60°)2 + ( 4 + 5sin 60°)2

= 9.98 N

(

)

Angle = tan−1 4 + 5sin 60° = 56.6° 3 + 5cos60°

6 Can answer parts a and b in either order.

Let the angle at the top right of the triangle be X and the direction of the resultant force be Y.

sin X sin128 = 18 28

sin X = 0.507

X = sin−1 (0.507) = 30.4°

Y = 180 − (128 + 30.4) = 21.6°

The direction of the resultant force is 21.6° to Ox.

D2 = 282 + 182 − 2 × 28 × 18 × cos 21.6°

D = 170.55 = 13.1 N

The magnitude of the resultant force is 13.1 N.

7 a R(→) (3 + 3 cos 40° − 5 cos 40° − 5 sin 40°) = (3 − 2 cos 40° − 5 sin 40°) N

R(↑) (3 sin 40° + 5 sin 50° − 5 sin 40°) = (5 sin 50° − 2 sin 40°) N Angle = tan−1  5sin 50° − 2sin 40° = – 55.5. 3 − 2cos 40° − 5sin 40°

(

)

Hence the resultant force acts at approximately 124° to Ox measured anticlockwise.

b Magnitude =

( 3 − 2cos 40° − 5sin 40°)2 + ( 5sin 50° − 2sin 40°)2 = 3.09 N.

4

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WORKED SOLUTIONS

8 a Let resultant force = F N 2 = x2 + (2x)2 – x × 2x × cos a F  2 = 5x2 – 2x2 cos a F

From F = µR, µ = F . R

µ = 80sin 28° = tan 28° = 0.532 80cos 28°

2 = x2 (5 – 2 cos a) F

2 R( )

b 2x = x 5 − 2 cos α

R − 6000 cos 8° = 0

R = 6000 cos 8°

4 = 5 – 2 cos a

2 cos a = 1

1 cos a = 2

a = 60°

9 a R(→) S cos θ = 2S sin θ tan  θ=

1 2

b When tan θ =

1 1 2 , sin θ = and cos θ = 2 5 5

F = µR

= 0.12 × 6000 cos 8° = 713 N

R( )

D − F − 6000 sin 8° = 0

D = F + 6000 sin 8°

= 713 + 6000 sin 8°

= 1548 N

The driving force is 1550 N to 3 s.f.

3 a R(→)

R(↑) T = S sin θ + 2S cos θ

1 2 T = S × + 2S × 5 5 S 4S 5S + = =S 5 5 5 5 10 a a2 = 5002 + 1002 – 2 × 500 × 100 cos 70° T =

a = 475.182

sin 70° sin θ = 475.182 100

sin θ = 0.197 75

θ = 11

Bearing = 50° + 11° = 061° sin(50° − α ) = sin110° 100 500

sin(50° – a) = 0.187 94

50° – a = 11°

a = 039°

T cos 20° − F = 0

F = T cos 20°

R(↑)

R + T sin 20° − 50 = 0

R = 50 − T sin 20°

F = µR

T cos 20° = 1 (50 − T sin 20°) 4

4T cos 20° = 50 − T sin 20°

4T cos 20° + T sin 20° = 50

T(4 cos 20° + sin 20°) = 50 50 T= 4cos 20° + sin 20°

= 12.2 N

b There would be no friction. 4 Let the total weight of the cyclist and bicycle be W  N: R( )

W sin α − F = 0

F = W sin α

R( )

1 R( )

R − W cos α = 0

F − 80 sin 28° = 0

R = W cos α

F = 80 sin 28°

R( )

R − 80 cos 28° = 0

R = 80 cos 28°

From F = µR, µ = F . R W sin α = tan α 0.3 = W c osα

Exercise 1.4A

b Let bearing be a

2 = 5 − 2 cos α

F = x 5 − 2 cos α

1

α = tan−1 0.3 = 16.7°

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1 Forces and equilibrium

F = 32 sin 25° − P cos 25°

R( )

R − P sin 25° − 32 cos 25° = 0

R = 32 cos 25° + P sin 25°

F = µR

32 sin 25° − P cos 25° = 0.15(32 cos 25° + P sin 25°)

32 sin 25° − 4.8 cos 25° = P cos 25° + 0.15P sin 25°

8 a R( )

1 W cos θ = F + W sin θ 2

R( ) R +

F + P cos 25° − 32 sin 25° = 0

5 R( )

F =

1 W sin θ = W cos θ 2

1 W cos θ – W sin θ 2

R = W cos θ –

1 W sin θ 2

m=

1 W cos θ − W sin θ F = 2 R W cos θ − 1 W sin θ 2

= P(cos 25° + 0.15 sin 25°) P = 32sin 25° − 4.8cos 25° cos 25° + 0.15sin 25°

m=

cosθ − 2sin θ 2 cos θ − sin θ

m=

1 − 2tan θ 2 − tan θ

b m =

1 − 2tan θ 2 − tan θ

= 4.8 cos 25° + 0.15P sin 25°

= 9.46 N

6 Let the angle of slope be θ : R( )

560 − F − W sin θ = 0

2m – m tan θ = 1 – 2 tan θ

F = 560 − W sin θ

2m – 1 = m tan θ – 2 tan θ

R( )

2m – 1 = tan θ (m – 2)

R − W cos θ = 0

R = W cos θ

2µ − 1 tan θ = µ−2

F = µR

560 − W sin θ = 2 × W cos θ 7 3920 – 7W sin θ = 2W cos θ

2µ − 1 >0 Since tan θ > 0, µ−2   2m – 1 > 0

3920 = 7W sin θ + 2W cos θ

m>

3920 = W(7 sin θ + 2 cos θ) 3920 W= 7sin θ + 2cosθ sin θ = 1 10

c

3 1 − 2tan θ = 10 2 − tan θ

6 – 3 tan θ = 10 – 20 tan θ

2

2

1 2

4 tan θ =   17

cos θ = 10 − 1 = 99 10 10

= 1457 N

= 1460 N correct to 3 s.f.

P = 30 sin 30° – 0.1(30 cos 30°)

7 a P = 8500 sin 50° + F R = 8500 cos 50° m = F = 15 000 − 8500 sin 50° = 1.55 R 8500 cos50°

b m =

P − 8500 sin 50° F 1 = R 8500 cos 50 °

P – 8500 sin 50°  8500 cos 50°

R( ) R = 30 cos 30°

F = mR 30 sin 30° – P = 0.1(30 cos 30°)

P = 30 ×

b R( ) P = 30 sin 30° + F

R( ) R = 30 cos 30°

F = mR P – 30 sin 30° = 0.1(30 cos 30°) P = 30 sin 30° + 0.1(30 cos 30°)

P  8500(cos 50° + sin 50°) P  12 000 N

1 3 –3× = 12.4 N 2 2

99 10

3920 1 7 × 10 +2×

W=

9 a R( ) P + F = 30 sin 30°

P = 30 ×

1 3 +3× = 17.6 N 2 2

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WORKED SOLUTIONS

max P – min P = 3  0×

F = 210 sin 30° − T cos 15°

R( )

R + T sin 15° − 210 cos 30° = 0

R = 210 cos 30° − T sin 15°

F = µR

210 sin 30° − T cos 15° = 2 (210 cos 30° − T sin 15°) 9    1890 sin 30° − 9T cos 15° = 420 cos 30° − 2T sin 15° 1890 sin 30° − 420 cos 30° = 9T cos 15° − 2T sin 15°

= T(9 cos 15° − 2 sin 15°)

T = 1890sin 30° − 420cos30° 9cos15° − 2sin15°

= 71.1 N

40 − F − 170 sin α = 0

F = 40 − 170 sin α

R( )

R − 170 cos α = 0

R = 170 cos α

F = µR

T cos 15° + F − 210 sin 30° = 0

10 a R( )

40 − 170 sin α = µ(170 cos α) From F = µR, µ = F R 40 − 170 sin α µ= = 4 − 17sin α 17cos α 170 cos α b tan α = 13 84

13 sin α = = 13 85 132 + 84 2 cos α = 84 85 4 − 17 × 13 340 − 221 17 × 20 − 17 × 13 85 = µ= = 17 × 84 17 × 84 17 × 84 85 = 17 × 7 = 1 17 × 84 12

c Let the weight of the shopping be W N:

100 − F − (170 + W) × 13 = 0 85 F = 100 − 13 (170 + W) 85 R( ) R − (170 + W) × 84 = 0 85 R = 84 (170 + W) 85 F = µR

R( )

100 − 13 (170 + W) = 1 × 84 (170 + W) 85 12 85

102 000 – 26 520 – 156W = 14 280 + 84W

61200 = 240W

W = 255

There is 255 N of shopping in the trolley.

1 W 2

1 W cos θ = F 2

R +

1 W sin θ = W 2

 m=

1 W cos θ F cos θ = = 2 R W − 1 W sin θ 2 − sin θ 2

5 2 b If cos θ = , sin θ = 3 3      m=

5 3 = 5 = 5 4 2− 2 6−2 3

13 a R( ) X cos a = F + W sin a

R( ) R = X sin a + W cos a

m=

cos a =

3 sin a cos a + cos2 a = 3 cos a – sin a

3 sin a cos a + sin a = 3 cos a – cos2 a

sin a(3 cos a + 1) = cos a(3 – cos a)

sin2 a(3 cos a + 1)2 = cos2 a(3 – cos a)2

(1 – cos2 a)(3 cos a + 1)2 = cos2 a(3 – cos a)2

(1 – cos2 a)(9 cos2 a + 6 cos a + 1) = cos2 a(9 – 6 cos a + cos2 a)

9 cos2 a + 6 cos a + 1 – 9 cos4 a – 6 cos3 a – cos2 a = 9 cos2 a – 6 cos3 a + cos4 a

12 a T =

1 3 1 +3× – 30 × 2 2 2 3 +3× 2 3 3 max P – min P = 3 × +3× =3 3 2 2

11 R( )

  c max P – min P = 30 × 1 + 3 × 3  2 2    3 1 – 30 × 2 − 3 × 2 

1

F X cosα − W sin α = R X sin α + W cosα 3W cosα − W sin α 3W sin α + W cosα

10 cos4 a + cos2 a = 6 cos a + 1

7

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1 Forces and equilibrium

b f(a) = 10 cos4a + cos2 a – 6 cos a – 1 = 0

f(30°) = 0.1788

f(31°) = –0.0099

Since f(30°) > 0 and f(31°) < 0 and f(a ) is continuous, 30° < a < 31°

Exam-style questions

T1 =

The tensions are 8.62 N and 8.59 N.

F − W sin θ = 0

F = W sin θ

R( )

R − W cos θ = 0

R = W cos θ

From F = µR, µ =

µ = W sin θ = tan θ W cosθ

sin θ = 5 13

cos θ =

cos 7° cos 5° sin 5°

+ sin 7°

= 8.62 N

T2 cos7° = 8.62 × cos7° = 8.59 N cos5° cos5°

R – 70 – P sin 8° = 0 R = 70 + P sin 8° b R(→)

R( )

T2 =

4 a R(↑)

1 Let the angle of slope be θ and the weight of the wardrobe be W N.

1.8

P cos 8° – F = 0 F = P cos 8°

c Substitute into F = µ R: P cos 8° = 1 × (70 + P sin 8°) 12 12P cos 8° = 70 + P sin 8°

F . R

12P cos 8° – P sin 8° = 70

P(12 cos 8° – sin 8°) = 70

= 5.96 N

H2 = 338.3 H = 18.4 N

2 a R(↑)

b Sine rule:

P cos 35° – 25 = 0

P = 30.5 N

b R(→) Q – P cos 55° = 0

Q = 30.5 cos 55° = 17.5 N

R(→)

T2 cos 7° − T1 cos 5 = 0

R(↑)

T1 sin 5° + T2 sin 7° – 1.8 = 0

Make T1 the subject of equation 1 :

T1 =

sin X = 0.666

X = sin–1(0.666) = 41.8°

Angle with positive x-axis = –138.2°.

6 R( )

3 Let the tension in the string at 5° be T1 and the tension in the string at 7° be T2.

sin X sin 50° = 18.4 16

70 12cos8° − sin 8°

5 a Cosine rule: H2 = 162 + 242 – 2 × 16 × 24 × cos 50

5 = 5 13 − 52 12 tan θ = 5 ÷ 5 = 12 13 12 13 2

P=

1

G cos β − 6 sin β = 0

G cos β = 6 sin β

G = 6 tan β = 6 × 3 = 4.5 N 4

7 a R(→)

(6 + 10 cos 50°) N

R(↑)

(9 + 10 sin 50°) N

Magnitude =

Substitute for T1 in equation 2 :

T2 cos7° sin 5° + T2 sin 7° − 1.8 = 0 cos5°

b Angle = tan−1  9 + 10sin 50° = 53.3°. 6 + 10cos50°

T2 cos7° cos5°

T2

° sin 5° + sin 7° ) = 1.8 ( cos7 cos5°

2

(6 + 10 cos50°)2 + (9 + 10sin 50°)2

= 20.8 N.

(

)

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WORKED SOLUTIONS

8 R( )

1

X Sine rule: 10.4 = sin 40° sin14.5°

F = 40 cos 35° − W sin 35° R( )

X = 10.4 × sin 14.5° = 4.07 N sin 40°

R − 40 sin 35° − W cos 35° = 0

c R2 = (5 + 7 cos60°)2 + (4.07 + 7sin 60°)2 = 13.2 N

R = 40 sin 35° + W cos 35°

F = µR

40 cos 35° − F − W sin 35° = 0

12 a R( )

F – F cos 55° – 36 sin 85° = 0

F(1 – cos 55°) = 36 sin 85° F = 36sin 85° = 84.1 N 1 − cos55° →

b R( )

X + 36 cos 85° – F cos 35° = 0

X = 84.1 cos 35° – 36 cos 85° = 65.8 N →

13 a R( ) F + 8 cos 20° = W sin 20°

40 cos 35° − W sin 35° = 1 (40 sin 35° + W cos 35°) 3 40 cos 35° − W sin 35° = 40  sin 35° + 1 W cos 35° 3 3 1 40 cos 35° − 40  sin 35° = W sin 35° + W cos 35° 3 3 40 cos 35° − 40  sin 35° = W(sin 35° + 1  cos 35°) 3 3 40 cos 35° − 40 3 sin 35° W= = 29.7 N sin 35° + 13 cos 35° 9 a R( )

F + 20 cos θ – 50 sin θ = 0 F = 50 sin θ – 20 cos θ R( )

F = mR

R – 20 sin θ − 50 cos θ = 0

R = 20 sin θ + 50 cos θ

F = µR

50 sin θ – 20 cos θ = 0.1(20 sin θ + 50 cos θ)

W sin 20° – 8 cos 20° = 0.18W cos 20° + 1.44 sin 20° W sin 20° – 0.18W cos 20° = 1.44 sin 20° + 8 cos 20° W(sin 20° – 0.18 cos 20°) = 1.44 sin 20° + 8 cos 20°

= 2 sin θ + 5 cos θ

b R( ) F + P cos 30° = W sin 30° F = mR W sin 30° – P cos 30° = 0.18(W cos 30° + P sin 30°) W sin 30° – P cos 30° = 0.18W cos 30° + 0.18P sin 30°

F = 50 sin 27.5° – 20 cos 27.5° = 5.36 N

10 2T sin θ = 24

P =

5

11 a R(↑) T = 20N

5 + 7 cos 60°

R=

14 a R( ) →

R(→)

46.3(sin 30° − 0.18 cos 30°) = 16.7 N 0.18 sin 30° + cos 30°

(7sin 60°) + (5 + 7 cos60°) = 109 = 10.4 N 2

2

b Sine rule: sin A = sin120° 10.4 7

sin A = 0.581

A = sin–1 0.581 = 35.5°

50° – 35.5° = 14.5°

Obtuse angle = 180° – (30° + (180° – (120° + 35.5°))

Third angle = 180° – (14.5° + 125.5) = 40°.

= 125.5°.

T cos θ + F − W sin θ = 0

F = W sin θ − T cos θ

R( )

R + T sin θ − W cos θ = 0

R = W cos θ − T sin θ

F = µR

0.18P sin 30° + P cos 30° = W sin 30° – 0.18W cos 30°

P(0.18 sin 30° + cos 30°) = W sin 30° – 0.18W cos 30°

12 = 12 = 20 N sin θ 3

7 sin 60°

R( ) R = W cos 30° + P sin 30° →

( )

1.44 sin 20° + 8 cos 20° = 46.3 N sin 20° − 0.18 cos 20°

W =

T=

W sin 20° – 8 cos 20° = 0.18(W cos 20° + 8 sin 20°)

48 sin θ = 25 cos θ tan  θ = 25 48 b θ = tan–1 25 = 27.5° 48

R( ) R = W cos 20 + 8 sin 20°

W sin θ − T cos θ = µ(W cos θ − T sin θ)

= µW cos θ − µT sin θ

µT sin θ − T cos θ = µW cos θ − W sin θ

T(µ sin θ − cos θ) = µW cos θ − W sin θ

9

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1 Forces and equilibrium

T = µW cosθ − W sin θ µ sin θ − cosθ  =

W ( µ cosθ − sin θ ) µ sin θ − cosθ

b 15 = 80 ( µ cos 20° − sin 20° ) µ sin 20 − cos 20

R( )

X – F – W sin θ = 0 42 – F – 3 W = 0 5 On the point of slipping down

R( )

X + F – W sin θ = 0 30 + F – 3 W = 0  5

15µ sin 20° − 15 cos 20° = 80µ cos 20° − 80 sin 20°

15µ sin 20° − 80µ cos 20° = 15 cos 20° − 80 sin 20°

µ(15 sin 20° − 80 cos 20°) = 15 cos 20° − 80 sin 20°

µ = 15cos 20° − 80sin 20° 15sin 20° − 80 cos 20°

= 0.189

50 sin α – X cos α – F = 0

R( )

R – X sin α – 50 cos α = 0

On the point of sliding, F = µR

50 sin α – X cos α = µ(X sin α + 50 cos α)

= µX sin α + 50µ cos α

50 sin α – 50µ cos α = µX sin α + X cos α

50 tan α – 50µ = µX tan α + X X = 50(tan α − µ) 1 + µ tan α

Add 1 and 2 : 72 – 6 W = 0 5 W = 5 × 72 = 60 N 6 b F = 3 W – 30 = 36 – 30 = 6 N 5

R( )

R – 60 cos θ = 0 R = 4 × 60 = 48 N 5 F From F = µR, µ = R 6 1 = µ= 48 8

18 R( ) F + P cos θ = W sin θ R( ) R = W cos θ + P sin θ

X(1 + µ tan α) = 50(tan α – µ)

2

R( )

F = mR

P cos 25° – F − 12 sin 15° = 0

2 (W cos θ + P sin θ) 7 2 5P sin θ – P cos θ = (5P cos θ + P sin θ) 7

F = P cos 25° − 12 sin 15°

35 sin θ – 7 cos θ = 10 cos θ + 2 sin θ

R( )

33 sin θ = 17 cos θ

R – P sin 25° − 12 cos 15° = 0

R = P sin 25° + 12 cos 15°

F = µR

P cos 25° − 12 sin 15° = 0.32(P sin 25° + 12 cos 15°)

P cos 25° − 0.32P sin 25° = 12 sin 15° + 3.84 cos 15°

P(cos 25° − 0.32 sin 25°) = 12 sin 15° + 3.84 cos 15° P = 12sin15° + 3.84 cos15° cos 25° − 0.32sin 25°

16 R( )

W sin θ – P cos θ =

17 33 19 a R( ) F + P cos 60° = W sin 30° tan θ =

= 0.32P sin 25° + 3.84 cos 15°

= 8.84 N

R( ) R = W cos 30° + P sin 60°

m=

m=

1W − 1 P 2 2 = 3W + 3 P 2 2

m=

3 W −P 3 W +P

1

15 Let the friction force be F N and the reaction force be R N

17 a Let the angle of inclination be θ. Hence sin θ = 3 5 and cos θ = 4 5 On the point of slipping up

W sin 30° − P cos 60° F = R W cos 30° + P sin 60°

(

W −P 3W + 3 P

)

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WORKED SOLUTIONS

b m > 0

(

)

3 W −P >0 3 W +P

W > P

12 5 , cos a = 13 13

T cos a = F

20 a 63 in the ratio 4 : 3 = 36 : 27

0.28(125) − 5 1 = 0.96(125) 4

b m =

22 a When sin a =

W – P > 0

Therefore the triangle has sides of 27 cm, 36 cm and 45 cm.

This is a 3 : 4 : 5 triangle and is right-angled.

Let the angle at A be θ.

Hence sin θ =

R(↑) TA cos θ = TB sin θ + 2.2g

TB =

3 4 and cos θ = . 5 5

2 T 3 A

3 TA × 4 = 2 TA × + 22 5 3 5

2 T = 22 5 A

TA = 55 N

R + T sin a = W F = mR T cos a = m(W – T sin a)

(

5 5 12 T= W – T 23 13 13

5 5 60 W– T T= 23 299 13

175 5 T= W 299 23

4025T = 1495W

)

35T = 13W b When sin a =

5 12 , cos a = 13 13

13 W 35

T =

( )

( )

12 13 5 13 W = m  W − W  13 35 13 35  

F = 55 × 3 + 2 × 55 × 4 5 3 5 F = 62.3 N

12 6 = m 35 7

21 a When sin θ = 0.28, cos θ = 0.96

m=

R( ) F + P = W sin θ

R( ) R = W cos θ

m=

F 0.28W − 5 m= = R 0.96W

Slipping up

R( ) P = F + W sin θ

R( ) R = W cos θ

F P − W sin θ m= = W cos θ R

m=

0.28W − 5 65 − 0.28W = 0.96W 0.96W

0.28 W – 5 = 65 – 0.28 W

0.56 W = 70

Slipping down

F W sin θ − P = W cos θ R

F 65 − 0.28W = R 0.96W

W = 125 N

23 a When sin a =

3 4 , cos a = 5 5

R( ) P = F + W sin a

R( ) R = W cos a

m=

3 F P − W sin α P − 5 W = = W cos α 4W R 5

m=

5P − 3W 4W

When sin b =

8 F P − W sin β P − 17 W m= = = 15 W R W cos β 17

m=

5P − 3W = 17P − 8W 15W 4W

15(5P – 3W) = 4(17P – 8W)

2 5

b R(→) F = TA sin θ + TB cos θ

1

8 15 , cos b = 17 17

17P − 8W 15W

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1 Forces and equilibrium

75P – 45W = 68P – 32W

7P = 13W

Mathematics in life and work

( )

0.3 = 1.494° = 1.5° to 1 d.p. 1 tan–1 11 .5

13W 7

b m =

5P − 3W = 4W

( )

5 13W − 3W 7 4W

2 Let the pushing force be P and parallel to the ramp

R( )

P – F – 1600 sin 1.494° = 0

F = P – 1600 sin 1.494°

P =

m=

65W − 21W 28W

m=

11 7

R( )

R − 1600 cos 1.494° = 0

24 a R(→) T cos θ = F

R = 1600 cos 1.494°

Substitute into F = µR:

F = mR

P – 1600 sin 1.494° = 0.14 × 1600 cos 1.494°

T cos θ = m(W – T sin θ)

T cos θ + mT sin θ = mW

T(cos θ + m sin θ) = mW

3 200 = 1600 sin 1.494° + µ × 1600 cos 1.494° µ = 200 − 1600sin1.494° = 0.0990 1600 cos1.494°

R(↑) R + T sin θ = W

T cos θ = mW – mT sin θ

T =

µW cos θ + µ sin θ

b k1W =

k1 =

P = 1600 sin 1.494° + 0.14 × 1600 cos 1.494° = 266 N

µW 4 + 3µ 5 5

µ 4 + 3µ 5 5

k2W =

µW 3 + 4µ 5 5

µ 3 + 4µ 5 5 3 4 k1 5 + 5 µ 3 + 4µ 5 = = = k2 4 + 3 µ 4 + 3µ 6 5 5

k2 =

6(3 + 4m) = 5(4 + 3m)

18 + 24m  = 20 + 15m

9m = 2 2 m= 9

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2

WORKED SOLUTIONS

2 Kinematics of motion in a straight line 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 noresponsibility 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 v2 = u2 + 2as =

402

+ 2 × 18 × 49

= 3364 v = 58 2 2 b s = v − u 2a 2 2 = 148 − 48 2 × 9.8 = 1000 2 12 × (19 + 34) × 22 = 583 cm2

3 a 4(T + 2) = 3T + 14

4T + 8 = 3T + 14

T=6

b 2T2 − 5T − 12 = 0

(2T + 3)(T − 4) = 0

T = 4 or − 1.5 4 a f′(x) = 12x − 7 b ∫ f(x) dx = 2x3 − 7 x2 + 8x + c 2

Exercise 2.1A 1 Overall distance travelled = 30 + 20 = 50 m

Displacement from original position = −30 + 20

= −10 m

2 a 73 + 73 = 146 m b 73 – 73 = 0 3 Overall distance travelled by the ball = (40 − 15) + 40

= 25 + 40 = 65 m

Displacement of the ball from original position = −15 m

4 a 2(7 + 4) = 22 m b Since the snail returns to its initial position, the overall displacement = 0 m.

5 a tan–1( 16 ) = 45° 16

Bearing = 90 + 45 = 135°

b Displacement = 16 2 + 16 2 = 22.6 km c 32 – 22.6 = 9.4 km 6 a 32 + 24 = 56 m b By walking west then south, the boy is tracing out the two perpendicular sides of a right-angled triangle.

The hypotenuse = 322 + 24 2 = 40 m

c Speed = distance ÷ time = 56 ÷ 28 = 2 m s−1 7 a The triangle is equilateral, so 20 km 20 b 2 = 30 km h–1 3

8 a 38 – (–2) = 40 5 – (–4) = 9 Total distance = 40 + 9 = 49 b Displacement = 40 2 + 92 = 41

tan–1( 9 ) = 77° Bearing = 360 – 77 = 283° 40

c 49 – 41 = 8 8 seconds 9 a Journey taken by bus = 15 + 36 = 51 km

Direct route = 152 + 36 2 = 39 km

Difference = 51 – 39 = 12 km

b Displacement = 39 km

tan–1( 15 ) = 67°

Bearing = 60 + 67 = 127°

36

Exercise 2.2A 1 a s = ut + 12 at2

= 70 × 10 +

= 550 m

1 2

× −3 × 102

b s =  u + v  t  2 

(

15+ 29 2

)×9

=

= 198 m

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2 Kinematics of motion in a straight line

c v = u + at

b s = ut + 1 at2 2 = 0 × 40 + 1 × 0.5 × 402 2 = 400 m

a = v −u t = 38 − 3 7 = 5 m s−2

d v2 = u2 + 2as

5 a u = 50 m s−1, s = 1500 m, v = 0 m s−1 v2 = u2 + 2as

2 2 s = v −u 2a

2 2 = 28 − 22 2×6

= 25 m

2 2 = 0 − 50 2 × 1500 = − 5 m s−2 6 v = u + at t = v −u a

e s = vt − 12 at2

0 = −28t −

1 2

× −7t2

7 2 t = 28t 2 t2 = 8t

t2 − 8t = 0 t (t − 8) = 0 t = 0 or 8 s 2 a u = 24 m s−1, a = 5 m s−2, s = 10 m v2 = u2 + 2as

= 242 + 2 × 5 × 10 = 676

v = 26 m s−1 b v = u + at t = v −u a

= 26 − 24 5

= 0.4 s v = u + at

= 60 − 8 × 10

= −20 m s−1

= 0 − 60 −8

= 7.5 s

4 a u = 0 m s−1, t = 40 s, a = 0.5 m s−2 v = u + at = 0 + 0.5 × 40 = 20 m s−1

= 0 − 550 −6

= 60 s

6 a u = 0 m s−1, a = 4 m s−2, s = 128 m s = ut + 1 at2 2 128 = 0 × t + 1 × 4 × t2 2 = 2t2 t2 = 64 t = 8 s b u = 180 m s−1, a = −3 m s−2, s = 0 m s = ut + 1 at2 2 0 = 180 × t + 1 × −3 × t2 2 3 2 2 t − 180t = 0   t2 − 120t = 0   t(t − 120) = 0 t = 0 or 120 s It takes 2 minutes. c u = 5 m s−1, a = 3 m s−2, s = 84 m 84 = 5T + 1 × 3T 2 2 3 2 T  + 5T − 84 = 0 2 3T2 + 10T − 168 = 0

b v = 0 m s−1 v = u + at t = v −u a

b No air resistance, etc.

3 a u = 60 m s−1, a = −8 m s−2, t = 10 s

2 2 a = v −u 2s

(3T + 28)(T − 6) = 0

T = 6 s

7 a = 0.25 m s−2, s = 930 m, t = 60 s 1 930 = u × 60 + × 0.25 × 602 2 = 60u + 450

60u = 480

u = 8 m s−1

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2

WORKED SOLUTIONS

v = u + at

c PS: u = 13 m s−1, a = 2.8 m s−2,

= 8 + 0.25 × 60

s = 830 + 820 = 1650 m

= 23 m s−1

v2 = u2 + 2as

= 132 + 2 × 2.8 × 1650

= 9409

v = 97 m s−1

= u + at

t = v −u a

= 97 − 13 2.8

= 30 s

Time from Q to S = 30 − 5 = 25 s.

8 60 km h−1 = 60 000 metres per hour = 50  m s−1 3 50 −1 −1 u = 0 m s , v = m s , t = 8 s 3

s =  u + v  t  2 

 0 + 50  3 =  × 8 2 

= 200  m 3 a 200 − 200 = 400  m 3 3 s = ut

10 a u = 20 m s−1, a = −4 m s−2, v = 0 m s−1 v2 = u2 + 2as

400 = 50 t 3 3 t=8 Total time = 8 + 8 = 16 s.

b 1000 − 200 = 2800  m 3 3 s = ut 2800 = 50 t 3 3 t = 56 Total time = 8 + 56 = 64 s. 9 a PQ: t = 5 s, s = 100 m s = ut + 1 at2 2 100 = 5u + 25 a 2 40 = 2u + 5a

PR: t = 20 s, s = 820 m s = ut + 1 at2 2 820 = 20u + 200a

82 = 2u + 20a

Subtract 1 from 2 :

2 2 s = v −u 2a

2 2 = 0 − 20 2 × −4

Since the debris is 80 m away, he will stop 30 m from the debris.

= 50 m

b u = 20 m s−1, a = 2 m s−2, t = 2 s  v = u + at

= 20 + 2 × 2

= 24 m s−1 s = ut + 1 at2 2 = 20 × 2 + 1 × 2 × 22 2 = 44 m

Distance from debris = 80 − 44 = 36 m.

1

u = 24 m s−1, a = −4 m s−2, s = 36 m

2

v2 = u2 + 2as

42 = 15a

= 242 + 2 × −4 × 36

a = 2.8 m s−2

= 288

v = 288 = 17.0 m s−1

b PQ: t = 5 s, s = 100 m, a = 2.8 m s−2

Could find v from t, s and a, but simpler to find and use u.

11 Let the displacement of A from its starting point be sA m when t = T s.

Substitute in 1 :

40 = 2u + 5 × 2.8

26 = 2u

u = 13 m s−1

v = u + at

= 13 + 2.8 × 5

= 27 m s−1

= 2u + 14

Let the displacement of B from its starting point be sB m when t = T s.

A: s = sA m, t = T s, u = 0 m s−1, a = 6 m s−2 s = ut + 1 at2 2

sA = 0 × T + 1 × 6 × T2 2 2 = 3T

15

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2 Kinematics of motion in a straight line

B: s = sB m, t = T s, u = 0 m s−1, a = −2 m s−2 ut + 1 at2 2

s=

K: u = 0 m s−1, a = 0.8 m s−2, v = 4 m s−1

v = u + at

4 = 0 + 0.8t

t = 5 s

L: u = 3 m s−1, a = 0.5 m s−2, v = 7 m s−1

sB = 0 × T + 1 × −2 × T2 2 2 = −T

Since B started 100 m from A: sA − sB = 100

v = u + at

3T2 − (−T2) = 100

7 = 3 + 0.5t

4T2 = 100

t = 8 s

T2 = 25

T=5

When L has a velocity of 7 m s−1, J has been moving for 19 s (6 + 5 + 8) and K for 13 s (5 + 8)

a When T = 5, sA = 3T2 = 3 × 52 = 75 m.

b T = 5

J: u = 2 m s−1, a = 1 m s−2, t = 19 s s = ut + 1 at2 2

c For A: t = 5 s, u =

0 m s−1,

a=

6 m s−2

= 2 × 19 + 1 × 1 × 192 2 = 218.5 m

v = u + at

=0+6×5

= 30 m s−1

For B: t = 5 s, u = 0 m s−1, a = −2 m s−2

v = u + at

=0−2×5

= −10 m s−1

Distance between J and K = 218.5 − 67.6 = 150.9 m.

So B’s speed is

10 m s−1.

12 u = 13 m s−1, a = −2 m s−2, s = ± 30 m. s = ut + 1 at2 2 When s = +30 m: 30 = 13 × t + 1 × −2 × t2 2 t2 − 13t + 30 = 0

(t − 3)(t − 10) = 0

When s = −30 m: −30 = 13 × t + 1 × −2 × t2 2 t2 − 13t − 30 = 0

(t − 15)(t + 2) = 0

t = 15 s or t = −2 s

Times when ball is 30 m from its starting point are 3 s, 10 s and 15 s.

t = 3 s or t = 10 s

13 J: u = 2 m s−1, a = 1 m s−2, s = 30 m       s = ut + 1 at2 2

K: u = 0 m s−1, a = 0.8 m s−2, t = 13 s s = ut + 1 at2 2 = 0 × 13 + 1 × 0.8 × 132 2 = 67.6 m

14 a Let U m s−1 be the velocity of the body at E and V m s−1 be the velocity of the body at F.

FG: u = V m s−1, t = 3 s, s = 66 m s = ut + 1 at2 2

66 = 3V + 1 × a × 32 2

44 = 2V + 3a

EF:  v = V m s−1, t = 7 s, u = U m s−1

v = u + at

V = U + 7a

Substituting V from (2) into (1):

44 = 2(U + 7a) + 3a

EH: t = 14 s, v = 0 m s−1, u = U m s−1

v = u + at

0 = U + 14a

Double 4 :

0 = 2U + 28a

= 2U + 17a

30 = 2 × t + 1 × 1 × t2 2 t2 + 4t − 60 = 0

Subtract 3 from 5 :

−44 = 11a

(t + 10)(t − 6) = 0

a = −4 m s−2

t = 6 s

1

2

3

4 5

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WORKED SOLUTIONS

b Substitute a = −4 into (4):

0 = U + 14 × −4

U = 56 m s−1

15 Let T s be the time taken between the garage (G) and the restaurant (R).

Hence the time taken between the restaurant and the library (L) is 2T s.

Let V m s−1 be the velocity at the restaurant.

RL: v = 9 m s−1, a = 0.15 m s−2, s = 240 m, t = 2T s s = vt − 1 at2 2

240 = 9(2T) − 1 × 0.15 × (2T)2 2 = 18T − 0.3T2

0.3T2

t2 = 1.2 5

t = 1.2 = 0.490 s 5

3 u = 7 m s−1, a = 10 m s−2, v = 15 m s−1

v2 = u2 + 2as

2 2 s = v −u 2a

2 2 = 15 − 7 2 × 10

= 8.8 m

4 a   u = 0 m s−1, a = 10 m s−2, s = 48 m

v2 = u2 + 2as

− 18T + 240 = 0

= 02 + 2 × 10 × 48

T2 − 60T + 800 = 0

= 960

v = 960 = 31.0 m s−1

(T − 20)(T − 40) = 0 9 m s−1,

0.15 m s−2,

V m s−1

RL: v =

v = u + at

If T = 20: 9 = V + 0.15 × 40, so V = 3 m s−1.

If T = 40: 9 = u + 0.15 × 80, so V = −3 m s−1.

Since the cyclist is not cycling backwards, T = 20 s and V = 3 m s−1. 9 m s−1,

a=

t = 2T s, u =

0.15 m s−2,

b v = u + at

t = v − u a

= 31.0 − 0 10

= 3.10 s

GL: v =

s = vt − 1 at2 2

Alternatively:

= 40 × 10 + 1 × −10 × 102 2 = −100 m

The cliff is 100 m high.

GR: v = 3 m s−1, t = 20 s, a = 0.15 m s−2 s = vt − 1 at2 2 = 3 × 20 − 1 × 0.15 × 202 2 = 30 m

Total distance = 30 + 240 = 270 m.

a=

t = 60 s

= 9 × 60 − 1 × 0.15 × 602 2 = 270 m

5 u = 40 m s−1, a = −10 m s−2 a t = 10 s s = ut + 1 at2 2

b v = 0 m s−1

v2 = u2 + 2as

2 2 s = v −u 2a

Exercise 2.3A

2 2 = 0 − 40 2 × −10

1 u = 0 m s−1, a = 10 m s−2, t = 1.8 s   s = ut + 1 at2 2 = 0 × t + 1 × 10 × 1.82 2 = 16.2 m

Maximum height = 100 + 80 = 180 m.

v = u + at

2 u = 0 m s−1, a = 10 m s−2, s = 1.2 m s = ut + 1 at2 2 1.2 = 0 × t + 1 × 10 × t2 2 5t2 = 1.2

t = v − u a

= 0 − 40 −10

= 4 s

2

= 80 m

c v = 0 m s−1

17

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2 Kinematics of motion in a straight line

7 a u = 0 m s−1, a = g m s−2, s = h m, t = t s s = ut + 1 at2 2

d t = 10 s

v = u + at

= 40 − 10 × 10

= −60

Speed = 60 m s−1.

e s = 0 m

h = 0 × t + 1 gt2 2

= 1 gt2 2 2h = gt2

s = ut + 1 at2 2

0 = 40 × t + 1 × −10 × t2 2 2 = 40t − 5t

t2 = 2h g

t=

= 5t(8 − t)

t = 0 s or 8 s

b u = 0 m s−1, a = 3g m s−2, s = 12h m, t = t s s = ut + 1 at2 2

6 u = 25 m s−1, a = −10 m s−2 a s = 0 m

s = ut + 1 at2 2

0 = 25 × t + 1 × −10 × t2 2 = 25t − 5t2

t = 0 s or 5 s b    v = 0

v2 = u2 + 2as

2 2 s = v −u 2a

2 2 = 0 − 25 2 × −10

= 31.3 m

It would take two times longer.

H = 5T2

= 252 + 2 × −10 × 23.4

= 156.25

v = 156.25 =

Dropped stone: u = 0 m s−1, a = 10 m s−2, s = H m, t = T s    s = ut + 1 at2 2 H = 0 × T + 1 × 10T2 2

12.5 m s−1

t = 2 2h g

8 Let the height of the castle be H m and the time taken by the dropped stone be T s.

c s = 3 × 31.3 = 23.4 m 4 v2 = u2 + 2as

12h = 0 × t + 1 × 3gt2 2 3 2 = gt 2 8h = gt2 t2 = 8h g

= 5t(5 − t)

The ball takes 5 seconds to return.

2h g

1 14 m s−1,

Thrown stone: u = a= t = (T − 1) s s = ut + 1 at2 2 H = 14 × (T − 1) + 1 × 10(T − 1)2 2 = 14(T − 1) + 5(T 2 − 2T + 1) H = 14T − 14 + 5T2 − 10T + 5

v = u + at

Put 1 = 2 :

12.5 = 25 − 10t

5T 2 = 14T − 14 + 5T 2 − 10T + 5

10t = 12.5

0 = 14T − 14 − 10T + 5

t = 1.25 s

4T = 9 T = 9 = 2.25 s 4 H = 5T 2

= 5 × 2.252

= 25.3 m

d Maximum height is an overestimate because of the effect of air resistance. e No difference since, according to equations, weight has no effect.

10 m s−2,

s = H m,

2

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2

WORKED SOLUTIONS

9 The ball is travelling in the opposite direction to the direction in which it was launched when it hits the ground, so its velocity is negative. Sawda’s answer is correct.

Katie’s solution is more efficient, but she needs to ensure that she chooses the correct sign for v.

10 a u = 70 m s−1, a = −10 m s−2, t = T s, s > 165 m

70T − 5T2 > 165

0 > 5T2 − 70T + 165

5T2 − 70T + 165 < 0

T2 − 14T + 33 < 0

b (T − 3)(T − 11) < 0

After rope snaps: u = 3.2 m s−1, a = −10 m s−2,

s = −6.4 m.

v2 = u2 + 2as

= 3.22 + 2 × −10 × −6.4

= 138.24

v = 138.24 = −11.8 m s−1

v = −11.8 m s−1, u = 3.2 m s−1, a = −10 m s−2

v = u + at

−11.8 = 3.2 − 10t

10t = 15.0

t = 1.50 s

3 < T < 11

14 Halfway down the building is 18.5 m (37 ÷ 2).

The object is higher than 165 m for 8 seconds.

Time for dropped stone to hit the ground: u = 0 m s−1, a = 10 m s−2, s = 37 m. s = ut + 1 at2 2 37 = 0 × t + 1 × 10t2 2 t2 = 37 = 7.4 5

11 s = −11.4 m, a = −10 m s−2, t = 4.3 s s = ut + 1 at2 2

−11.4 = u × 4.3 + 1 × −10 × 4.32 2 = 4.3u – 92.45

4.3u = 81.05

t = 7.4 = 2.720 s

u = 18.8 m s−1

Time for dropped stone to get halfway: u = 0 m s−1, a = 10 m s−2, s = 18.5 m. s = ut + 1 at2 2 18.5 = 0 × t + 1 × 10t2 2 t2 = 18.5 = 3.7 5

v = 0 m s−1, a = −10 m s−2, u = 18.8 m s−1, t = T s

v = u + at

0 = 18.8 − 10T T = 18.8 = 1.88 s 10

12 u = 43 m s−1, a = −10 m s−2, s = 90 m   s = ut + 1 at2 2

90 = 43 × t + 1 × −10 × t2 2 5t2 − 43t + 90 = 0 2

43 ± (−43) − 4 × 5 × 90 or (5t – 18)(t – 5) = 0 2×5 = 18 or 5 5 The ball is more than 90 m above the ground for 5 − 18 = 7  s. 5 5 t =

The thrown stone takes 0.7967 s to get to the ground (2.720 − 1.923).

Thrown stone: a = 10 m s−2, s = 37 m, t = 0.7967 s. s = ut + 1 at2 2 37 = u × 0.797 + 1 × 10 × 0.79672 2 = 0.797u + 3.174

33.826 = 0.7967u

u = 42.5 m s−1

t = 3.7 = 1.923 s

15 a Minimum value of U:

13 Pulled by rope: u = 0 m s−1, a = 0.8 m s−2, t = 4 s

v = 0 m s−1, s = 45 m, a = −10 m s−2

v = u + at

v2 = u2 + 2as

= 0 + 0.8 × 4

0 = U2 + 2 × −10 × 45

= 3.2 m s−1  s = ut + 1 at2 2 = 0 × 4 + 1 × 0.8 × 42 2 = 6.4 m

U2 = 900

U = 900 = 30 m s−1

Maximum value of U:

v = 5 m s−1, s = 45 m, a = −10 m s−2

v2 = u2 + 2as

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2 KINEMATICS OF MOTION IN A STRAIGHT LINE

25 = U2 + 2 × −10 × 45 Displacement (km)

U2 = 925 U = 925 = 30.4 m s−1 b u = 5 m s−1, v = −5 m s−1, a = −10 m s−2 v = u + at −5 = 5 − 10t 10t = 10

30

t = 1s 16 Dropped: u = 0 m s−1, s = −H m s−1, a = −10 m s−2, t = T s s = ut + 1 at2 2 –H = 0 × T + 1 × −10 × T2 2 = −5T2

0 11:20 a.m.

12 noon

1 p.m. 1:36 p.m.

Time (min) b 136 minutes after 11:20 is 13:36 or 1:36 p.m.

Launched: u = U m s−1, s = −H m s−1, a = −10 m s−2, t = (T + X) s s = ut + 1 at2 2 –H = U(T + X) + 1 × −10 × (T + X)2 2 = UT + UX − 5(T + X)2

3

a & b First particle: One (0 s–12 s): velocity = 3 m s−1, time = 12 s, displacement = 3 × 12 = 36 m Two (12 s–27 s): velocity = −2 m s−1, time = 15 s, displacement = −2 × 15 = −30 m Second particle: velocity = 1.6 m s−1 starting three seconds later.

−5T2 = UT + UX − 5T2 – 10XT – 5X2 0 = UT + UX – 10XT – 5X2 10XT – UT = UX –

5X2

T(10X – U) = X(U – 5X) T=

X(U − 5X ) 10X − U

 U − 5X  H = 5T2 = 5X 2  10X − U 

Particle 2

Particle 1

(d, t)

0

12

27

Time (s) c The graphs cross whilst the first particle is returning.

a −7 ÷ 20 = −0.35 m s−2 So the deceleration is

Equation of line for first particle: 0.35 m s−2.

b 10 × 12 + 1 (12 + 5) × 20 + 10 × 5 = 340 m 2 2

36

2

Exercise 2.4A 1

Displacement (m)

Hence −5T2 = UT + UX − 5(T + X)2

a One: velocity = 45 km h−1, time = 40 min. Since 40 min = 2 h, displacement = 30 km 3 Two: stationary for 60 min Three: velocity = −50 km h−1, displacement = −30 km. Time = −30 ÷ −50 = 3 h = 36 min 5

d − 36 = −2(t − 12) 2t + d = 60 Equation of line for second particle: d − 0 = 1.6(t − 3) d = 1.6t − 4.8 Substitute: 2t + 1.6t − 4.8 = 60 3.6t = 64.8 t = 18 s d = 1.6 × 18 − 4.8 = 24 m

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2

WORKED SOLUTIONS

a Car A: One: velocity = 75 km h−1, time = 20 min. Since 20 min = 1 h, displacement = 25 km 3 Two: stationary for 10 min

V

Velocity (m s–1)

4

Three: velocity = 51 km h−1, displacement = 17 km. Time = 17 ÷ 51 = 1 h = 20 min 3 Car B: One: velocity = −72 km h−1, time = 10 min Since 10 min = 1 h, displacement = −12 km 6 Two: stationary for 20 min

0

Three: velocity = −72 km h−1, displacement = −30 km 5 Time = −30 ÷ −72 = 12 h = 25 min

15V = 1 (X + 100)V 8 120 = X + 100

B

X = 20

30

c

25

t = 50 s, a = −0.9 m s−2, u = V m s−1, v = 0 m s−1 v = u + at u = v − at V = 0 − (−0.9) × 50

A 10

= 45 m s−1 20

30

60

6

Time (min) b The graphs cross whilst both cars are on their final stages. Gradients are given as kilometres per minute. Equation of line for first car: d − 25 = 0.85(t − 30)

a Velocity (m s–1)

0

police car

24

motorcycle

18

1

d = 0.85t − 0.5

0

Equation of line for second car:

10

2

d = −1.2t + 66 Equate 1 = 2 : 0.85t − 0.5 = −1.2t + 66

T

t = 32 (nearest integer) Time = 17:17 or 5:17 p.m. a One (0 s–30 s): u = 0 m s−1, v = V m s−1, t = 30 s Two (30 s–(X + 30) s): steady speed of V m s−1 for X seconds V m s−1,

b Let T be the time the police car catches up with the motorcyclist The distance travelled by each vehicle is the same, so the areas are the same. 18T = 1 (T − 10 + T − 20) × 24 2 = 12(2T − 30)

2.05t = 66.5

Three ((X + 30) s–100 s): u = until 100 s

20

Time (s)

d − 30 = −1.2(t − 30)

5

100

b 1 × 30 × V = 1  1 ( X + 100)V  42 2 

42

Displacement (km)

X + 30 Time (s)

30

v=

0 m s−1,

= 24T − 360 360 = 6T T = 60 s c 18 × 60 = 1080 m

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2 KINEMATICS OF MOTION IN A STRAIGHT LINE

a

9

Displacement (m)

D Q

For second stage, u = V m s−1, t = T2 s. There is no acceleration so substitute into s = ut:

P

30

s = VT2 0

For third stage, u = V m s−1, t = T3 s, v = 0 m s−1. Substitute into s = 1 (u + v)t 2 1 1 s = (V + 0) × T3 = VT3 2 2

T

Time (s)

b For P , D ÷ T = 1.25, so D = 1.25T For Q, d − 30 = −0.5(t − T).

The total distance is given by the sum of these distances. Total distance = 1 VT1 + VT2 + 1 VT3 2 2 = 1 V(T1 + 2T2 + T3) 2

When t = 0, d = D, so D − 30 = 0.5T Substitute D = 1.25T into D − 30 = 0.5T. 1.25T − 30 = 0.5T 0.75T = 30

Since T = T1 + T2 + T3, 1 total distance = V(T + T2), which is the area 2 of the trapezium with parallel sides of T and T2 and a perpendicular height of V.

T = 40 c D = 1.25 × 40 = 50 8

a First athlete: One: u = 0 m s−1, t = 10 s, a = 0.8 m s−2

10 a First vehicle:

v = u + at = 0 + 0.8 × 10 = 8 m s−1

One (0 s–20 s): u = 0 m s−1, a = 1.5 m s−2, t = 20 s: v = u + at = 0 + 1.5 × 20 = 30 m s−1

Two: steady speed of 8 m s−1 until 105 s Second athlete: One: u =

0 m s−1,

t = 40 s, a =

Two (20 s–90 s): steady speed of 30 m s−1 for 70 s

0.225 m s−2

v = u + at = 0 + 0.225 × 40 = 9 m s−1

Three: to rest

Two: different acceleration until 102 s

Second vehicle:

V Velocity (m s–1)

For first stage, u = 0 m s−1, t = T1 s, v = V m s−1. Substitute into s = 1 (u + v)t: 2 1 1 s = (0 + V) × T1 = VT1 2 2

One (0 s–50 s): u = 0 m s−1, a = 0.8 m s−2, t = 50 s: v = u + at = 0 + 0.8 × 50 = 40 m s−1

te

d athle

Secon

9

First athlete

8

Two (50 s–70 s): steady speed of 40 m s−1 for 20 s Three: to rest

0

10

40

102 105

Time (s) 1 b First athlete: Area = 2 (105 + 95) × 8 = 800 m c Second athlete: 1 1 × 40 × 9 + (9 + V) × 62 = 800 2 2 180 + 31(9 + V) = 800 31V + 279 + 180 = 800

Velocity (m s–1)

7

Vehicle B

40

Vehicle A

30

0

20

50

70

90

T

Time (s)

31V = 341 V = 11 m s−1

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2

WORKED SOLUTIONS

b Same distance for each vehicle. 1 (T + 70) × 30 = 1 (T + 20) × 40 2 2

e Shorter. The skydiver will not descend in freefall at g due to air resistance. 12 One: steady speed of V m s−1 for 45 seconds

15(T + 70) = 20(T + 20) 15T + 1050 = 20T + 400

Two: u = V m s−1, a = −1.6 m s−2, v = 0 m s−1, t = T s

5T = 650

a

T = 130 s 1 c 2 (130 + 70) × 30 Velocity (m s–1)

V

= 3000 m d First vehicle: a = −30 ÷ 40 = −0.75 m s−2 So deceleration is 0.75 m s−2. Second vehicle: a = −40 ÷ 60 = −0.67 m s−2 So deceleration is 0.67 m s−2. 11 a Velocity (m s–1)

450

45

45

t

b a = −1.6 = − V , so V = 1.6T T Area = 1 (T + 45 + 45) × V = 2300 2 1 (T + 90) × 1.6T = 2300 2 1.6T(T + 90) = 4600

8 0

T + 45

225

1.6T2 + 144T − 4600 = 0

Time (s)

T2 + 90T − 2875 = 0 c (T + 45)2 − 2025 − 2875 = 0

b

(T + 45)2 = 4900 Distance (m)

T + 45 = ±70 T = 25 s or −115 s Hence T = 25 s V = 1.6 × 25 = 40

0

45

225

Time (s)

Exercise 2.5A 1

a v=

ds = 13 − 8t dt

= 13 − 8 × 2 = −3 m s−1 Acceleration (m s–2)

c

b a = dv = −8 m s−2 (which does not depend on dt time)

10

2 0

45

225

Time (s)

a v = dr = 15t2 − 14t dt a = dv = 30t − 14 dt b v = 15(2)2 − 14(2) = 32 m s−1 c a = 30(3) − 14 = 76 m s−2

d 1 × 45 × 450 + 180 × 8 = 11565 m 2

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2 Kinematics of motion in a straight line

3 a s = ∫ v dt = ∫ (t − 2) dt = 1 t2 − 2t + c 2

When t = 6, s = 11 11 = 1 (6)2 − 2(6) + c 2 = 18 − 12 + c

= 4 × 1 + 64 × 1 2 16 = 2 + 4 = 6 m s−1

c=5 s = 1 t2 − 2t + 5 2

3

− a = dv = −2t 2 − 128t−3 dt

= −2 × 4

b When t = 8: s = 1 (8)2 − 2(8) + 5 = 21 m 2 c 1 t2 − 2t + 5 = 53 2 t2 − 4t + 10 = 106

−3 2

− 128 × 4−3

= −2 × 1 − 128 × 1 8 64 1 = − 1 − 2 = −2 m s−2 4 4 7 a v = ∫ adt = ∫ (6t − 20) dt = 3t2 − 20t + c

t2 − 4t − 96 = 0

When t = 12, v = 223

(t − 12)(t + 8) = 0

223 = 3(12)2 − 20(12) + c

t = 12 s

c = 31

v = 12 − 2 = 10 m s−1

v = 3t2 − 20t + 31

4 a v =

b s = ∫ v dt = ∫ (3t 2 − 20t + 31) dt = t3 − 10t2 + 31t + c2

ds = 3t2 − 16t + 5 = 0 dt

(3t − 1)(t − 5) = 0 t = 1 s or 5 s 3 b 3t2 − 16t + 5 > 0 when t < 1 s and when t > 5 s 3 d v = 6t − 16 > 0 c a = dt t > 8 s 3 d 6t − 16 = 2

5 18t − t2

t2

8(3)2

+ 5(3) + 10 = −20 m

= 65

− 18t + 65 = 0

s = t3 − 10t2 + 31t − 30

c s = (t − 2)(t − 3)(t − 5) = 0 t = 2 s, 3 s or 5 s 3t2 − 20t + 12 = 0

1

16 + q = 33

q = 17

−8 m s−2

6 a Substitute t = 4: s = 8 4 − 64 = 8 × 2 − 16 = 0 4 1 b s = 8t 2 − 64t−1 − ds = 4t 2 + 64t−2 v= dt

ds = 4t3 − 24t2 + 44t − 20 = 4 dt

4t3 − 24t2 + 44t − 24 = 0

t3 − 6t2 + 11t − 6 = 0

When t = 5, a = 18 − 2 × 5 = 8 m s−2 When t = 13, a = 18 − 2 × 13 =

8 a 44 − 8 × 43 + 22 × 42 − 20 × 4 + q = 33

a = dv = 18 − 2t dt

−1 2

b v =

t = 5 s or 13 s

=4×4

c2 = −30

(3t − 2)(t − 6) = 0 t = 2 s or 6 s 3

(t − 5)(t − 13) = 0

v = 3(3)2 − 16(3) + 5 = −16 m s−1 s=

40 = (7)3 − 10(7)2 + 31(7) + c2

t=3 (3)3

When t = 7, s = 40

d 3t2 − 20t + 31 = 19

6t = 18

(t − 1)(t − 2)(t − 3) = 0

t = 1 s, 2 s or 3 s

When t = 1, s = 14 − 8 × 13 + 22 × 12 − 20 × 1 + 17 = 12 m

When t = 2, s = 24 − 8 × 23 + 22 × 22 − 20 × 2 + 17 = 17 m

When t = 3, s = 34 − 8 × 33 + 22 × 32 − 20 × 3 + 17 = 20 m.

+ 64 × 4−2

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2

WORKED SOLUTIONS

a s = 9 (2)2 − 1 (2)3 = 46  m 2 3 3 9 b When t = 4, s = (4)2 − 1 (4)3 = 152  m 2 3 3

c a = dv = 12t2 − 48t + 44 = 188 dt

12t2 − 48t − 144 = 0

t2 − 4t − 12 = 0

(t − 6)(t + 2) = 0

t = 6 s or −2 s

When t = 5, s = 9 (5)2 − 1 (5)3 = 425  m 2 3 6

Distance = 425 − 152 = 121  m 6 3 6 c Note that the particle is at instantaneous rest when t = 9. When t = 8, s = 9 (8)2 − 1 (8)3 = 352  m 2 3 3

When t = 6, s = 64 − 8 × 63 + 22 × 62 − 20 × 6 + 17 = 257 m.

9 a 6t − 30 = 0 t = 5 s

v = ∫ a dt = ∫ (6t − 30) dt = 3t2 − 30t + c

When t = 9, s = 9 (9)2 − 1 (9)3 = 243  m 2 3 2

When t = 0, v = 72

72 = 3(0)2 − 30(0) + c

c = 72

When t = 10, s = 9 (10)2 − 1 (10)3 = 350  m 2 3 3 243 350 243 352 Distance = + = 9 m − − 2 3 2 3

v = 3t2 − 30t + 72

When t = 5, v =

3(5)2

− 30(5) + 72 =

−3 m s−1

t2 − 10t + 24 = 0

t = 4 s or 6 s c s = ∫ v dt = ∫ (3t 2 − 30t + 72) dt =

t3

15t2

+ 72t + c2

When t = 0, s = 0

0 = (0)3 − 15(0)2 + 72(0) + c2

c2 = 0

s = t3 − 15t2 + 72t

When t = 4, s = 112 m

When t = 6, s = 108 m

When t = 10, s = 220 m

Total distance = 112 + 4 + 112 = 228 m

10 v = ∫ a dt = ∫ (9 − 2t ) dt = 9t −

t2

+c

0 = 9(0) − (0)2 + c c=0

1

( 12 t + 3) dt = 14 t + 3t + c

∫ 2 (t + 6) dt

=

2

When t = 0, v = 2 2 = 1 (0)2 + 3(0) + c 4 c=2 v = 1 t2 + 3t + 2 4

When t = 4, v = 1 (4)2 + 3(4) + 2 = 18 m s−1 4

For 4 < t  8, v = ∫ a dt =

When t = 4, v = 18

18 = −160(4)−2 + c2

18 = −10 + c2

c2 = 28

v = 28 −160t−2

dt = ∫ 320t ∫ 320 t3

−3

dt

i When t = 3, v = 1 (3)2 + 3(3) + 2 = 13.25 m s−1 4 ii When t = 5, v = 28 −160(5)−2 = 21.6 m s−1 b For 0  t  4, s = ∫ v dt = 1 t3 + 3 t2 + 2t + c 3 12 2

t2

s = ∫ v dt = ∫ (9t − t 2) dt = 9 t2 − 1 t3 + c2 2 3 When t = 0, s = 0 0 = 9 (0)2 − 1 (0)3 + c2 2 3 c2 = 0 s = 9 t2 − 1 t3 2 3

)

= −160t−2 + c2

When t = 0, v = 0

v = 9t −

) (

11 a For 0  t  4, v = ∫ a dt =

b 3t2 − 30t + 72 = 0 (t − 4)(t − 6) = 0

(

( 14 t

2

)

+ 3t + 2 dt =

When t = 0, s = 0 0 = 1 (0)3 + 3 (0)2 + 2(0) + c3 12 2 c=0 s = 1 t3 + 3 t2 + 2t 12 2 When t = 4, s = 1 (4)3 + 3 (4)2 + 2(4) = 112  m 12 2 3

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2 KINEMATICS OF MOTION IN A STRAIGHT LINE

60 = 8(6)2 − 1 (6)3 − 44(6) + c2 3 c2 = 108 s = 8t2 − 1 t3 − 44t + 108 3 When t = 5, s = (5)2 + 4(5) = 45 m When t = 9, s = 8(9)2 − 1 (9)3 − 44(9) + 108 3 = 117 m

For 4 < t  8, s = ∫ v dt = ∫ (28 − 160t −2) dt = 28t + 160t−1 + c4 When t = 4, s = 112 3 112 = 28(4) + 160(4)−1 + c 4 3 344 c4 = − 3 s = 28t + 160t−1 − 344 3

Alternatively: Could use area of trapezium to find distance when t = 5. Area = 1 (4 + 14) × 5 = 45 m. 2 Could also use area of trapezium to find distance when t = 6. Area = 1 (4 + 16) × 6 = 60 m. 2

When t = 8, s = 28(8) + 160(8)−1 − 344 = 388 3 3 = 129 m. 12 a

Velocity (m s–1)

20

13 Stage one: u = 6 m s−1, v = 33 m s−1, t = 18 s s = u +v t 2

16

( ) = ( 6+233 ) × 18 = 351 m

4

0

6

Stage two: For t > 18, a = 1 (t − 12) 4 1 v = ∫ a dt = ∫ 1 (t − 12) dt = ∫ 4 t − 3 dt = 1 t2 − 3t + c 8 4

(

10

When t = 18, v = 33 33 = 1 (18)2 − 3(18) + c 8

Time (s) b Maximum speed of P is 20 m s−1 when t = 8 s.

c = 93 2

c i For 0  t  6, a = dv = 2 dt When t = 5, a =

v = 1 t2 − 3t + 93 2 8 1 93 dt s = ∫ v dt = ∫ t 2 − 3t + 8 2

2 m s−2

(

ii For 6 < t  10, v = 20 − (t − 8)2 = 20 − (t2 − 16t + 64) = 16t − t2 − 44

When t = 9, a = 16 − 2 × 9 = −2 m s−2 d For 0  t  6, s = ∫ v dt = ∫ (2t + 4) dt = t2 + 4t + c When t = 0, s = 0 0 = (0)2 + 4(0) + c c=0 When t = 6, s = (6)2 + 4(6) = 60 m For 6 < t  10, s = ∫ v dt = ∫ (16t − t 2 − 44) dt = 8t2 − 1 t3 − 44t + c2 3 When t = 6, s = 60

)

= 1 t3 − 3 t2 + 93 t + c2 24 2 2 When t = 18, s = 351 351 = 1 (18)3 − 3 (18)2 + 93 (18) + c2 24 2 2 c2 = −243 s = 1 t3 − 3 t2 + 93 t − 243 2 2 24 When t = 30, s = 1 (30)3 − 3 (30)2 + 93 (30) − 243 24 2 2 = 927 m

a = dv = 16 − 2t dt

s = t2 + 4t

)

Exam-style questions 1

a u = 1.5, t = 6, s = 90 s = ut + 1 at2 2 90 = 1.5 × 6 + 1 × a × 62 2 a = 4.5 m s−2

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2

WORKED SOLUTIONS

b u = 1.5, v = 37.5, a = 4.5

9p = 216

v2

p = 24

=

u2

+ 2as

c=2

2 2 s = v − u 2a 2

= 37.5 − 1.5 2 × 4.5

v = 12t2 − 10t + 2 s = ∫ v dt = ∫ (12t 2 − 10 t + 2)dt = 4t3 − 5t2 + 2t + c2

2

When t = 0, s = 0

= 156 m

0 = 4(0)3 − 5(0)2 + 2(0) + c2

2 a a = −10, v = 0, s = 80

c2 = 0

v2 = u2 + 2as

s = 4t3 − 5t2 + 2t

u2 = v2 − 2as =

02

When t = 3, s = 4(3)3 − 5(3)2 + 2(3) = 69 m

− 2 × −10 × 80

5 a s = –20 m, u = 37.5 ms–1, a = –10 m s–2

= 1600

s = ut + 12 at2

u = 40 m s−1 b s = 0, a = −10, u = 40 s = ut + 1 at2 2 0 = 40t + 1 × −10 × t2 2 5t2 − 40t = 0

1 2

× –10 × t2

–20 = 37.5t +

5t2

– 37.5t – 20 = 0

2t2

– 15t – 8 = 0

b (2t + 1)(t – 8) = 0 t = – 12 or 8

t(5t − 40) = 0

t = 8 s

t = 40 ÷ 5 = 8 s

c v = 0 ms–1, u = 37.5 ms–1, a = –10 ms–2 2 2 2 2 s = v − u = 0 − 37.5 2s 2(−10)

3 a s = t3 − 14t2 + 48t = 0 t(t − 6)(t − 8) = 0

s = 70.3125

t = 6 s or 8 s b v =

ds = 3t2 − 28t + 48 dt

6 a PQ: s = 40, t = 5 s = ut + 1 at2 2 40 = 5u + 12.5a

a = dv = 6t − 28 = 0 dt 14 t = 3

3

2

When t = 2, v = 30 30 = 1 p(2)2 − 10(2) + c 2 1

When t = 2.5, v = 52 52 = 1 p(2.5)2 − 10(2.5) + c 2 2

Multiply (1) by 8: 16p + 8c = 400

1

PR: s = 60, t = 15 s = ut + 1 at2 2 60 = 15u + 112.5a

4 v = ∫ a dt = ∫ (pt − 10) dt = 1 pt2 − 10t + c 2

25p + 8c = 616

16 = 2u + 5a

s =  14  − 14 14  + 48 14  = 20.7 m  3  3  3

2p + c = 50

Total distance = 2 × 70.3125 + 20 = 161 m

8 = 2u + 15a

Subtract 1 from 2 :

−8 = 10a

2

a = −0.8 m s−2

Substitute:

16 = 2u + 5 × −0.8

20 = 2u

u = 10 m s−1 b a = −0.8 m s−2

3

Subtract 3 from 2 :

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2 KINEMATICS OF MOTION IN A STRAIGHT LINE

7

36T = 18T + 1.5T2

a

0 = 1.5T2 − 18T

u

T = 12 s 10 a 24

0

3

6

Time (s) Velocity (m s–1)

Velocity (m s–1)

= T(1.5T − 18)

–u b v = 0, s = 45, t = 3 s = 1 (u + v)t 2 45 = 1 (u + 0) × 3 2 u = 30 m s−1

0

dv = 6t − 12 = 0 dt t=2 Integrate v to find:

T = 225 s

Motorcycle takes 95 − 20 = 75 s.

s = t3 − 6t2 + 11t + c

d Motorcycle: 1920 = 1 × 75 × V 2 V = 51.2 m s−1

0=0+0+0+c c=0 s = t3 − 6t2 + 11t 9

375 = T + 150 c Car: 1920 = 1 (T1 + T1 − 30) × 24 2 T1 = 95 s

a=

11 There are four stages to this race. 6(2)2

= 182 + 2 × 3 × 42

For the first stage, u = 0, a = 4, t = 2. s = ut + 1 at2 2 = 0 × 2 + 1 × 4 × 22 2 = 8m

= 576

v = u + at = 0 + 4 × 2 = 8 m s−1

When t = 2, s =

+ 11(2) = 6 m

a u = 18, a = 3, s = 42 v2 = u2 + 2as

v=

24 m s−1

b For B: u = 18, a = 3, s = 42, v = 24 v = u + at 24 = 18 + 3t t=2 For A: u = 36, t = 2, a = 0 s = ut = 72 m c B overtakes A when time (T) is same and distance is same. For A: s = 18T + 1.5T2 For B: s = 36T

T

b 4500 = 1 (T + 150) × 24 2

For minimum velocity, a = 0

(2)3

180

Time (s)

c Air resistance 8

30

For the second stage, u = 8, a = 1 and t = 3. 3 1 2 s = ut + at 2 = 8 × 3 + 1 × 1 × 32 2 3 51 = m 2 v = u + at = 8 + 1 × 3 = 9 m s−1 3 For the third stage, u = 9, a = 3 and t = 13. 13 1 2 s = ut + at 2 = 9 × 13 + 1 × 3 × 132 2 13 273 = m 2 v = u + at = 9 + 3 × 13 = 12 m s−1 13

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2

WORKED SOLUTIONS

For the fourth stage, u = 12 m s−1, a = −2.4 m s−2 and v = 0 m s−1

For 6 < t  30, u = 12 m s−1, v = 0 and a = − 1 m s−2. 2 Use v2 = u2 + 2as

2 2 2 2 s = v − u = 0 − 12 = 30 m 2a 2 × −2.4

2 2 s = v −u 2a

v − u 0 − 12 = = 5s t= a −2.4

2 2 = 0 − 121 2×−2

The total distance is given by 8 + 51 + 273 + 30 2 2 = 200 m.

= 144 m

The total time is given by 2 + 3 + 13 + 5 = 23 s. Average speed =

You also need to find the displacement from 30 s to 40 s. For t > 30, u = 0 m s−1, t = 10 s and a = −1 m s−2. 2 Use s = ut + 1 at2: 2 1 s = 0 × 10 + × − 1 × 102 2 2 = −25 m

total distance 200 = 8.70 m s−1. = total time 23

12 a For 0  t  6, v = (8t − t2) m s−1 a = dv = 8 − 2t dt = 8 − 2 × 5 = 8 − 10 = −2 m s−2 b For 0  t  6, v = 8t − t2 = 16 − (t − 4)2. The velocity of P is positive for 0  t  6. For t > 6, v = 15 − 1 t 2 The particle is at rest when v = 0. 15 − 1 t = 0 2 30 − t = 0

The total distance is given by 72 + 144 + 25 = 241 m. 13 a 180 = 1 (6 + 18) × t 2 t = 15 B

t = 30

To find the displacement during the first six seconds, integrate v with respect to t. s = ∫ v dt = ∫ (8t − t 2) dt = 4t2 − 1 t3 + c 3 When t = 0, s = 0. 0 = 4 × 02 − 1 × 03 + c 3 Therefore, c = 0 Hence, s = 4t2 − 1 t3. 3 When t = 6, s = 4 × 62 − 1 × 63 = 144 − 72 = 72 m. 3 1 For t > 6, v = 15 − t 2 a = dv = − 1 2 dt The acceleration is constant. Since the acceleration is constant, you can use the equations of uniformly accelerated motion.

V Velocity (ms–1)

The velocity of P is positive for 6 < t  30, but negative for t > 30.

18

6

0

15

33

66

Time (s) b 15 + 18 + 33 = 66 s c 180 + 1 (6 + V) × 18 + 33V = 1200 2 180 + 54 + 9V + 33V = 1200 42V = 966 V = 23 m s−1 14 a v = 1 t2 − 6t + c 2 10 = 0 + 0 + c

v = 8t − t2

c = 10

When t = 6, v = 8 × 6 − 62 = 48 − 36 = 12 m s−1.

v = 1 t2 − 6t + 10 2

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2 Kinematics of motion in a straight line

= 3U + 27 4 AC:

b 1 t2 − 6t + 10 = 0 2 t2 − 12t + 20 = 0

u = U m s−1, s = 4x m s−1, a = 1.5 m s−2, t = 10 s 1 s = ut + at2 2

(t − 2)(t − 10) = 0

t = 2 or 10 s c s = 1 t3 − 3t2 + 10t + c2 6 =  1 (10)3 − 3(10)2 + 10(10) −  1 (2)3 − 3(2)2 + 10(2) 6  6 

Hence 4  3U + 27  = 10U + 75  4

= − 128  m 3

4x = 10U + 1 × 1.5 × 102 2 = 10U + 75

12U + 27 = 10U + 75

So the distance is  128   m. 3

2U = 48

15 First ball:

U = 24

u = 35 m s−1, s = −H m s−1, a = −10 m s−2, t = T s s = ut + 1 at2 2 –H = 35T + 1 × −10 × T2 2 = 35T − 5T2 Second ball:

4x = 10(24) + 75

= 240 + 75 = 315 m

17 a s = t3 + kt2 v = 3t2 + 2kt 3 × 62 + 2k × 6 = 0

k = –9

u = 24 m s−1, s = −H m s−1, a = −10 m s−2, t = (T – 2) s s = ut + 1 at2 2 –H = 24(T – 2) + 1 × −10 × (T – 2)2 2 = 24T – 48 −5(T – 2)2

a = 6t + 2k = 6 × 13 + 2 × –9 = 60 m s–2 b s = t2(t – 9)

During first 6 s, s = [62(6 – 9)] – [0] = –108

From 6 s to 10 s, s = [102(10 – 9)] – [62(6 – 9)] = 100 – (–108) = 208

Total distance = 108 + 208 = 316 m

Hence 35T − 5T2 = 24T – 48 −5(T – 2)2 35T − 5T2 = 24T – 48 − 5T2 + 20T – 20 35T = 24T – 48 + 20T – 20 9T = 68 68 T= 9 H = 5T2 – 35T 2

= 5 68  – 35 68   9  9 = 21.0 m 16 AB:

Mathematics in life and work 1 Bus accelerates for 12 = 6 s 2 Bus decelerates for 20 – (6 + 5) = 9 s 12 Deceleration = 9 = 1.33 m s–2 1 2 2 (5 + 20) × 12 = 150 m 1 3 When t = 6, v = (6)(12 – 6) = 12 m s–1 3 6

u = U m s−1, v = (U + 4.5) m s−1, a = A m s−2, t = 3 s v = u + at U + 4.5 = U + 3A A = 1.5 m s−2 Let AB = x m and BC = 3x m AB: u = U m s−1, s = x m s−1, a = 1.5 m s−2, t = 3 s 1 s = ut + at2 2 1 x = 3U + × 1.5 × 32 2

(

)

6

s = 1 ∫ 12t − t 2 dt = 1 6t 2 − 1 t 3  3  3 0 30

1 6 6 2 − 1 6 3  ( ) 3( )  3  

=

= 48 m 150 – 48 = 102 m left

Let T be the time taken to decelerate. 1 102 = 2 × T × 12 T = 17

Total journey time = 6 + 17 = 23 s.

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3

WORKED SOLUTIONS

3 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 indegrees, unless a different level of accuracy is specified in the question.

Prerequisite knowledge 1 a x =

y−b 3

5 p = m × v, so 38.7 = m × 860 giving m = 45 g

b 5t = mx − 4x

= x(m − 4) 5t x= (m − 4)

g =x+n y

g −n=x y

or

x=

a 0.150 kg × 40 m s−1 = 6 kg m s−1 b about 3 mg × 2 m s−1 = 0.000 006 kg m s−1

p 2 b 2p 2p =p c 2 d 2p 3 e p × 4 = 2p 2

8 a

2 a (30 × 60 × 60) ÷ 1000 = 108 b (15 × 1000 000) ÷ 60 = 250 000 c (13 × 60 × 60 ) ÷ 1000 = 46.8 km h−1 ≈ 46.8 × 5 = 29.25 8 d 0.1 e 2000

9 Initially p = 8 × 5 = 40. After t seconds, 8 × 5 × 2t > 1000 2t > 25 24 = 16

3 −10 km h−1

25 = 32, so t = 5 seconds.

Exercise 3.1A 1 a 0.056 × 35 = 1.96 kg m s−1 b 160 × 10 = 1600 kg m s−1 c 30 km h−1 = 8 1  m s−1, so p = 180 × 8 1 3 3 = 1500 kg m s−1 d 860 × 5 = 4300 kg m s−1 e 10 km h−1 = 10 000  m s−1 so p = 80 × 2.778. 60 × 60 = 2 9.1 ×

7 These are all approximate.

d 10 m s−1 × 90 kg = 900 kg m s−1

g − ny y

10−31

6 3 × 120 000 = 360 000 kg m s−1

c 70 mph ≈ 30 m s−1, so p = 1000 × 30 = 30 000 kg m s−1

c g = y(x + n)

4 p = m × v, so p = 1300 × 3 = 3900 kg m s−1 the direction is down the driveway.

222 kg m s−1

× 2.2 × 106 = 2.0 × 10−24 kg m s–1

3 p = m × v, so 500 000 = 100 000 × v giving v = 5 m s−1

Alternative method: t > log2 25 = 4.6438, giving t = 5 seconds. 10 a v = u + at v = 30 + 3 × 5 = 45 m s–1

So p = mv = 1300 × 45 = 58 500 kg m s–1

b v = u + 15

So p = 1300 × (u + 15) = (1300u + 19 500) kg m s–1

11 Initially p = mu t then v = 2s – u 2 t 2s  So final momentum = m − u  t  Using s = (u + v)

So change in momentum is 2s 2ms m − u  − mu = − 2mu t  t 

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3 Momentum

Exercise 3.2 A

v2 = 0 + 2 × 3 × 6 v2 = 36

1 a 2.8 × 2 + 3 × 12 = 2.8 × u + 3 × 5 giving u = 9.5 m s−1

v = 6 m s–1 Momentum gives:

b 2 × 15 + 10 × 12 = 2 × u + 10 × 16, giving u = −5 m s−1

200 × 6 = 200 × 0 + 400 × v1

c m × 12 + 0.1 × 3 = m × 7 + 0.1 × 13, giving m = 0.2 kg

v1 = 200 ×

6 400

v1 = 3 m s–1

d 3 × 4 + 5 × −4.4 = 3 × (−12.5) + 5 × u, giving u = 5.5 m s−1

10 Speed of Gabriel =

2 0 = 40 × −1.5 + 30 × v, giving v = 2 m s−1

a If they weigh the same then they have the same speed, but different directions, so 4 seconds.

3 150 × 3 + 120 × 0 = 150 × u + 120 × 2, giving u = 1.4 m s−1 4 120 × 1.5 + 100 × −1.8 = 120 × −1.2 + 100 × u, giving u = 1.44 m s−1. B is now travelling backwards. 5 m × 0.8 + m × 0 = m × u + m × 1.3, giving 0.8 = u + 1.3 so u = −0.5, i.e. it rebounds at a speed of 0.5 m s−1.

b 2 mvI = 2.5 m vI =

2.5 = 1.25 2

so t =

10 = 8 seconds 1.25

Or alternatively use ratios to deduce that the time will have doubled.

6 100 × 0 + 5.5 × 0 = 100 × u + 5.5 × 140, giving u = −7.7 m s−1

11 0.1 × 3 = –0.1 × v1+ 0.1 v2

7 a 1200 kg and 5 m s−1 meets 1000 kg and 4 m s−1 b 1000 kg and 5 m s−1 meets 1200 kg and 4 m s−1

v1=

c 1200 kg and 5 m s−1 meets 1000 kg and −4 m s−1

5 = 2.5 ms−1 towards the start. 2

d 1000 kg and 5 m s−1 meets 1200 kg and −4 m s−1

v2 = 3 – 2.5 =

Giving:

Exercise 3.3A

a v = (1200 × 5 + 1000 × 4) = 8.33 m s−1 1200

b 20 × 5 + 80 × u = 100 × 1.8, giving u = 1 m s−1

c v = (1200 × 5 − 1000 × 4) = 1.67 m s−1 1200

c 15 × 5 + m × −9 = (m + 15) × −6, giving m = 55 kg

d v = (1000 × 5 − 1200 × 4) = 0.167 m s−1 1200 8 As they have the same mass mu = mv1 + mv2 ⇒ u = v1 + v2 so ignore m.

d 120 × 0.1 + 1380 × u = 1500 × 0.031, giving u = 0.025 m s−1 2 10 × 2 + 35 × 0 = 45 × v, giving v = 4  m s−1 9 3 Due to conservation of momentum, if m increases then v must decrease, so the truck will slow down.

Because v3 rebounds it is negative. Because v3 does not catch up v1 it means v1 must be negative. v2 = –v3 + v4 If C does not catch up A ⇒ v1 > v3 v1 > v4 – v2 (using 2 ) v1 > v4 – (u + v1)

1 or 0.5 m s–1 away from the start. 2

1 a  6.8 × 11 + 6.7 × 2 = 13.5 × v, giving v = 6.53 m s−1 (2 d.p.)

b v = (1000 × 5 + 1200 × 4) = 8.17 m s−1 1200

u = –v1 + v2

10 = 2.5 ms–1 4

1 2

4 63 × 0 + 2 × 13 = 65 × v, so v = 0.4 m s−1 5 0.01 × 540 + 2 × 0 = 2.01v, giving v = 2.686 567 = 2.69 m s−1(3 s.f.) 6 a 4  m × 2 u = 6 m × v

so

4 u=v 3

v1 > v4 – u – v1

b 8 mu + 2 mu = 6 mv

u > v4 – 2v1

5 So v = u 3 c The particles will not meet.

9 s = 3, u = 0, v = v, a = 6 v2 = u2 + 2 as

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3

WORKED SOLUTIONS

7 mAuA + mB uB = (mA + mB)v

= mAv + mB v

mAuA − mAv = mB v − mB uB mA(uA − v) = mB v − mB uB mA =

mBv − mBuB mB(v − uB) or u uA − v A −v

8 1000 × 14 = 600 × v + 400 × 47, giving v = −8 m s–1 9 a 0.5 × 20 = 32 × v ⇒ v = 0.313 m s–1 b Using v2 = u2 + 2as ⇒ v2 = 0 + 2 × 0.25 × 2 = 1 ⇒ v = 1 m s–1

 36 000  3 a 1500 ×  3600  = 15 000 kg m s−1 36 000  = 13 000 kg m s−1  1300 ×   3600 

b Total of the momentum is 28 000, 28 000 = 18.7 m s−1. giving v = 1500 4 a 2m × 3u + 3m × 2u = 2m × vA + 3m × 4u

6mu + 6mu = 2mvA + 12mu

Thus vA = 0.

b 2m × 3u + 3m × 2u = 2m × vA + 3m × 3u

So he lands on the cart with an initial velocity of 1 m s–1

6mu + 6mu = 2mvA + 9mu

2vA = 3u, so vA = 1.5u

1 × 20 = 32 × v

c 2m × 3u + 3m × 2u = 2m × v + 3m × v

So v = 0.625 m s–1

6mu + 6mu = 5mv So v = 12 u = 2.4u. 5

c Let v1 = speed of landing on the cart:

20v1 = 32 × 2

d 2m × 3u + 3m × 2u = 2m × (−3u) + 3m × vB

v1 = 3.2

6mu + 6mu = −6mu + 3mvB

Using v2 = u2 + 2 as

3vB = 18u

3.22

vB = 6u

= 0 + 2 × 0.25 × s

Giving a run-up of s = 20.48 m.

It is not realistic for Sebastian to run for this far at the same acceleration. 10 a mAu = mAvA + mB vB mAu – mAvA = mB vB

m Au − m Av A = vB mB

b 2mA = mB

m Au − m Av A = vB 2m A

u − vA = vB 2 u – vA = 2vB

vA = u – 2vB But vA > 0 So u – 2vB > 0 u > 2vB

Exam-style Questions 1 p = 0.5 × 4.5 = 2.25 kg m s−1 If the ball stops then 2.25 = 3v, so v = 0.75. If v > 0.75 the ball will rebound. 2 a 35 km h−1 = 35 × 1000 = 9.7222 m s−1 3600 p = 54 = 9.722 22 × m, giving m = 5.55 kg b v = 35 × 2 = 70 km h−1

5 a 1 × 12 = 12 kg m s−1 b 12 = 2vB, so vB = 6 m s−1 c 2 × 6 = 3 vC , so vC = 4 m s−1 d (4 × 3 =) 12 kg m s−1 which is the same as the initial momentum of A. 6 a s = ut + 0.5at2

s = 0 + 0.5 × 0.4 × 32

s = 1.8 < 2 m

b v = u + at

= u + 0.4 × 3 = 1.2 Initially m × 1.2 = mv, so B moves with a velocity of 1.2 as well.

7 a 13m × u + m × (−u) = 13m × vb + m × vg

12mu = 13mvb + mvg

12u = 13vb + vg

12u − 13vb = vg

b m × u + M × U = 0 (as velocity before firing is zero). m × u = −M × U m = −U M u

Because we are comparing speed, not velocity: m = −U M u

As required.

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3 Momentum

8 a 45 × 12 + 45 × 0 = 90 × v, giving v = 6 m s−1

13 36 km h–1 = 10 m s–1

b 90 × 6 + 45 × 0 = 135 × v, giving v = 4

135 × 4 + 45 × 0 = 180 × v, giving v = 3 180 × 3 + 45 × 0 = 225 × v, giving v = 2.4 m s−1, as required.

9 a 4.5 × 4 + m × (−6) = 0, giving m = 3 kg b 4.5 × 4 + m × −6 = (4.5 + m) × −2.4, giving m = 8 kg

6u = vx + 2vy

vx = 6u − 2vy

v’ = 20.7 m s–1 = 74.6 km h–1 14 Let the velocity of ball A when they meet be uA and the velocity of ball B when they meet uB So 2uA + 3uB = 5 × 3

So X has reduced the velocity from 4u to u, which is a 75% reduction.

Using v = u + at uB = 0 + 0.1 × 10 = 1 Using 1 u A =

v = 50 cm s–1

c 30 = 4v

15 − 3 =6 2

For A: u = 0, v = 6, a = a, t = 10 Using v = u + at 6 = 0 + 10a

v = 7.5 m s–1

a = 0.6 ms–2

12 Let start of platform = point A

Using s = ut +

End of platform = point B sA =

Car = point C A–B: s = 160, u = u, a = a, t = 5

sB =

A–C: s = 285, u = u, a = a, t = 10 Using s = ut +

1

For B: u = 0, v = uB, a = 0.1, t = 10

b Unchanged as the start and end balls are the same as in part a.

5 ⇒ v = 22 m s–1 2

10 000 × 22 + 1200 × 10 = 11 200 × v

11 a 50 = 1 × v (as momentum at the start = momentum at the end)

t 2

Initial momentum = final momentum

b vy = 2.5u, so vx = 6u − 2 × 2.5u = u.

s = (u + v)

22 = 10 + a × 5 ⇒ a = 2.4 m s–2

10 a m × 4u + 2m × u = m × vx + 2m × vy

s = 80, u = 10, v = v, t = 5

v = u + at

d 10.5 × −1 + 12 × v = 22.5 × −4, giving v = −6.63 m s−1 6mu = mvx + 2mvy

So for the bus:

80 = (10 + v) ×

c 4.5 × 4 + 18 × −6 = 22.5 × v, giving v = −4 i.e. 4 m s−1 in the direction of B

In the 5 seconds the car will travel an extra 10 × 5 = 50 m.

A–C: 285 = 10u + 50a

1 × 0.6 × 102 = 30 m 2 1 × 0.1 × 102 = 5 m 2

So distance between them at the start = 30 – 5 = 25 m

1 2 at 2

A–B: 160 = 5u + 12.5a ⇒ 320 = 10u + 25a

1 2 at 2

1

15 A moving to B

2

F = ma → 18 000 – 2000 = 25 000 a

2 – 1 35 = –25a so a =

–1.4 m s–2

a = 0.64 m s–2

and u = 35.5 m s–1

s = 50, u = 0, v = v, a = 0.64

As the train hits the car: v = u + at gives v = 35.5 – 1.4 × 10 = 21.5

v2 = u2 + 2as

So initial momentum = 21.5 × 20 000 = 430 000

v = 8 m s–1 when A meets B.

Conservation of momentum means that 430 000 = 22 000 V

At A joining to B

V = 19.5 m s–2

v2 = 0 + 2 × 0.64 × 50 = 64

25 000 × 8 = (25 000 + 15 000) v1 So v1 = 5 m s–1 velocity of both A and B joined.

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WORKED SOLUTIONS

3

AB moving to C: F = ma → 18 000 – 2500 = 40 000a a = 0.3875 m s–2 s = 50, u = 5, v = v2, a = 0.3875 (v2)2 = u2 + 2as (v2)2 = 25 + 2 × 0.3875 × 50 = 63.75 v2= 63.75 m s–1 when AB meets C. At AB joins to C: 40 000 × 63.75 = 55 000 v3 So v3 = 5.81 m s–1

Mathematics in life and work 1 160 000u = 80 000(−u) + 80 000v1 240 000u = 80 000v1 v1 = 3u as required. 2 80 000 × (3u) = 40 000(−3u) + 40 000v2 360 000u = 40 000v2 v2 = 9u 40 000 × (9u) = 20 000(−9u) + 20 000v3 540 000u = 20 000v3 v3 = 27u 3 160 000 × 140 = 140 000(vf ) + 20 000 × 27 × 140 140 000(vf ) = −53 200 000 vf = −380 km s−1, v 3 = 3780 km s−1

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4 Newton’s laws of motion

4 Newton’s laws of 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

Exercise 4.1A

1 a F = ma

20 = 4x

Write 800 g in SI units.

x = 5

F = 0.8 × 3.1 = 2.48 N

y = 10 + 5 = 15

b F = ma

2 x cos θ – y sin θ = y cos θ + x sin θ x cos θ – x sin θ = y cos θ +  ysin θ x(cos θ – sin θ) = y cos θ + y sin θ x=

m = F = 5200 = 1300 kg a 4 2 a W = mg = 8.4 × 10 = 84 N

y cos θ + y sin θ cos θ − sin θ

b W = mg

3 R(→):

735 = m × 10

F = T cos 26°

m = 735 = 73.5 kg 10 W = mg

R( ):

= 73.5 × 3.75 = 276 N

T cos 26° – F = 0

R + T sin 26° – 10 = 0

3 v = 22 m s−1, u = 4 m s−1, t = 15 s = u + at

R = 10 – T sin 26° Substitute into F = µR: T cos 26 = 1 (10 – T sin 26°) 10

a= v −u t

10T cos 26° = 10 – T sin 26°

= 22 − 4 = 1.2 m s−2 15

10T cos 26° + T sin 26° = 10

Resultant force, F = 1270 − 190 = 1080 N.

T(10 cos 26° + sin 26°) = 10

F = ma 1080 = m × 1.2

10 = 1.06 N 10cos26° + sin 26°

4 u = 4 m s−1, s = 55 m, t = 5 s a s = ut + 1 at2 2 55 = 4 × 5 + 12 × a × 52 55 = 20 + 25 a 2 25 35 = 2 a a = 2.8 m s−2 b v = u + at = 4 + 2.8 × 5 = 4 + 14 = 18 m s−1

1080 m= F = = 900 kg 1.2 a 4 R( ) R – 75g = 0 →

T=

R = 75g F = µR

= 1 × 75g = 25 × 10 = 250 N 3 Resultant force, F = (X − 250) N. a F = ma X − 250 = 75 × 0 X = 250 N

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WORKED SOLUTIONS

b F = ma

a = −0.968 75 m s−2

X − 250 = 75 × 2

u = 30 m s−1, v = 0 m s−1, a = −0.968 75 m s−2

X = 400 N

2 2 s= v −u 2a

0 m s−1,

c u =

s = 2 m, t = 0.5 s 1 s = ut + 2 at2 2 = 0 × 0.5 + 12 × a × 0.52 = 0.125a a = 16 m s−2

0 2 − 30 2 2 × − 0.96875

=

= 465 m

7 5900 – 1100 = 1600 a a = 3 m s–2

F = ma

v = u + at

X − 250 = 75 × 16

v = 0 + 3(11) = 33 m s–1

X = 1450 N

Let braking force be B N –B – 1100 = 1600 a

5 R – 1500g = 0 R = 1500g

s = 240 m, u = 33 m s–1, v = 0 m s–1

F = mR

a=

= 0.035 × 1500g = 52.5 × 10 = 525 N Resultant force, F = 1125 – 525 = 600 N

v 2 − u 2 0 2 − 332 = 2s 2(240)

a = –2.27 m s–2 –B – 1100 = 1600(–2.27)

F = ma

B = 2530 N

600 = 1500a

8 a D – F = ma

a = 600 ÷ 1500 = 0.4 ms–2 u=

4 m s−1,

s = 385 m, a =

0.4 m s−2

R = mg F = mR = m mg

v2 = u2 + 2as = 42 + 2 × 0.4 × 385

D – m mg = ma

= 324

v = 18 m s−1

= u + at 18 = 4 + 0.4t

14 = 0.4t

t = 35 s 6  u = 15 a=    =

4

3 mg = ma 8 2 89 – mg = ma 3 3 2 54 – mg = 89 – mg 8 3 54 –

1296 – 90m = 2136 – 160m

m = 12 kg

ms–1,

2

v −u 2s

v = 30

ms–1,

s = 400 m b a =

2

2

54 − 3 mg 54 − 3 × 12 × 10 8 8 = m 12

a = 0.75 m s–2 2

30 − 15 2 × 400

9 a T cos a – F = ma

= 0.84375 m s–2

Let the resistance to motion = X N

Resultant force, F = (1240 – X) N. F = ma

20 T – F = 0.2 m 29 20 21 When cos a = , sin a = 29 29

T sin a + R = mg 21 T + R = 10m 29

1240 – X = 960 × 0.84375

F = mR

= 810

X = 430 N

(

20 17 10m – 21 T T – 0.2m = 29 29 40

F = ma

800T – 232m = 4930m – 357T

−930 = 960a

1157T = 5162m

Resultant force, F = (−430 − 500) = −930 N.

) 37

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4 Newton’s laws of motion

13T = 58m

1 25 = 0 × t + 2 × 9.75 × t2 = 4.875t2

T sin a + R = mg

20 T + 180 = 10m 29

20T + 5220 = 290m

4T + 1044 = 58m

13T = 4T + 1044

t = 2.265 s Difference = 0.028 s. 2 Resultant force, F = (0.35g − 2.8) N. F = ma

T = 116 N

0.35g − 2.8 = 0.35a

b m = 13 × 116 = 26 kg 58

a = 2 m s−2

10 a When sin q =

u = 3.7 m s−1, s = 1.9 m, a = 2 m s−2

3 4 , cos q = 5 5

v2 = u2 + 2as

60 cos q – F = 20a

R + 60 sin q  = 20g

60 cos q – 20a = 0.25(20g – 60 sin q)

60

48 – 20a = 41

= 21.29

()

4.61 = 3.7 + 2t

()

4 – 20a = 0.25 20g – 60 3  5 5  

t = 0.457 s 3 Resultant force when lowered, F = (Mg − 12) N. F = ma

s = 1440 m, u = 4

m s–1,

a = 0.35

Mg − 12 = 0.4M

m s–2

M × 10 − 0.4M = 12

1 2 at 2

1 1440 = 4t + (0.35)t2 2

1440 = 4t +

9.6M = 12 M = 1.25 Let the compression when raised = C N.

7 2 t 40

Resultant force when raised, F = (C − 1.25g) N.

7t2

57 600 = 160t +

7t2 + 160t = 57 600

F = ma C − 1.25g = 1.25 × 0.4

b 7t2 + 160t – 57 600 = 0

= 0.5

C = 13 N

(7t + 720)(t – 80) = 0

4 Resultant force, F = (250 − 20g) N.

t = 80

F = ma

Exercise 4.2A

250 − 20g = 20a

1 s = 25 m, u =

0 m s−1

Without air resistance, a = s = ut + 1 at2 2 25 = 0 × t + 12 × 10 × t2

= u + at

a = 0.35 m s–2

= 3.72 + 2 × 2 × 1.9

v = 4.61 m s−1

F = mR

s = ut +

250 − 200 = 20a a = 2.5 m s−2

10 m s−2

u = 2 m s−1, v = 5 m s−1, a = 2.5 m s−2 v2 = u2 + 2as 2 2 s= v −u 2a

= 5t 2

t = 2.236 s With air resistance, 0.12g − 0.03 = 0.12a. a = 9.75 m s−2 1 s = ut + 2 at2

2 2 = 5 − 2 2 × 2.5

= 4.2 m

5 a u = 0 m s−1, v = 4 m, s = 2.5 m v2 = u2 + 2as

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WORKED SOLUTIONS

2 2 a = v − u 2s

2 2 = 4 −0 2 × 2.5

= 3.2 m s−2

To find the mass of the weight:

W = mg

3 = m × 10

m = 0.3 kg

Resultant force, F = (T − 3) N.

F = ma T − 3 = 0.3 × 3.2 T = 3.96 N b u = 4 m s−1, a = −10 m s−2, v = 0 m s−1 2 2 s = v − u 2a 2 2 = 0 −4 2 × −10

b s = –18 m, u = 13 ms–1, a = –10 ms–2 v2 = u2 + 2as v2 = 132 + 2(–10)(–18) v2 = 529 v = –23 ms–1

Speed = 23 ms–1

7 a 50g – 10 = 50a a = 9.8 ms–2 s = 10 m, u = 0 ms–1, a = 9.8 m s–2 v2 = u2 + 2as v2 = 02 + 2(9.8)(10) v2 = 196 v = 14 m s–1

50g – 2500 = 50a

a = –40 m s–2 v = 0 m s–1, u = 14 m s–1, a = –40 m s–2 s =

= 0.8 m Maximum height = 2.5 + 0.8 = 3.3 m.

c From maximum height to when it hits the floor, u = 0 m s−1, a = 10 m s−2, s = 3.3 m. v2 = u2 + 2as

= 02 + 2 × 10 × 3.3

= 66

v = 8.12 m s−1 6 a s = 2 m, u = 7 m s–1, a = –10 m s–2 v2

=

u2

+ 2as

v 2 − u 2 0 2 − 14 2 = 2a 2(−40)

s = –2.45 m

Maximum distance = 2.45 m

b 60g – 10 = 60a 59 m s–2 6

a =

v2 = 02 + 2 v2 =

v2 = 9

a = –

60g – 2500 = 60a

v = 3 m s–1

3 = 7 – 10t

t = 0.4 s

420 – 28g = 28a

a = 5 m s–2 s = 16 m, u = 3 m s–1, a = 5 m s–2 v2 = u2 + 2as v2 = 32 + 2(5)(16) v2 = 169 v = 13 ms–1 v = u + at

13 = 3 + 5t

( 596 )(10)

590 3

v2 = 72 + 2(–10)(2)

v = u + at

s =

95 m s–2 3

590 2 v 2 − u2 0 − 3 = 2a 2 − 95 3

( )

s = –3.11 m

Distance = 3.11 – 2.45 = 0.66 m

8 s = 45 m, u = 0 m s–1, a = 10 m s–2 v2 = u2 + 2as v2 = 02 + 2(10)(45) v2 = 900 v = 30 m s–1 Let mass of lighter rock be M kg

t = 2 s

Lighter rock

Mg – 3250 = Ma

Total time = 0.4 + 2 = 2.4 s

4

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4 Newton’s laws of motion

Mg − 3250 M

u = 30 m s–1, v = 0 m s–1, a =

v −u 5−0 −2 t = 2.5 = 2 m s

a= Mg − 3250 m s–2 M

v 2 − u2 45M 0 2 − 30 2 s= = = 2a Mg − 3250  325 − M  2 M  

a=

R( ): 3g sin θ − F = 3a F = 3g × 3 − 3 × 2 5 = 12 N R( ):

(M + 6)g – 3250 = (M + 6)a

R − 3g cos θ = 0 R = 3g × 4 = 24 N 5 F = µR

u = 30 m s–1, v = 0 m s–1, a = s=

(M + 6)g − 3250 m s–2 M +6

v 2 − u2 45(M + 6) 0 2 − 30 2 = = 2a  ( M + 6 ) g − 3250  325 − (M + 6) 2  M +6 

45(M + 6) 45M = + 0.25 325 − (M + 6) 325 − M 45(M + 6)(325 – M)

µ = F = 12 = 1 R 24 2 4 a R( ): →

(M + 6)g − 3250 M +6

R − 35 3  sin 30° − 5g cos 30° = 0

R = 35  3 sin 30° + 5g cos 30° = 73.6 N

F = µR = 1 × 73.6 = 21.3 N 2 3

= 45M(319 – M) + 0.25(325 – M) (319 – M)

M 2 – 644M – 247 325 = 0 644 ± (−644)2 − 4(1)(−247325) M= 2(1)

R( ):

a=

Heavier rock

35 3  cos 30° − F − 5g sin 30° = 5a 5a = 35  3 cos 30° − 21.3 − 5g sin 30°

M = 914 kg

= 6.25

a = 1.25 m s−2

a = 1.25 m s−2, u = 0 m s−1 and s = 6.3 m

Substitute in s = ut + 1 at2: 2 1 2 6.3 = 0 + 2 × 1.25t t 2 = 10.1

a = g sin 26° – 0.35g cos 26°

t = 3.17 s

a = 1.24 m s–2

b Substitute in v = u + at:

Exercise 4.3A →

1 a R( ) 0.9g sin 26° – F = 0.9a R( ) R = 0.9g cos 26° →

F = mR

0.9g sin 26° – 0.9a = m(0.9g cos 26°)

b 1.24

m s–2

2 26 – 4g sin q  = 4 × 0.5 sin q  =

24 3 = 4g 5

3 Let angle of slope be θ. sin θ  = 3 5

v = 0 + 1.25 × 3.17

= 3.96 m s−1

5 a  Let the angle between the rope and the ground be θ. 44 sin θ = 125

2 2 cos θ = 125 − 44 = 117 125 125

R(→):

To find acceleration, use v = u + at.

u = 0 m s−1, v = 5 m s−1, t = 2.5 s

10 cos θ − F = 0 F = 10 × 117 125 = 9.36 N

cos θ  =

52 − 32 = 4 5 5

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4

WORKED SOLUTIONS

a = −7.80 m s−2

R( ):

R + 10 sin θ − 5g = 0

a = −7.80 m s−2, u = 4.40 m s−1 and v = 0 m s−1

R = 5g − 10 × 44 = 46.48 N 125

Substitute in v 2 = u2 + 2as:

µ = F = 9.36 = 0.201 R 46.48

2 2 s= v −u 2a

2 s = 0 − 4.40 2 × −7.80

s = 1.24 m

b R( ):

R = 5g − 15 × 44 = 44.72 N 125 F = µR = 0.201 × 44.72 = 9.01 N

R(→):

15 cos θ − F = 5a

5a = 15 × 117 − 9.01 = 5.03 125

a=

7 a R( ): →

R + 15 sin θ − 5g = 0

R − 0.5g cos 40° = 0

R = 0.5g cos 40°

F = µR

= 0.6 × 0.5g cos 40° = 2.30 N

1.01 m s−2

b R( ):

6 a When inclination is 25°:

0.5g sin 40° − F = 0.5a 0.5g sin 40° − 2.30 = 0.5a

R( ):

R − 2g cos 25° = 0

0.5a = 0.916

R = 2g cos 25°

a = 1.83 m s−2

R( ):

Since a is positive, motion will occur.

F − 2g sin 25° = 0

F = 2g sin 25°

µ = F = 2g sin 25° = 0.466 R 2g cos 25°

8 a With 4 N force: R − 4 sin 23° − 2g cos 23° = 0

R = 4 sin 23° + 2g cos 23°

When inclination is 20°:

R( ):

R( ):

2g sin 23° − 4 cos 23° − F = 0

R − 2g cos 20° = 0

F = 2g sin 23° − 4 cos 23°

R = 2g cos 20°

F = µR

2g sin 23° − 4 cos 23° µ= F = = 0.207 R 4 sin 23° + 2g cos 23°

With X  N force:

R( ):

R − X sin 23° − 2g cos 23° = 0

R = X sin 23° + 2g cos 23°

R( ):

X cos 23° − 2g sin 23° − F = 0

F = X cos 23° − 2g sin 23°

= µR

X cos 23° − 2g sin 23° = 0.207(X sin 23° + 2g cos 23°)

20 − F − 2g sin 20° = 2a

2a = 20 − 8.76 − 2g sin 20° = 4.40

a = 2.20 m s−2

R( ):

b With 20 N force, t = 2 s, a = 2.20 m s−2 and u = 0 m s−1.

Substitute in v = u + at:

v = 0 + 2.20 × 2

v = 4.40 m s−1

Without 20 N force, F is still 8.76 N.

R( ):

−F − 2g sin 20° = 2a

2a = −8.76 − 2g sin 20° = −15.6

1

= 0.466 × 2g cos 20° = 8.76 N

R( ):

= 0.207X sin 23° + 0.207 × 2g cos 23° X cos 23° − 0.207X sin 23° = 2g sin 23° + 0.207 × 2g cos 23°  X =

2g sin 23° + 0.207 × 2g cos 23° = 13.8 N cos 23° − 0.207 × sin 23°

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4 Newton’s laws of motion

s = 8 m, u = 0 ms–1, a = 0.363 ms–2

b R( ): R − 2g cos 23° = 0

R = 2g cos 23°

s = ut +

1 2 at 2

R( ):

1 (0.363)t2 2

X − 2g sin 23° − F = 2a

t2 = 44.08

13.8 − 2g sin 23° − 0.207 × 2g cos 23° = 2a

t = 6.64 s

2a = 2.17

a = 1.11 m s−2

9 When sin a = When sin b =

8=0+

Total time = 6.64 +

8 = 9.31 s 3

3 4 , cos a = 5 5

Exercise 4.4A

7 24 , cos b = 25 25

1 Resultant force on H = (3 − T) N. Substitute into F = ma:

Let acceleration on second plane be a m s–2.

3 − T = 0.35a

2g sin a – F = 2(10a) = 20a

Resultant force on G = T N.

R = 2g cos a

Substitute into F = ma:

2g sin b – F = 2a

T = 0.15a

R = 2g cos b

() ()

( ) ( )

2g 3 − 20a 2g 7 − 2a 5 25 F m= = = R 4 2g 2g 24 5 25 6g − 100a 14g − 50a = 8 48 48(6g – 100a) = 8(14g – 50a) 2 a = m s–2 5

() () ()

2(10) 3 − 20 2 5 5 1 m= = 4 4 2(10) 5 10 a s = 8 m, u = 6 m s–1, v = 0 m s–1 a =

v 2 − u 2 02 − 62 = = –2.25 m s–2 2(8) 2s

a = 6 m s−2 T = 0.15a = 0.15 × 6 = 0.9 N 2 Resultant force on P = (X − T − 4mg) N. Substitute into F = ma: X − T − 4mg = 4m × 4 g 7 Resultant force on Q = (T − 3mg) N. T − 3mg = 3m × 4 g 7

X − 7mg = 7m × 4 g 7

–mg sin 7.5 – F = –2.25m

= 11mg N

–mg sin 7.5 + 2.25 m = m(mg cos 7.5°)

− g sin 7.5° + 2.25 m= = 0.095 g cos7.5° b Time taken to go up. 2s 2(8) 8 = = s t = u+v 6+0 3

3 Resultant force on car = (1710 − T − 300k) N. Substitute into F = ma: 1710 − T − 300k = 300a Substitute into F = ma: T − 840k = 840a

mg sin 7.5° – m mg cos 7.5° = ma

1710 = 1140(k + a)

k + a = 1710 1140

a = 0.363

2

Add 1 and 2 : 1710 − 1140k = 1140a

ms–2

1

Resultant force on caravan = (T − 840k) N.

mg sin 7.5° – F = ma g sin 7.5° – 0.095g cos 7.5° = a

2

From 2 , T = 3mg + 3m × 4 g = 33 mg. 7 7 Add 1 and 2 :

X = 7mg + 4mg

1

Substitute into F = ma:

Let mass of block be m kg.

F = mR

2

3 = 0.5a

R = mg cos 7.5

1

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WORKED SOLUTIONS

The lighter particle does not collide with the heavier particle since it only travels 0.21 m after the string goes slack.

From 2 , T = 840(k + a) N. T = 840 × 1710 1140

b 0.5 − 0.21 = 0.29 m

= 1260 N 4 a  Let the 7 kg particle be A and the 4 kg particle be B.

5 Resultant force on X = (100 − T − 5g) N. Substitute into F = ma:

R( ) for B:

100 − T − 5g = 5a

RB − 7g = 0

Resultant force on Y = (T − 3g) N.

RB = 7g

Substitute into F = ma:

FB = µRB = 7 × 7g 20 R(→) for B:

T − 3g = 3a

Add: 100 − 8g = 8a

35 − FB − T = 7a

35 − 7 × 7g − T = 7a 20

10.5 − T = 7a

R( ) for A:

v2 = u2 + 2as

RA − 4g = 0

RA = 4g

2 2 s= v −u 2a

FA = µRA = 4 × 4g 20

R(→) for A:

T − FA = 4a

T − 4 × 4g = 4a 20

T − 8 = 4a

a = 2.5 m s−2 Whilst connected:

1

u = 0 m s−1, a = 2.5 m s−2, v = 9 m s−1

2 2 = 9 − 0 2 × 2.5

= 16.2 m When cord snaps: u = 9 m s−1, a = −10 m s−2, s = −16.2 m v2 = u2 + 2as 2

1 + 2

10.5 − 8 = 11a

a = 0.227 m s−2

= 92 + 2 × −10 × −16.2

= 405

v = −20.1 m s−1 v = u + at 0 m s−1,

0.227 m s−2.

t= v −u a

Whilst taut, u =

Substitute in v = u + at:

v = 0 + 0.227 × 4

= −20.1 − 9 −10

= 0.909 m s−1

= 2.91 s

Whilst slack:

R(→) for A:

v = u + at

−8 = 4a

a = −2 m s−2

a = v − u t

u = 0.909 m s−1, v = 0 m s−1, a = −2 m s−2

Substitute in 2

4

2

s= v −u 2a

2 = 0 − 0.909 2 × −2

= 0.21 m

t = 4 s, a =

6 a u = 12 m s−1, v = 32 m s−1, t = 8 s

v2

=

u2

+ 2as:

= 32 − 12 8 = 2.5 m s−2 Resultant force on minibus = (3000 − T − 475) = (2525 − T) N. Substitute into F = ma: 2525 − T = 950 × 2.5 = 2375

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4 Newton’s laws of motion

8 208 – 6g – T = 6a

Let mass of trailer = M kg.

Resultant force on trailer = (T − 25) N.

T – 14g = 14a

Substitute into F = ma:

T − 25 = 2.5M

208 – 20g = 20a a = 0.4 m s–2

t = 5 s, u = 0 m s–1, a = 0.4 m s–2

v = u + at

2500 = 2.5M + 2375

M = 50

v = 0 + 0.4(5) = 2 m s–1

b Resultant force on trailer when uncoupled =  −25 N.

u = 2 m s–1, v = 0 m s–1, a = –10 m s–2

0 = 2 + (–10)t

v = u + at

Substitute into F = ma:

−25 = 50a a =

t = 0.2 s

−0.5 m s−2

9 a Let the car be C and the caravan be V.

u = 32 m s−1, v = 0 m s−1, a = −0.5 m s−2

On the 7° plane

R( ) RC = 900g cos 7°

R( ) RV = 1200g cos 7°

= 0 − 32 2 × −0.5

Since F = mR, FC = m(900g cos 7°) and FV = m(1200g cos 7°)

= 1024 m

R( ) D – T – FC – 900g sin 7° = 900a

R( ) T – FV – 1200g sin 7° = 1200a

2

2

7 a Let the driving force = D N.

Resultant force on car = (D − T − X) N.

Substitute into F = ma:

→ →

2 2 s = v − u 2a

→ →

v2 = u2 + 2as

D – FC – FV – 2100g sin 7° = 2100a D – m(900g cos 7°) – m(1200g cos 7°) – 2100g sin 7° = 2100(0.65)

D − T − X = Ma

D = m(2100g cos 7°) + 2100g sin 7° + 1365

Resultant force on caravan = (T − 400) N.

Substitute into F = ma:

D = m(2100g cos 10°) + 2100g sin 10° + 315

T − 400 = 750a

D − 400 − X = (M + 750)a

Substitute D = 1450 N and a = 0.6 m s−2:

1450 − 400 − X = (M + 750) × 0.6

1050 − X = 0.6M + 450

600 = X + 0.6M

Substitute D = 2450 N and a = 1.4 m s−2:

2450 − 400 − X = (M + 750) × 1.4

2050 − X = 1.4M + 1050

1000 = X + 1.4M

Subtract 1 from 2 .

400 = 0.8M

M = 500 b Substitute M = 500 into 1 :

600 = X + 0.6 × 500

600 = X + 300

X = 300

1

On the 10° plane

m(2100g cos 7°) + 2100g sin 7° + 1365 = m(2100g cos 10°) + 2100g sin 10° + 315

m(2100g cos 7°) – m(2100g cos 10°) = 2100g sin 10° + 315 – 2100g sin 7° – 1365

2100g m(cos 7° – cos 10°) = 2100g(sin 10° – sin 7°) – 1050

m=

2100g(sin10° − sin 7°) − 1050 = 0.230 2100g(cos7° − cos10°)

b D = (0.230)(2100g cos 7°) + 2100g sin 7° + 1365 D = 8720 N 2

10 a s = 2790 m, t = 180 s, v = 20 m s–1 s = vt –

1 2 at 2

2790 = 20 × 180 –

1 a(180)2 2

a = 0.05 m s–2

Since mass of B is M kg, mass of A is and mass of C is

5 M kg 3

2 M kg. 3

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4

WORKED SOLUTIONS

R(↑) RA =

Since F = mR, FA = and FC =

Exercise 4.4B

5 2 Mg, RB = Mg and RC = Mg 3 3

( )

1 5 1 Mg , FB = (Mg) 10 3 10

(

1 2 Mg 10 3

)

1 a Resultant force = (R − 55g) N. F = ma R − 55g = 55 × 2.4 R = 55g + 55 × 2.4

Let tension between A and B be T1 N and tension between B and C be T2 N.

= 55(10 + 2.4) = 682 N b Resultant force = (55g − R) N.

5 M(0.05) 3

R(→) D – T1 – FA =

R(→) T1 – T2 – FB = M(0.05)

2 R(→) T2 – FC = M(0.05) 3

R = 55g − 55 × 1.6

D – FA – FB – FC =

8400 –

F = ma

= 55(10 − 1.6)

10 M(0.05) 3

( )

= 462 N

(

1 1 Mg = M 3 6

10 1 8400 – M = M 6 3

7 M = 8400 2

M = 2400 5 M(0.05) 3

( )

1 5 5 Mg = M(0.05) 10 3 3

8400 – T1 –

8400 – T1 – 4000 = 200

T1 = 4200 N 2 c R(→) –FC = Ma 3

( ) 1 2 2 – Mg ) = Ma 10 ( 3 3

1 2 2 Mg = Ma – 10 3 3

–1600 = 1600a

a = –1 m s–2 a = –1 m s–2, u = 20 m s–1, v = 0 m s–1 0 = 20 – 1t

= 517 N 2 a  For the dog and the floor of the lift, resultant force = (25g − R) N.

= 55(10 − 0.6)

F = ma

b D – T1 – FA =

v = u + at

F = ma R = 55g − 55 × 0.6

8400 –

c Resultant force = (R − 55g) N. R − 55g = 55 × −0.6

)

1 5 1 1 2 Mg – (Mg) – Mg = 10 3 10 10 3

10 3 M(0.05)

55g − R = 55 × 1.6

25g − R = 25 × 1.8

R = 25g − 25 × 1.8 = 25(10 − 1.8) = 205 N b  For the system as a whole, resultant force = (165g − T) N. F = ma

165g − T = 165 × 1.8

T = 165g − 165 × 1.8 = 165(10 − 1.8) = 1353 N = 1350 N to 3 s.f. 3 Total weight = 75g + 65g + 550g = 690g N. Resultant force = (7616 − 690g) N. F = ma 7616 − 690g = 690a 716 = 690a a = 1.04 m s−2

t = 20 s

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4 Newton’s laws of motion

It is assumed that the lift is stationary when the adults enter. u = 0 m s−1, s = 24 m, a = 1.04 m s−2 s = ut + 12 at2 24 = 0 × t + 12 × 1.04 × t2 = 0.519t2

(9M + 760)(M – 40) = 0

M = 40 a =

1130 − 20(40) = 0.75 2(40) + 360

b Mass of Liu = 80 kg, mass of Zhao = 120 kg

Total mass = 80 + 120 + 360 = 560 kg

t 2 = 46.26

T – 560g = 560 × 0.6

t = 46.26 = 6.80 s

T = 5936 N = 5900 N

4 a Total mass = 5 × 66 + 280 = 610 kg T – 610g = 610 × 0.2

7 Total weight with a 160 kg load = 400g + 160g = 560g N. Resultant force = (T − 560g) N.

T = 6222 N = 6200 N

F = ma

b Mass of sixth person = 6 × 70 – 5 × 66 = 90 kg R – 90g = 90 × 0.15

T − 560g = 560 × 0.75 T = 560g + 560 × 0.75

R = 913.5 N = 910 N

= 560(10 + 0.75)

c Total mass = 6 × 70 + 280 = 700 kg

= 6020 N

T – 700g = 700 × 0.15

Total weight with a 200 kg load = 400g + 200g = 600g N.

T = 7105 N = 7100 N 5 Let mass of girl be M kg, mass of father be 2M kg and mass of lift be (2M + 100) kg (5M + 100)g – 2660 = (5M + 100) ×

1 2

50M + 1000 – 2660 = 2.5M + 50 M = 36 2M = 72 72g – R = 72 ×

1 2

R = 684 N 6 a  Let mass of Liu be 2M kg and mass of Zhao be 3M kg

4730 – (2M + 360)g = (2M + 360)a

1130 – 20M = (2M + 360)a

a =

Resultant force = (6020 − 600g) N. F = ma 6020 − 600g = 600a 20 = 600a a = 1 m s−2 30 u = 0 m s−1, t = 30 s, a = 1 m s−2 30 s = ut + 12 at2   = 0 × 30 + 1 × 1 × 302 2 30   = 15 m v = u + at   = 0 × 1 × 30 30   = 1 m s−1

1130 − 20M 2M + 360

3Mg – 1140 = 3M(a – 0.25)

30M – 1140 = 3Ma – 0.75M 30.75M − 1140 a = 3M

Once the cable snaps, the lift will just be affected by gravity. u = 1 m s−1, s = −15 m, a = −10 m s−2 v2 = u2 + 2as

1130 − 20M 30.75M − 1140 = 2M + 360 3M

= 12 + 2 × −15 × −10

= 301

3M(1130 – 20M) = (30.75M – 1140)(2M + 360)

v = 301 = −17.3 m s−1

60M2

61.5M2

3390M – – 410 400

0 = 121.5M2 + 5400M – 410 400

9M2 + 400M – 30 400 = 0

=

– 2280M + 11 070M

−17.3 = 1 − 10t 10t = 18.3

t = 1.83 s

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WORKED SOLUTIONS

8 a i and ii solved together.

Let mass of Margarita = m kg.

Resultant force on Margarita = (960 − mg) N.

960 − mg = ma

960 = mg + ma

Resultant force on whole system = (4320 − 280g − mg) N.

1

F = ma

4320 − 280g − mg = (280 + m)a

4320 − 280g − 280a = mg + ma

Let 1 = 2 .

960 = 4320 − 280g − 280a

280a = 560

mG = 700 g +a 700 10 + 2.5

= 56 kg

From 2 :

mJ =

900 g +a

=

900 10 + 2.5

= 72 kg

= 280a + ma 2

Exercise 4.5A 1 a  Reaction force is equal and opposite to the weight of G.

a = 2 m s−2

From 1 :

=

F = ma

R = 3mg

b For G: T = 3ma

From 1 :

m = 960 g +a = 960 10 + 2 = 80 kg b i, ii and iii solved together

Let weight of Guillaume = mGg kg and weight of Josef = mJg kg.

Resultant force on Guillaume = (700 − mGg) N.

For H: 2mg − T = 2ma

2mg = 5ma

a = 2 g m s−2 5 c T = 3ma = 3m × 2 g 5 6 = mg N 5 2 a For the 4 kg particle: 4g − T = 4a.

F = ma

For the 3 kg particle: T − 3g = 3a.

700 − mGg = mGa

700 = mGg + mGa

Resultant force on Josef = (900 − mJg) N.

1

g = 7a a = 1 g m s−2 7

F = ma

b T = 3g + 3a

900 − mJg = mJa

900 = mJg + mJa

Resultant force on whole system = (5100 − 280g − mGg − mJg) N.

= 3g + 3 × 1 g 7 24 g N = 7

2

F = ma

5100 − 280g − mGg − mJg = (280 + mG + mJ)a

5100 − 280g − 280a = mGg + mGa + mJg + mJa

Substituting from equations 1 and 2 :

5100 − 280g − 280a = 700 + 900

280a = 700

4

= 280a + mGa + mJa

c u = 0 m s−1, a = 1 g m s−2, t = 1.4 s 7 s = ut + 1 at2 2 = 0 × 1.4 + 1 × 1 g × 1.42 2 7 = 1.4 m Since one particle will have travelled 1.4 m upwards and the other 1.4 m downwards, the difference in height = 2.8 m

a = 2.5 m s−2

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4 Newton’s laws of motion

3 a Let the mass of P2 = M kg. For P2: Mg − T = Ma = M × 1 g 5 For P1: T − 4g = 4a = 4 × 1 g 5 Add: (M − 4)g = 1 (M + 4)g 5 M − 4 = 1 (M + 4) 5 5M − 20 = M + 4

For B: 7g − T = 7a.

7g = 10a

a = 7 m s−2 This acceleration will remain until B hits the ground.

b The force exerted on the pulley = 2T. T = 4g + 4a = 4g + 4 × 1 g = 48 N 5 Force exerted on pulley = 48 N × 2 = 96 N.

u = 0 m s−1, a = 7 m s−2, s = 2 m v2 = u2 + 2as

4 a R(↑ ) for J:

R − 2g = 0

R = 2g

F = µR = 0.3 × 2g = 0.6g N = 6 N

T − F = 2a

T − 0.6g = 2a

R(↓) for K: 3.2g − T = 3.2a 1 + 2 :

3.2g − 0.6g = 5.2a

2.6g = 5.2a

a = 1 g = 5 m s−2 2

c T − 0.6g = 2 × 5

T = 6 + 10

= 02 + 2 × 7 × 2

= 28

v = 2 7 m s−1

= u + at

t = v − u a

b R(→) for J:

P2 has a mass of 6 kg.

= 5 m

5 For A: T = 3a.

M = 6

2 2 s= v −u 2a 0 30 − = 2 × −3

4M = 24

Substitute in v2 = u2 + 2as:

1 2

= 2 7 − 0 7

= 0.756 s Once B has hit the ground, the string will go slack and A will continue at a constant speed for the remaining 3 m.

u = 2 7 m s−1, a = 0 m s−2, s = 3 m s = ut + 1 at2 2 3 = 2 7t + 0 t = 0.567 s Total time = 0.756 + 0.567 = 1.32 s. 6 Let x > y.

= 16 The tension is 16 N.

d Whilst taut, s = 3 m, a = 5 m s−2, u = 0 m s−1.

For the x kg particle: xg − T = xa. For the y kg particle: T − yg = ya.

Substitute in v2 = u2 + 2as:

v2

xg − yg = (x + y)a

= 30

v = 30 = 5.48 m s−1

Whilst slack, for J:

Substitute for a in T = ya + yg:

−0.6g = 2a

 g (x − y ) T = y + yg  x + y 

=0+2×5×3

−3 m s−2

a=

v = 0 m s−1, a = −3 m s−2, u = 30  m s−1

Hence a =

=

g (x − y ) . x+y

yg ( x − y ) + yg ( x + y ) x+y

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4

WORKED SOLUTIONS

=

xyg − gy 2 + xyg + g y 2 x+y

=

2xyg  N x+y

b T = 0.7 × 4.17 + 0.175g

This formula is symmetrical in x and y, so the result will also be true if y > x. 1 7 For the ball: 0.5g − T = 0.5a. Resolving vertically for the flowerpot: R – 0.3g = 0.

Hence using F = µR, F = 2 × 0.3g = 2 N. 3 Resolving horizontally for the flowerpot:

= 4.67 N

c Whilst taut, t = 3 s, a = 4.17 m s−2, u = 0 m s−1. Substitute in s = ut + 1 at2: 2 s = 0 + 1 × 4.17 × 32 2 = 18.75 m

2

T − 2 = 0.3a

X travels 18.8 m, correct to 3 s.f., before Y hits the ground.

d Whilst taut, t = 3 s, a = 4.17 m s−2, u = 0 m s−1.

Substitute in v = u + at:

0.5g − 2 = 0.8a

v = 0 + 4.17 × 3 = 12.5 m s−1

Whilst slack:

R( ) for X:

−0.175g = 0.7a a = − 1 g = −2.5 m s−2 4 u = 12.5 m s−1, a = −2.5 m s−2, v = 0 m s−1

= 0 + 3.75 × 0.2

Substitute in v2 = u2 + 2as:

= 0.75 m s−1

2 2 s= v −u 2a

a = 3.75 m s−2 The ball and the flowerpot will both move with an acceleration of 3.75 m s−2 until the string snaps, at which point the string will become slack. u = 0 m s−1, t = 0.2 s, a = 3.75 m s−2 v = u + at

When the string becomes slack, the flowerpot will be slowed down by the 2 N frictional force. −2 = 0.3a a = −6 2  m s−2 3

0.625g = 1.5a a = 5 g = 4.17 m s−2 12

2 = 0 − 12.5 2 × −2.5

= 31.25 m

X travels a further 31.3 m, correct to 3 s.f., before it comes to instantaneous rest.

u = 0.75 m s−1, v = 0 m, a = −6 2  m s−2 3 v = u + at

9 Let the mass of P = m kg and let the mass of Q = (m + 3) kg.

t= v −u a

Since Q is heavier, Q will fall and P will rise. For P: 45.5 − mg = ma

= 0 − 0.275 −6 3

For Q: (m + 3)g – 45.5 = (m + 3)a Add:

= 0.1125 s

(m + 3)g − mg = (2m + 3)a

Total time = 0.2 + 0.1125 = 0.3125 s = 0.313 s correct to 3 s.f.

mg + 3g − mg = (2m + 3)a 3g = (2m + 3)a

8 a Let the angle of slope be θ.

3g 2m + 3

R(  ) for X:

a=

T − 0.7g sin θ = 0.7a T − 1 × 0.7g = 0.7a 4

Also, 45.5 = mg + ma

T − 0.175g = 0.7a

R(↓) for Y:

0.8g − T = 0.8a

1 + 2 :

1 2

Substitute a =

3g into 45.5 = mg + ma: 2m + 3

45.5 = mg + m

3g 2m + 3

45.5(2m + 3) = mg(2m + 3) + 3mg

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4 Newton’s laws of motion

Exam-style questions

91m + 136.5 = 20m2 + 30m + 30m 20m2 − 31m – 136.5 = 0 m=

31 ±

( −31)2 − 4 × 20 × −136.5 2 × 20

1 a R(↑): R – Mg = 0 R = Mg

= 3.5 or −1.95

F = µR = 0.4Mg

m cannot be negative, so is equal to 3.5.

b R(→):

The mass of P is 3.5 kg and the mass of Q is 6.5 kg.

24 – F = M × 2

24 – 0.4Mg = 2M

a=

3g 3g = 3 m s−2 = 2m + 3 2 × 3.5 + 3

c 24 = 2M + 0.4Mg M(2 + 4) = 24

10 a R( ) for A:

M = 4

R = 1.8g cos 30°

F = µR

= 3 × 1.8g cos 30° = 0.3g 9 R( ) for A:

T − F − 1.8g sin 30° = 1.8a

T − 0.3g − 0.9g = 1.8a

T − 1.2g = 1.8a

R(↓) for B:

5.2g − T = 5.2a

2 a R( ): →

R – mg cos 30° = 0 R = 50 × 10 × cos 30° = 433 N

R − 1.8g cos 30° = 0

b R( ):

1 2

1 + 2 :

4g = 7a

a = 4 g m s−2 7

The tension in the string is 22.3 N, correct to 3 s.f. c Whilst taut, t = 2 s, a = 4 g m s−2, u = 0 m s−1. 7 Substitute in v = u + at: v = 0 + 4 g × 2 = 11.4 m s−1 7 Whilst slack: R( ) for A:

−1.2g = 1.8a a = − 2 g m s−2 3 u = 11.4 m s−1, a = − 2 g m s−2, v = 0 m s−1 3 Substitute in v = u + at:

c u = 0 m s–1, s = 15 m, a = 3.56 m s–2 s = ut + 1 at2 2 15 = 0 + 1 × 3.56t2 2 t =

b T − 1.2g = 1.8 × 4 g 7 T = 22.3 N

mg  sin 30° – F = ma 500 sin 30° – 1 × 433 = 50a 6 a = 3.56 m s–2

t= v −u a = 0 − 211.4 −3g

= 1.71 s

Total time = 2 + 1.71 = 3.71 s

1 2

15 = 2.90 s × 3.56

3 a For B: resultant force = (3g − T) N. F = ma

3g − T = 3a

For A: resultant force = T N.

F = ma T = 5a

3g = 8a a = 3 g m s−2 = 3.75 m s−2 8 T = 5 × 3.75 T = 18.8 N, correct to 3 s.f. b u = 0 m s−1, s = 1.3 m, a = 3.75 m s−2 s = ut + 1 at2 2 1.3 = 0 × t + 1 × 3.75 × t2 2 = 1.875t2 t2 = 1.3 1.875 t =

1.3 = 0.833 s 1.875

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4

WORKED SOLUTIONS

4 The student has made the mistake of misreading or assuming that m > 6 and therefore that the m kg particle will move downwards and the 6 kg particle upwards. The correct solution is given below: i For the 6 kg particle, resultant force = (6g − T) N.

b When accelerating, u = 0 m s−1, a = 0.8 m s−2, t = 15 s. v = u + at = 0 + 0.8 × 15 = 12 m s−1

F = ma

When decelerating, u = 12 m s−1, v = 0 m s−1, t = 10 s.

6g − T = 6 × 2 g = 4g 3 T = 2g = 20 N

v = u + at

0 = 12 + a × 10

ii For the m kg particle, resultant force = (T − mg) N.

−10a = 12

a = −1.2 m s−2

F = ma

T − mg = m × 2 g 3 2 2g − mg = mg 3 5 2g = mg 3

When the lift is decelerating, resultant force on man = (R − 90g) N.

F = ma R − 90g = 90 × −1.2 R = 90(10 − 1.2)

2 = 5m 3 m = 1.2

= 792 N c The tension is the same for both parts of the cable.

5 u = 0 m s−1, s = 24 m, t = 4 s s = ut + 1 at2 2 24 = 0 × 4 + 1 × a × 42 2 = 8a

7 Let the angle of slope be θ.

a R( ) for the car:

3200 − 400 − 400g sin θ − T = 400a 3200 − 400 − 400g × 0.2 − T = 400a

2000 − T = 400a

For A: resultant force = (X − 0.5g − T) N.

R( ) for the horsebox:

F = ma

T − 600 − 600g sin θ = 600a

T − 600 − 600g × 0.2 = 600a

For B: resultant force = (T − 0.2g) N.

T − 1800 = 600a

F = ma

T − 0.2g = 0.2 × 3 = 0.6

1

2

Add 1 and 2 : X − 0.7g = 2.1 X = 0.7g + 2.1 = 9.1 N 6 a  When the lift is accelerating, resultant force = (T − 250g) N.

2

1 + 2 :

200 = 1000a

a = 0.2 m s−2

b R( ) for the car without tension:

X − 0.5g − T = 0.5 × 3 = 1.5

1

a = 3 m s−2

2000 = 400a

a = 5 m s−2

c  Whilst coupled, s = 90 m, u = 0 m s−1, a = 0.2 m s−2.

F = ma

Substitute in v2 = u2 + 2as:

v2 = 0 + 2 × 0.2 × 90

T − 250g = 250 × 0.8

= 36

T = 250(10 + 0.8)

v = 6 m s−1

= 2700 N

Whilst uncoupled:

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4 Newton’s laws of motion

R( ) for the horsebox:

c T = 2 + 4.5a

−1800 = 600a

= 2 + 4.5 × 0.2

a = −3 m s−2

v=

0 m s−1,

= 2.9 N

u=

Substitute in

2 2 s= v −u 2a

6 m s−1,

v2

=

u2

a=

d There will now be a thrust force in the rod.

−3 m s−2

+ 2as:

F = ma T − 3 = 5.5a

2

= 0 − 6 2 × −3

= 6 m

Hence the horsebox travels 6 m after the uncoupling.

Resultant force for P2 = (T − 3) N.

Resultant force for P1 = (−T − 2) N.

F = ma

8 a u = 22 m s−1, s = −12.25 m, a = −10 m s−2

−T − 2 = 4.5a

−5 = 10a

u = 2 m s−1, a = −0.5 m s−2, v = 0 m s−1

V2 = 222 + 2 × −10 × −12.25

v2 = u2 + 2as

= 729

2 2 s = v − u 2a

V = 729 = 27 m s−1 b Resultant force = (7g − 3000) N.

2 2 = 0 −2 2 × −0.5

F = ma

= 4 m

e T − 3 = 5.5a

7g − 3000 = 7a

a = −418.57 m s−2

T = 3 + 5.5 × −0.5

u = 27 m s−1, v = 0 m s−1, a = −418.57 m s−2

= 3 − 2.75

v2 = u2 + 2as

= 0.25 N

2 2 s = v − u 2a

10 a R(→) for van:

2 2 = 0 − 27 2 × −418.57

2250 − 800 − T cos 12° = 900a

R(→) for cart:

= 0.871 m

T cos 12° − 400 = 500a

9 a Resultant force for P2 = (7 − T − 3) = (4 − T) N. F = ma 4 − T = 5.5a

Resultant force for P1 = (T − 2) N.

F = ma T − 2 = 4.5a

2 = 10a

a = 0.2 m s−2 b u = 0 m s−1, a = 0.2 m s−2, t = 10 s v = u + at = 0 + 0.2 × 10 = 2 m s−1

2

a = −0.5 m s−2

v2 = u2 + 2as

1

1 2

1 2

1 + 2 :

1050 = 1400a

a = 0.75 m s−2

b T cos 12° − 400 = 500 × 0.75

T cos 12° = 775 T = 775 = 792 N cos12° Hence the tension in the tow-bar is 790 N, correct to 2 s.f.

11 a Resultant force for car =  (1500 − T − 300) = (1200 − T) N. F = ma

1200 − T = 750a

Resultant force for trailer = (T − 100) N.

1

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4

WORKED SOLUTIONS

v = 3 m s−1

F = ma T − 100 = 250a

1100 = 1000a

2

When B returns:

u = 3 m s−1, s = 0 m, a = −10 m s−2 s = ut + 1 at2 2 0 = 3t + 1 × −10 × t2 2 = t(3 − 5t) t = 0 s or 3 s 5 Hence t = 0.6 s.

a = 1.1 m s−2 b T − 100 = 250a T = 100 + 250 × 1.1 = 375 N c Assumed that mass is concentrated at a single point.

13 a  C and D are modelled as particles so that each mass is concentrated at a single point.

d Resultant force for trailer =  (−125 − 100) = −225 N. F = ma

b Let the angle of slope be θ. tan θ = 0.75 = 3 4 3 sin θ = =3 32 + 4 2 5

−225 = 250a

a = −0.9 m s−2

Resultant force for car =  (−F + 125 − 300) = (−F − 175) = N. F = ma

4 cos θ = 5

−F − 175 = 750 × −0.9

R( ) for C:

−F − 175 = −675

T − 2g sin θ = 2a 3 T − 2g × 5 = 2a

T − 1.2g = 2a

R(↓) for D:

3g − T = 3a

1 + 2 :

1.8g = 5a

a = 9 g = 3.6 m s−2 25 T = 1.2g + 2 × 3.6 = 19.2 N

F = 500 N 12 a u = 0 m s−1, s = 3.75 m, t = 2.5 s s = ut + 1 at2 2 3.75 = 0 × 2.5 + 1 × a × 2.52 2 3.75 = 3.125a a =

1.2 m s−2

b Resultant force for A = (2.8g − T) N. F = ma

2.8g − T = 2.8 × 1.2 = 3.36

Let mass of B = m kg.

Resultant force for B = (T − mg) N.

1

F = ma T − mg = 1.2m

2

2

c Whilst taut, t = 2 s, u = 0 m s−1, a = 3.6 m s−2.

Substitute in v = u + at:

v = 0 + 3.6 × 2

= 7.2 m s−1

Whilst slack:

2.8g − mg = 1.2m + 3.36

R( ) for C:

2.8g – 3.36 = m(10 + 1.2)

−1.2g = 2a

24.64 = 11.2m

a = −0.6g m s−2

m = 2.2

s = 0 m, u = 7.2 m s−1, a = −0.6g m s−2

c  Assumed that acceleration is the same for both.

Substitute in s = ut + 1 at2: 2 0 = 7.2t + 1 × −0.6gt2 2 0 = t(7.2 − 3t)

d For B, u = 0 m s−1, s = 3.75 m, a = 1.2 m s−2.

1

v2 = u2 + 2as

= 02 + 2 × 1.2 × 3.75

t = 7.2 = 2.4 s 3

= 9

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4 Newton’s laws of motion

14 a R( ) RU = Mg cos q + T sin 30°

s = –3h m, u2 =

2 gh m2s–2, a = –g ms–2 3

R( ) Mg sin q – FU – T cos 30° = Ma

v2 =

2 2 R( ) T cos 30 + Mg sin q – FV = Ma 5 5     FU = 0.2RU

2 gh + 2(–g)(–3h) 3

v2 =

20gh 3

2 R( ) RV = Mg cos q – T sin 30° 5

v2 = u2 + 2as

FV = mRV

16 a R( ) for Q: R = (M + 5)g cos 45°

24 7 When cos q = , sin q = 25 25

7 Mg sin q – 0.2 (Mg cos q + T sin 30°) 5

(

– m 2 Mg cos θ − T sin 30° = 7 M  43 g 5 5 625

)

  8 3 Mg 7 Mg – 0.2  Mg × 24 + × 3 5 25 625 2  

– m  2 Mg × 24 − 8 3 Mg × 1  = 7 Mg   43 g   625   5 25 625 2  5

( ) ( 7 43 = ( 5 625 )

) ()

R( ) for P: T – Mg sin 45 = Ma = 2 2 M

R( ) for Q: (M + 5)g sin 45° – T – F = (M + 5)a = 2 2 (M + 5)

–Mg sin 45 – F + (M + 5)g sin 45°

  2   2  2 –Mg – 0.2 (M + 5)g  + (M + 5)g   2  2      2   = 2 2 (2M + 5)

–Mg – 0.2[(M + 5)g] + (M + 5)g = 4(2M + 5)

0.099 84 = 0.372 91m

–10M – 2(M + 5) + 10(M + 5) = 8M + 20

m = 0.27

M=2

b Mg sin q – FU – T cos 30° = Ma

b T – Mg sin 45° = 2 2 M

Mg

T = 2 2 M + Mg sin 45°

12 3 Mg = T 625 2

T =

T = 14 2 N

8 3 Mg 625

15 Let the mass of A be M kg. R(↑) for A: 2.4Mg – T – Mg = Ma R(↑) for B: T – 0.8Mg = 0.8Ma 1 g ms–2 3

s = h m, u = 0 ms–1, a = v2

=

u2

+ 2as

v2 = 02 + 2 v2 =

( 13 g )(h)

2 gh 3

When string breaks:

 2 Force = 2T cos 45° = 2(14 2 ) = 28 N  2    5 12 17 a When sin q = , cos q =   13   13

R( ) for block: 7g + 6.5g sin q – F – T = 6.5a

R( ) for block: R = 6.5g cos q

0.6Mg = 1.8Ma a=

 2 T = 2 2 (2) + 2g  2 

2

= 2 2 M + 2 2 (M + 5)

7 7 2  24 − 4 3  – 0.2 24 + 12 – m 5 25 5  25 625  25 625

43 g )= T cos 30° ( 257 )– 0.2( 2425 )Mg – M( 625

1

)

(

F = mR = 0.2[(M + 5)g cos 45°]

1 (6.5g cos q) 3

F = mR = 1 g ms–2 3

R(↑) for ball: T – Mg = Ma

1 7g+ 6.5g sin q – (6.5g cos q) – Mg     3 = (M + 6.5)a

7g + 6.5g

75 – 10M = (M + 6.5)a

(135 )– 13  6.5g (1213 ) – Mg = (M + 6.5)a

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WORKED SOLUTIONS

a =

75 − 10M M + 6.5 75 − 10M 0 M + 6.5

Since a  0,

75 – 10M  0

M  7.5

Maximum value of M = 7.5

b M = 3.5 a =

75 − 10(3.5) = 4 ms–2 3.5 + 6.5 ms–2,

s = 0.8 m, a = 4 s = ut +

ms–1

u=0

1 2 at 2

0.8 = 0 +

1 (4)t2 2

(5 − 7µ ) g

=

Whilst slack:

R(→) for P:

−µ × 2.1g = 2.1a

a = −µg

u2 =

6

(5 − 7µ ) g , v = 0 m s−1, a = −µg m s−2, s < 1.5 m

6 Substitute in v2 = u2 + 2as:

2 2 s = v −u 2a

0− 6 < 1.5 2 × −µ g

(5−7µ )g

(5 − 7µ ) g

> −3µg 6 −g(5 − 7µ) > −18µg

t = 0.632 s

−5g + 7gµ > −18µg

18 a R( ) for P:

25gµ > 5g

µ>1 5

Hence 1 < µ < 5 . 5 7

t2 = 0.4

R − 2.1g = 0

R = 2.1g

F = µR = µ × 2.1g

R(→) for P:

T − F = 2.1a

T − µ × 2.1g = 2.1a

R(↓) for Q:

19 a R( ) for car:

3200 – T – 100g – 400g sin β = 400a = 400 × 0.2 = 80

R( ) for milk float:

1

T – 100g – 600g sin β = 600 × 0.2 = 120 2

1.5g − T = 1.5a 1 + 2 :

3200 – 200g – 1000g sin β = 200

1.5g − µ × 2.1g = 3.6a

1000g sin β = 100

5g − 7gµ = 12a

sin β = 0.1

g(5 − 7µ) = 12a

β = 5.7°

a = ( 5 − 7µ ) g 12

b Let the reaction force exerted on the car by the hill be R N.

12

>0

R( ) for car: →

(5 − 7µ ) g

R – 400g cos β = 0

5 − 7µ > 0

5 > 7µ

7µ < 5

µ< 5 7

Whilst taut, u = 0 m s−1, a =

Substitute in v2 = u2 + 2as:

v2 = 0 + 2 ×

R = 400g × cos 5.7° = 3980 N

(5 − 7µ ) g 12

×1

(5 − 7µ ) g  m s−2, s = 1 m. 12

µ = F = 100g = 0.251 R 3980

Mathematics in life and work 1 Let the angle be θ and the weight of the car be W. If sin θ = 9 , then cos θ = 40 and tan θ = 9 . 41 41 40 R( ):

b Since a > 0,

4

F – W sin θ = 0

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4 Newton’s laws of motion

R( ): →

R – W cos θ = 0

When F = µR, µ = F . R

µ = W sin θ = tan θ = 9 40 W cosθ

2 R( ):

D – 1230g sin θ – F = 1230 × 0.5

F = D – 1230g sin θ – 615 R( ):

R – 1230g cos θ = 0 R = 1230g cos θ

Substitute into F = µR:

D – 1230g sin θ – 615 = 9 (1230g cos θ) 40 9 D – 12 300 × – 615 = 9 12300 × 40 41 40 41 D – 2700 – 615 = 2700

(

)

D = 6015 N = 6020 N (3 s. f.)

3 R( ) for car:

6015 – 1230g sin θ – 9 × 1230g cos θ – T = 1230a 40 R( ) for trailer:

T – 82g sin θ – 9 × 82g cos θ = 82a 40 Add: 6015 – 1312g sin θ – 9 × 1312g cos θ = 1312a 40 9 6015 – 13 120 × – 9 × 13 120 × 40 = 1312a 41 41 40

6015 – 2880 – 2880 = 1312a

255 = 1312a

a = 255 = 0.194 m s–2 1312 4 Coefficient of friction unlikely to be the same.

Driving force likely to be higher.

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WORKED SOLUTIONS

5

5 Energy, work and power 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 h = 17 sin 25° = 7.18 m 2 u = 0 m s–1, a = 10 m s–2, s = 60 m s = ut + 1 at2 2 1 60 = 0 + × 10 × t2 2 t2 = 12

Exercise 5.1B 1 a s = ut = 2.5 × 8 = 20 m b R(→): X cos 6° – F = 0 c R(↑):

F = X cos 6°

R – X sin 6° – 16g = 0 d F = µR

3 R( ):

X cos 6° = 0.3(160 + X sin 6°)

R – 3.5g cos 33° = 0

R = 3.5g cos 33°

X cos 6° – 0.3X sin 6° = 48

X(cos 6° – 0.3 sin 6°) = 48 48 X = cos6° − 0.3sin 6° = 49.8

t = 12 = 3.46 s

R( ):

P – F – 3.5g sin 33° = 0 P = 7 × 3.5g cos 33° + 3.5g sin 33° = 39.6 N 10

Exercise 5.1A b WD = Fd = 5200 × 1.3 = 6760 J c WD = Fd = 9.3 × 0.8 = 7.44 J 2 a WD = Fd cos q = 6 × 4 × cos 18 = 22.8 J b WD = Fd cos q = 17 × 0.4 × 4 = 5.44 J 5 c WD = Fd cos q = 380 000 × 7.4 × 35 = 2660 kJ 37 3 a F = WD = 400 = 160 N 2.5 d b d = WD = 1050 = 1.4 m 750 F

6 F = WD = 2295 = 425 N 5.4 d m = WD = 425 = 42.5 kg g 10 7 WD = Fd = 16 000 × 6.5 = 104 kJ

e WD = Fd = 49.8 × 20 × cos 6° = 991 J 2 a Let the angle of inclination of the plane be q. Hence cos q = 60 and sin q = 11 . 61 61 R( ): R – 244g cos q = 0 R = 244 × 10 × 60 = 2400 N 61 b F = µR     = 2 × 2400 15     = 320 N c WD = Fd = 320 × 12.2 = 3904 J d h = d sin q = 12.2 × 11 = 2.2 m 61 e WD = Fd = 244 × 10 × 2.2 = 5368 J f 3904 + 5368 = 9272 J = 9.27 kJ 3 a R( ):

4 WD = Fd = 900 × 10 × 18 = 16 200 J = 162 kJ 5 d = WD = 528 = 1.2 m F 440

= 48 + 0.3X sin 6°

1 a WD = Fd = 4 × 2.1 = 8.4 J

R = 160 + X sin 6°

R – 9.5g cos 8.3 = 0 R = 94.0 N F = µR = 0.22 × 94.0 = 20.68 N WD = Fd = 20.68 × 3.8 = 78.6 J b    h = d sin q = 3.8 sin 8.3 = 0.549 m

WD = Fd = 9.5 × 10 × 0.549 = 52.1 J

57

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5 Energy, work and power

4 s = ut = 0.2 × 60 = 12 m R( ):

2 a PE = mgh = 4 × 10 × 2.3 = 92 J This is a 92 J increase.

R – 25g cos 30 = 0

b PE = mgh = 0.9 × 10 × 12 = 108 J

R = 125 3  N F = µR = 3 × 125 3 15 = 25 N

c PE = mgh = 80 × 10 × 4 = 3200 J d Total mass = 520 + 2 × 75 = 670 kg.

PE = mgh = 670 × 10 × 35 = 234 500 J

This is a 234 500 J decrease.

Work done against friction = Fd = 25 × 12 = 300 J.

3 WD = mg

h = d sin q = 12 sin 30° = 6 m

Work done against gravity = Fd = 25 × 10 × 6 = 1500 J.

m = WD = 16 000 = 1600 kg g 10

Total work done = 300 + 1500 = 1800 J.

5 Work done against friction = Fd = µRd = µmgd cos q.

Work done against gravity = Fd = mgd sin q.

Hence WD = µmgd cos q + mgd sin q. WD − mgd sin θ 43 − 3.6 × 10 × 4.2sin10° = µ= = 0.112 3.6 × 10 × 4.2cos10° mgd cosθ

63 km h–1 = 63 000 = 17.5 m s–1 3600 45 km h–1 = 45 000 = 12.5 m s–1 3600 Change in KE = 1 m(v2 – u2) = 1 × 1600 × 2 2 (12.52 – 17.52) = –120 000.

6 Let the angle of inclination be q. Hence sin q = 3 5 and cos q = 4 . 5 Work done against friction = Fd = µRd = µmgd cos q.

v2 = u2 + 2as

Work done against gravity = Fd = mgd sin q.

= 4002 + 2 × –0.2 × 500

Hence WD = µmgd cos q + mgd sin q.

= 159 800

Change in KE = 1 m(v2 – u2) 2  1 = × 0.000 3 × (159 800 – 4002) = – 0.03 J. 2

WD = mgd(µ cos q + sin q ) 513 WD = d= mg ( µcosθ + sin θ ) 6 × 10 ( 83 ×

= 5139 = 513 = 9.5 m 54 60 ( 10 )

4 5

+

3 5

)

7 a Work done against gravity = Fd = mgd sin a.

3600 = 51 × 10 × 15 sin a = 7650 sin a sin a = 3600 = 8 7650 17

b If sin a = 8 , then cos a = 15 . 17 17 Work done against friction = Fd = µRd = µmgd cos a. WD = 0.24 × 51 × 10 × 15 × 15 = 1620 J 17

Exercise 5.2A 1 × 8 × 1.52 = 9 J 2 1 × 1200 × 102 = 60 000 J = 60 kJ 2 c Change in KE =  1 m(v2 – u2) = 1 × 0.45 × (92 – 32) 2 2 = 16.2 J. This is a 16.2 J increase.

1 a KE = 1 mv2 = 2 1 b KE = mv2 = 2

d Change in KE = 1 m(v2 – u2) 2 = 1 × 15 × (2.72 – 6.32) = –243 J. 2 This is a 243 J decrease. 58

This is a decrease of 120 kJ.

4 u = 400 m s–1, a = 0.2 m s–2, s = 500 m

5 a m = WD = 45 = 4.5 kg g 10 Change in KE = 1 m(v2 – u2). 2 1 × 4.5 × (V  2 – 192) = 828 2 V  2 – 192 = 368

V  2 = 368 + 192 = 729 V = 729 = 27 m s–1

b s = 23 m, u = 19 m s–1, v = 27 m s–1 v2 = u2 + 2as 2 2    a = v − u 2s

2 2 = 27 − 19 = 368 = 8 m s–2 46 2 × 23

Exercise 5.3A 1 1 a KE = 1 mv2 = × 0.4 × 182 = 64.8 J 2 2  b 64.8 J c Change in PE = mgh = 0.4 × 10 × h = 64.8.        h = 64.8 = 16.2 m 0.4 × 10

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WORKED SOLUTIONS

2 a Since v = 0, KE = 0 J. b h = 5.4 sin q = 5.4 × 3 = 3.24 m 5 Decrease in PE = mgh = 30 × 10 × 3.24 = 972 J.

6 a m = 0.04 kg, h = 60 m

Hence change in PE = –972 J.

Change in PE = mgh = 0.04 × 10 × 60 = 24 J.

The ball has lost 24 J of potential energy. 1 1 Change in KE = m(v2 – u2) = × 0.04 × (v2 – 02) 2 2 = 0.02v2.

c R( ): R – 30g cos q = 0 4 R = 30 × 10 × = 240 N 5

The ball has gained 0.02v2 J of kinetic energy.

Hence 24 = 0.02v2.

d F = µR

= 0.15 × 240 = 36 N

e WD = Fd = 36 × 5.4 = 194.4 J = 194 J (3 s.f.) 1 1 f Change in KE =  m(v2 – u2) = × 30 × (v2 – 02) 2 2 = 15v2.

v = 1200 = 34.6 m s–1

b All the potential energy is converted into kinetic energy, constant gravity over small changes in altitude. c s = –60 m, u = 0 m s–1, a = –10 m s–2

Change in PE = –972 J.

Work done = 194.4 J.

v2 = u2 + 2as

Using the conservation of energy:

= 02 + 2 × –10 × –60

Change in KE + Change in PE + Work done = 0

= 1200

15v2 – 972 + 194.4 = 0

v = 1200 = 34.6 m s–1

15v2 = 777.6

= 7.2 m s–1

3 a Change in KE = 1 m(v2 – u2) = 1 × 2.8 × (252 – 112) 2 2 = 705.6 J. b Change in PE = mgh = 2.8 × 10 × h = 705.6. h = 705.6 = 25.2 m 2.8 × 10 2 25.2 × = 7.2 m 7 4 a KE = 1 mv2 = 1 × 1.6 × 82 = 51.2 J 2 2

7 Change in KE = 1  m(v2 – u2) = 1 × 0.7 × (02 – 72) 2 2 = –17.15 J.

Increase in PE = mgh = 0.7 × 10 × 18 sin 7° = 126 sin 7°.

Using the conservation of energy:

Change in KE + Change in PE + Work done = 0

–17.15 + 126 sin 7° + WD = 0

R( ):

WD = 17.15 – 126 sin 7°   F = WD = 17.15 − 126sin 7 18 d →

v2 = 51.84

R – 0.7g cos 7° = 0

b 51.2 J

R = 0.7 × 10 cos 7° = 7 cos 7°

c WD = Fd       F = WD = 51.2 = 16  N 48 15 d

µ =

R(↑):

R – 1.6g = 0

R = 1.6g = 16 N 16 µ = F = 15 = 1 R 16 15 1 5 Change in KE = 1 m(v2 – u2) = × 0.0025 × (02 – 3002) 2 2 = –112.5 J.

v2 = 1200

WD = Fd = 2250d

112.5 = 2250d d = 112.5 = 0.05 m or 5 cm 2250

F 17.15 − 126sin 7 = 0.0143 = R 18 ( 7cos7 )

8 a Change in KE = 0 – 0 = 0 J.

Decrease in PE = mgh = 66 × 10 × (d – 6)sin q. 660 × (d – 6) × 1 = 132(d – 6)  J 5 Change in PE = –132(d – 6)  J.

The work done against the resistive force, WD = Fd = 12(2d + 6).

Using the conservation of energy:

Change in KE + Change in PE + Work done = 0

0 – 132(d – 6) + 12(2d + 6) = 0

–132d + 792 + 24d + 72 = 0

864 = 108d d=8

59

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1 1 m(v2 – u2) = × 66 × (v2 – 02) 2 2 = 33v2 J.

Change in KE =

b Change in KE = 

1 1 m(v2 – u2) = – (8.5)(142) = –833 J 2 2

WD = Fd = 15F

Decrease in PE = mgh = mgd sin q 1 = 66 × 10 × 8 × = 1056 J. 5 Change in PE = –1056 J.

Using the conservation of energy:

Halfway along the horizontal, he has travelled 1 (d + d) = 12 m in total. 2  The work done against the resistive force, WD = Fd = 12 × 12 = 144 J.

F = 4.53 N

Change in KE + Change in PE + Work done = 0 –833 + 765 + 15F = 0

Using the conservation of energy:

Change in KE + Change in PE + Work done = 0

3 4 , cos q = 5 5 4 R(  ): R = mg cos q = 5 mg 3 4 3 F = mR = mg mg = 20 16 5

33v2

Change in PE = mgh = 1.6mg

11 When sin b = →

(

– 1056 + 144 = 0

33v2 = 912 v2 = 304 11

v=

1 1 m(v2 – u2) = m(v2 – (2v + 1)2) 2 2   3 1.6  3 1.6  2  mg mg WD = Fd = = mg = 20  sin β  20  3  5  5 

Change in KE =

304 = 5.26 m s–1 11

1 1 m(v2 – u2) = × 2 × (v2 – 02) 2 2 = v2 J.

9 a Change in KE =

)

Using the conservation of energy: Change in KE + Change in PE + Work done = 0

Decrease in PE = mgh = 2 × 10 × 13 = 260 J.

1 2 m(v 2 – (2v + 1)2) + 1.6mg + mg = 0 2 5

Change in PE = –260 J.

v 2 – (4v 2 + 4v + 1) + 4g = 0

The work done against the resistive force, WD = Fd = 4 × 20 = 80 J.

v 2 – 4v 2 – 4v – 1 + 40 = 0

Using the conservation of energy:

Change in KE + Change in PE + Work done = 0

v2 = 180

v=

180 = 13.4 m s–1

(3v + 13)(v – 3) = 0 v=3 12 Change in KE = 1 m(v 2 – u2) = 1 × 0.6 × (142 – 02) 2 2   = 58.8 J.

Given that the height of the roof is 3 m, the height of the rest of the house is 12 – 3 = 9 m.

Decrease in PE = mgh = 0.6 × 10 × 12 = 72 J.

Change in PE = –72 J.

R( ):

b Change in KE = 0 J.

Decrease in PE = mgh =  2 × 10 × (13 – H) = (260 – 20H)  J.

Change in PE = (20H – 260)  J.

The work done against the resistive force, WD = Fd = 4 × 30 = 120 J.

Using the conservation of energy:

Change in KE + Change in PE + Work done = 0

0 + (20H – 260) + 120 = 0

20H = 140

H = 7 m

3 3 10 When sin q = , tan q = 5 4

()

3 h = 12 tan q = 12 =9m 4 Change in PE = mgh = 8.5g(9) = 765 J

v2 – 260 + 80 = 0

3v 2 + 4v – 39 = 0

R – 0.6g cos q = 0

If tan q =

2 , then cos q = 4

= 2 2. 3 R = 0.6 × 10 × 2 2 = 4 2  N 3 2 F = µR = × 4 2 = 4 N 2

4 42 + 2

= 4 = 4 18 3 2

Let the distance the slate slides down the roof be d m.

The work done against the resistive force, WD = Fd = 4d J.

Using the conservation of energy:

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WORKED SOLUTIONS

5

Change in KE + Change in PE + Work done = 0

b For AB:

58.8 – 72 + 4d = 0

4d = 13.2

Change in KE = 1 m(v2 – u2) = 1 m(62 – 02) = 18m J. 2 2

Let d be the length AB

Decrease in PE =  mgh = m × 10 × d sin 20° = 10md sin 20° J.

Change in PE = –10md sin 20° J.

R( ):

d = 3.3 m

13 a Since both the initial velocity and final velocity are zero, the change in KE = 0 J.

Decrease in PE = mgh = 48 × 10 × 0.5 = 240 J.

Hence change in PE = –240 J. 1 1 WD = Fd = F(0.5 + × 2π × 3 + 6 + × 2π × 3) 4 4 Using the conservation of energy:

Change in KE + Change in PE + Work done = 0 1 1 0 – 240 + F(0.5 + × 2π × 3 + 6 + × 2π × 3) = 0 4 4 F(6.5 + 3π) = 240 F = 240 = 15.1 N 6.5 + 3π

R – mg cos 20° = 0 R = m × 10 × cos 20° = 10m cos 20° N F = µR = 0.3 × 10m cos 20° = 3m cos 20° N

The work done against friction, WD = Fd = 3m cos 20° × d = 3md cos 20° J.

Using the conservation of energy:

Change in KE + Change in PE + Work done = 0

b Again the change in KE = 0 J.

18m – 10md sin 20° + 3md cos 20° = 0

Let x be the distance travelled up the second vert.

10d sin 20° – 3d cos 20° = 18

Decrease in PE = mgh =  48 × 10 × (0.5 – x) = 240 – 480x.

Hence change in PE = (480x – 240)  J.

WD = Fd = 10(6.5 + 3π + x)

c For AB, s = 29.9 m, u = 0 m s–1, v = 6 m s–1.

Using the conservation of energy:

Change in KE + Change in PE + Work done = 0

s =  u + v  t  2 

0 + (480x – 240) + 10(6.5 + 3π + x) = 0

480x – 240 + 65 + 30π + 10x = 0

490x = 175 – 30π

x = 175 − 30π = 0.165 m 490 14 a For BC:

1 1 m(v2 – u2) = m(02 – u2) 2 2 1 = – mu2 J. 2 Change in PE = 0 J.

R(↑):

Change in KE =

R – mg = 0 R = mg F = µR = 0.3 × m × 10 = 3m N

The work done against friction, WD = Fd =  3m × 6 = 18m J. Using the conservation of energy:

Change in KE + Change in PE + Work done = 0 1 – mu2 + 0 + 18m = 0 2 1 mu2 = 18m 2 u2 = 36 u = 36 = 6 m s–1

d(10 sin 20° – 3 cos 20°) = 18 18 d= = 29.9 m 10sin 20° − 3cos 20°

t = 2s = 2 × 29.9 = 9.98 s 0+6 u +v For BC, s = 6 m, u = 6 m s–1, v = 0 m s–1. t = 2s = 2 × 6 = 2 s u +v 6 + 0

Total time = 9.98 + 2 = 12.0 s

Exercise 5.4A 1 a P = Fv = 1500 × 14 = 21 000 W b P = Fv = 10 × 0.2 = 2 W c 54 km h–1 = 15 m s–1 D = P = 7200 = 480 N 15 v

d v = P = 9900 = 5.5 m s–1 D 1800 e P = WD = 20 000 = 2000 W 10 t f WD = Pt = 6100 × 29 = 177 000 J 2 a R(↑): R – 305g = 0 R = 305g = 305 × 10 = 3050 N 2 b F = µR = × 3050 = 1220 N 5

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c WD = Fd = 1220 × 100 = 122 000 J d P = WD = 122 000 = 24 400 W = 24.4 kW t 5 e R(→):

b 14 800 – 390 cos 6° – 13 000 sin 6° = 1300a 6

1300a = 719.9

a = 0.554 m s–2

a = 2.4 m s–2 3 R(→): D – F = ma

P – µR = ma v

R(↑):

5 a R( ):

D – F = ma 24 400 – 1220 = 305a 12.5 732 = 305a

P – 800 – 3200g × 3 = 0 1.8 16 P = 1.8(800 + 6000) = 12 240 W = 12.2 kW b R(→): 12 240 – 800 = 3200a 1.8 3200a = 6000 a = 1.875 = 1.88 m s–2

6 a R(↑):

R – 1200g = 0

R – mg = 0

R = 1200g = 1200 × 10 = 12 000 N

R = mg

7200 – 500 = 1200a v

a 7200 – 500 = 1200 × 1.25 = 1500 v 7200 = 2000v v = 3.6 m s–1 b 7200 – 500 = 1200 × 0.25 = 300 v 7200 = 800v 0 m s–2

c a = 7200 – 500 = 0 v 7200 = 500v

v = 14.4 m s–1

4 a R( ): R – 1300g cos 6° = 0 R = 13 000 cos 6° N F = µR = 0.03 × 13 000 cos 6° = 390 cos 6° N R( ):

D – 390 cos 6° – 13 000 sin 6° = 0 P – 390 cos 6° – 13 000 sin 6° = 0 v 14 800 – 390 cos 6° – 13 000 sin 6° = 0 v 14 800 = 8.47 m s–1 v = 390 cos6° + 13 000sin 6°

b WD = Pt = 8400 × 30 = 252 kJ c R( ):

v = 9 m s–1

D – F = 0 8400 – 2 m = 0 20 5 2 m = 420 5  m = 1050 kg

R – 1050g cos 3° = 0 R = 1050 × 10 × cos 3° = 10 500 cos 3° F = µR = 1 × 10 500 cos 3° = 420 cos 3° 25

R( ):

F = µR = µmg = 1 × m × 10 = 2 m 5 25 R(→):

8400 – 420 cos 3° – 10 500 sin 3° = 0 v 8400 v = = 8.67 m s–1 420 cos3° + 10 500sin 3° d 8400 – 420 cos 3° – 10 500 sin 3° = 1050a 5 a = 0.677 m s–2 7 a R( ) for first car:

9400 – T – 400 – 800g sin q = 800a 5

R( ) for second car:

1480 – T – 800g sin q = 800a

7200 – 1 × 12 000 = 1200a v 24

T – 400 – 1000g sin q = 1000a

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WORKED SOLUTIONS

1080 – 1800g sin q = 1800a

1080 – 18000 sin q = 1800 × 0.4 = 720

sin  q = 0.02

18 000 sin q = 360

b 9400 – T – 400 – 800g sin q = 0 v

T – 400 – 1000g sin q = 0

Hence 9400 – 800 – 1800g sin q = 0. v 9400 – 800 – 18000 × 0.02 = 0 v 9400 = 1160 v v = 9400 = 8.10 m s–1 1160 c

9400 – 400 – 8000 × 0.02 = 800a 9400 1160

1160 – 400 – 160 = 800a

a = 0.75 ms–2 →

8 a R( ): R = 180g cos 4°

9 a D – kV = 1600a 28 800 – 12k = 1600(1.26) 12

2400 – 12k = 2016

k = 32 F = 32V b

28 800 – 32V = 1600(0.5) = 800 V

28 800 – 32V 2 = 800V

V 2 + 25V – 900 = 0

(V + 45)(V – 20) = 0

V = 20 10 a D – 40V – 1200g sin 3° = 0

75 000 – 40V – 1200g sin 3° = 0 V

1875 – V – 30g sin 3° = 0 V

1875 – V   2 – 300V sin 3° = 0

V 2 + 300V sin 3 – 1875 = 0 V = 

−300sin 3° ± (300sin 3°)2 − 4(1)(−1875) 2

= 36.2 ms–1

F = mR = m(180g cos 4°) R( ): D – 180g sin 4° – F = 180a

b D – 30V – 1200g sin q = 0

P – 180g sin 4° – m(180g cos 4°) = 180a v

6800 – 180g sin 4° – m(180g cos 4°) = 180(0.2) = 36 18

6800 − 180g sin 4° − 36 m = 18 = 0.120 180g cos 4°

b 6800 – 180g sin 4° – 0.120(180g cos 4°) = 0 v     v =

6800 = 19.9 m s–1 180g sin 4° + 0.120(180g cos 4°)

Let further distance be d m.

Change in PE = 180g d sin 4° J.

Change in KE = 

1 1 m(v2 – u2) = – (180)(19.92) 2 2 = –35 630 J

WD = Fd = 0.120(180g cos 4°)d

Using the conservation of energy:

Change in KE + Change in PE + Work done = 0

–35 630 + 180g d sin 4° + 0.120(180g cos 4°)d = 0

d =

35 630 = 104 m 180g sin 4° + 0.120(180g cos 4°)

5

( )

45 000 1 – 30V – 1200g =0 20 V 45 000 – 30V – 600 = 0 V 1500 – V  2 – 20V = 0

V  2 + 20V – 1500 = 0

(V + 50)(V – 30) = 0

V = 30 m s–1

Exam-style questions 1 a Loss in PE = mgh = 0.003 × 10 × 26 = 0.78 J b Increase in KE =

1 1 mv2 = (0.003)v2 = 0.78 2 2

v2 = 520 v = 22.8 m s–1 2 a  Change in KE = 0.8v2 = 7.2 v2 = 7.2 = 9 0.8

1 m(v2 – u2) = 1 × 1.6 × (02 – v2) 2 2 = –7.2 J.

v = 3 m s–1

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b 7.2 J

c WD = Fd

7.2 = F × 5 F = 7.2 = 1.44 N 5

Resultant force for car = (T − 300) N. F = ma T − 300 = 900a

1700 = 2500a

a = 0.68 m s−2

3 a R(→): D – 3 = 0

b T = 300 + 900a

D = 3 N

= 300 + 900 × 0.68

= 912 N

b

P =3 v

c The car and truck have the same acceleration. 1 1 d Change in KE = m(v2 – u2) = × 900 × (02 – 252) 2 2 = –281 250 J.

v = 7.5 = 2.5 m s–1 3 c P – 3 = ma v

7.5 – 3 = 1.3a 1.2 1.3a = 3.25

a = 2.5 m s–2 4 Change in KE = 1 m(v2 – u2) = 1 m(02 – 242) 2 2 = –288m J.

No change in PE.

Work done against friction = Fd = 300d J.

Using the conservation of energy:

Change in KE + Change in PE + Work done = 0

–281 250 + 300d = 0

300d = 281 250

d = 938 m

7 a R(↑):

The ball has lost 288m J of kinetic energy.

Change in PE = mgh = m × 10 × h = 10mh.

The ball has gained 10mh J of potential energy.

R = 2250 N

Hence 288m = 10mh. h = 288m = 28.8 m 10m

m = 2.4 kg b Change in PE = mgh = 2.4 × 10 × h = 360. h = 15 c u = 0 m s–1, v = 28 m s–1, a = 10 m s–1 v = u + at t = v − u = 28 − 0 = 2.8 s 10 a

d v2 = u2 + 2as

2 2 2 2 s = v − u = 28 − 0 = 39.2 m 2 × 10 2a

6 a  Resultant force for truck =  (2500 − 500 − T) = (2000 − T) N.

F = ma

2000 − T = 1600a

R(→):

D–F=0

2160 – 0.16 × 2250 = 0 v 2160 v = = 6 m s–1 0.16 × 2250 b R( ):

Maximum height = 28.8 + 6 = 34.8 m. 5 a Change in KE = 1 m(v2 – u2) = 1 m(282 – 222) 2 2 = 360.

R – 2250 cos 5° = 0 R = 2250 cos 5°

R( ):

R – 225g = 0

D – F – 2250 sin 5° = 0 2160 – 0.2 × 2250 cos 5° – 2250 sin 5° = 0 v 2160 v = = 3.35 m s–1 0.2 × 2250 cos5° + 2250sin 5° 8 Let the angle of slope be θ. 4 a cos θ = 0.8 = 5

sin θ =

52 − 42 3 = 5 5

Loss of potential energy = mgh = mgd sin θ. mgd sin θ = 0.25 × 10 × 2.5 × 3 = 3.75 J 5

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WORKED SOLUTIONS

R − 0.25g cos θ = 0 4 R = 0.25g × = 0.2g N 5 F = µR 3 = × 0.2g  = 3 g N 7 35 WD = Fd = 3 × 10 × 2.5 = 2.14 J 35 c Change in KE = 1 m(v2 – u2) = 1 × 0.25 × (v2 – 02) 2 2 = 0.125v2 J.

Decrease in PE = –3.75 J.

Using the conservation of energy:

Change in KE + Change in PE + Work done = 0

0.125v2 – 3.75 + 2.14 = 0

0.125v2 = 1.61 v = 3.59 m s–1

9 a Work done against gravity = mgh = mgd sin q. mgd sin q = 2.5 × 10 × 6.5 × 57 = 5.7 J 1625

b R( ): R – mg cos q = 0 1624 . If sin q = 57 , then cos q = 1625 1625 F = µR = 5 × 2.5 × 10 × 1624 = 7 1625 13 232 Work done against friction = Fd = 7 × 6.5 13 = 3.5 J.

Total work done = 5.7 + 3.5 = 9.2 J.

c R( ): mg sin q – F = ma mg sin q – µmg cos q = ma 57 44 a = 10 × – 5 × 10 × 1624 =  m s–2 1625 232 1625 325 u = 0 m s–1, a = 44  m s–2, s = 6.5 m 325 v2 = u2 + 2as = 0 + 2 × 44 × 6.5 325 = 44 25 v = 1.33 m s–1 10 a mAuA + mBuB = mAvA + mBvB

2mA + 3 × –4 = –2.5mA + 3 × 3.5

4.5mA = 22.5

mA = 5 kg 1 1 m u 2 + mBuB2 2 2 A A 1 1 2 = × 5 × 2 + × 3 × (–4)2 = 34 J. 2 2

b Initial KE = 

1 1 m v 2 + mBvB2 2 A A 2 1 1 = × 5 × (–2.5)2 + × 3 × (3.5)2 = 34 J. 2 2

Final KE = 

Since initial KE = final KE, there is no loss of KE.

1 1 11 a Change in KE for P = m(v2 – u2) = × 4 × (v2 – 02) 2 2 = 2v2 J. 1 1 Change in KE for Q = m(v2 – u2) = × 6 × (v2 – 02) 2 2 = 3v2 J. Total increase in KE = 5v2 J.

Increase in PE for P = mgh = mgd sin q

Decrease in PE for Q = mgh = 6 × 10 × 2 = 120 J.

Change in PE = 40 – 120 = – 80 J.

Work done against resistive force = Fd =  30 × 2 = 60 J. Using the conservation of energy:

Change in KE + Change in PE + Work done = 0

5v2 – 80 + 60 = 0

= 4 × 10 × 2 × sin 30° = 40 J.

v2 = 4

v = 2 m s–1

b R( ):

b R( ):

– 30 – 4g cos 60° = 4a

4a = –50 a = –12.5 m s–2

u=

2 m s–1,

v = 0 m s–1, a = –12.5 m s–2

v2 = u2 + 2as 2 2     s = v − u 2a

2 2 = 0 − 2 = 0.16 m 2 × −12.5

Total distance = 2 + 0.16 = 2.16 m.

12 a R(→) for lorry:

3300 − F − T cos 9° = 1600a

R(↑) for lorry: R – 1600g = 0 1 3300 − × 1600g − T cos 9° = 1600a 16

R(→) for cement mixer:

2300 − T cos 9° = 1600a

1

T cos 9° − F = 800a

R(↑) for cement mixer: R – 800g = 0 1 T cos 9° − × 800g = 800a 16

T cos 9° − 500 = 800a

2

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5 Energy, work and power

1 + 2  :

b From Y to Z:

1800 = 2400a

a = 0.75 m s−2

1 1 Change in KE = m(v2 – u2) = × 200 × (v2 – 222) 2 2 = (100v2 – 48 400) J. Change in PE = 0 J.

b T cos 9° = 500 + 800 × 0.75

The work done against the resistive force, WD = Fd = 232 × 138 = 32 016 J.

= 1100 T = 1100 = 1114 N cos9

Using the conservation of energy:

The tension is 1110 N, correct to 3 s.f.

Change in KE + Change in PE + Work done = 0

c

Whilst coupled, u = 0 m s−1, s = 24 m, a = 0.75 m s−2.

(100v2 – 48 400) + 32 016 = 0

Substitute in v2 = u2 + 2as:

v2 = 0 + 2 × 0.75 × 24

= 36

v = 6 m s−1

P = Fv = 3300 × 6 = 19 800 W

d R(→) for lorry: 19 800 − 1000 = 0 v v = 19.8 m vs–1

Increase in PE = mgh =  200 × 10 × d sin 30° = 1000d J.

Using the conservation of energy:

Change in KE + Change in PE + Work done = 0

a = −0.625 m s−2.

–16 384 + 1000d + 232d = 0

Substitute in v2 = u2 + 2as:

1232d = 16 384

−500 = 800a Whilst uncoupled, u = 6 m s−1, v = 0 m s−1,

2 2 s= v −u 2a

= 28.8 m

13 a From X to Y:

1 1 Change in KE = m(v2 – u2) = × 200 × (v2 – 02) 2 2 = 100v2 J.

d = 13.3 m

14 a R( ) for the tractor: D – F – T – 25 000 sin 10° = 0

R( ) for the tractor:

2 = 0 − 6 2 × −0.625

The work done against the resistive force, WD = Fd = 232d J.

a = −0.625 m s−2

c From Z to the end: 1 Change in KE =  m(v2 – u2) 2  1 = × 200 × (02 – 163.84) = –16 384 J. 2

e R(→) for cement mixer:

v = 163.84 = 12.8 m s–1

v2 = 163.84

R – 25 000 cos 10° = 0

48 000 – 0.125 × 25 000 cos 10° – T 4.2

Let the weight of the combine harvester be Mg N. R( ) for the combine harvester:

– 25 000 sin 10° = 0

Decrease in PE = mgh = 200 × 10 × 30 = 60 000 J.

Change in PE = –60 000 J.

The work done against the resistive force, WD = Fd = 232 × 50 = 11 600 J.

T – F – Mg sin 10° = 0

Using the conservation of energy:

Change in KE + Change in PE + Work done = 0

100v2 – 60 000 + 11 600 = 0

v = 484 = 22 m s–1

R( ) for the combine harvester:

v2 = 484

1

R – Mg cos 10° = 0

T – 0.125 × Mg cos 10° – Mg sin 10° = 0

2

1 + 2 48 000 – 0.125 × 25 000 cos 10° – 25 000 sin 10° 4.2 – 0.125 × Mg cos 10° – Mg sin 10° = 0 48 000 Mg (0.125 cos 10° + sin 10°) = – 0.125 × 4.2

25 000 cos 10° – 25 000 sin 10°

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5

WORKED SOLUTIONS

Using the conservation of energy:

Change in KE + Change in PE + Work done = 0

3120 – 60m + 8m = 0

b T = 13 500(0.125 cos 10° + sin 10°) = 4010 N

52m = 3120

c R( ) for the combine harvester:

Mg =

− 0.125 × 25 000 cos10° − 25 000sin10° 0.125cos10° + sin10°

= 13 500 N

48 000 4.2

– 0.125Mg cos 10° – Mg sin 10° = Ma

a = –1.25 cos 10° – 10 sin 10° = –2.97 m s–2

u = 4.2 m s–1, v = 0 m s–1, a = –2.97 m s–2

v = u + at t = v −u a 0 = − 4.2 −2.97

= 1.42 s

m = 60 kg 1 c Change in KE = m(v2 – u2) = 1 × 60 × (v2 – 02) 2 2 = 3120 J.

16 a WD = Fd = 200 × 8 = 1600 J b WD = mgh = 9 × 10 × 8 + 7 × 10 × 8 = 1280 J c KE = 1600 – 1280 = 320 J

Change in KE = 0 J.

For A to B:

R( ):

Decrease in PE = mgh = mgd sin q. 3 mgd sin q = m × 10 × 10 × = 60 m J. 5 Change in PE = –60m J.

R – mg cos q = 0 4 3 When sin q = , cos q = . 5 5 4 R = m × 10 × = 8m 5 1 4 × 8m = m F = µR = 10 5  For B to C:

R(↑):

R – mg = 0 R = m × 10 = 10m 1 F = µR = × 10m = m 10 The work done against friction, WD = Fd = 4  m × 10 + m × d = m(8 + d). 5 Using the conservation of energy:

Change in KE + Change in PE + Work done = 0

0 – 60m + m(8 + d) = 0

v = 104 = 10.2 m s–1

15 a From A to C:

v2 = 104

1 m(v2 – u2) 2 1 = × 9 × (v2 – 02) = 9 v2 J. 2 2 1 Change in KE for Q = m(v2 – u2) 2 7 1 = × 7 × (v2 – 02) = v2 J. 2 2 9 2 7 2 Overall change in KE = v + v = 320. 2 2 8v2 = 320 Change in KE for P =

v2 = 40 v = 6.32 m s–1

1 1 d Change in KE =  2 m(v2 – u2) = × 7 × (v2 – 40) 2

7 2  =  2 v – 140 J. Decrease in PE = mgh = 7 × 10 × 13 = 910 J.

Change in PE = – 910 J.

 7 v 2 – 140 – 910 = 0  2

7 v2 = 1050 2 v2 = 300 v = 17.3 m s–1

17 a D – (28v + 20) = 1800a

60 200 – (28V + 20) = 1800(0.4) = 720 V

d + 8 = 60

60 200 – V(28V + 20) = 720V

d = 52 m

15050 – 7V 2 – 5V = 180V

b From C to D:

7V 2 + 185V = 15 050

Change in KE = 3120 J.

Change in PE = – 60m J.

The work done against the resistive force, WD = Fd = 8m J.

b 7V 2 + 185V – 15 050 = 0

(7V + 430)(V – 35) = 0

V = 35

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5 Energy, work and power

( )

c

60 200 – (28v + 20) = 0 v

Change in PE = –Mg 3 h 2

15 050 – 7v2 – 5v = 0

Change in KE =

7v2

v =

+ 5v – 15 050 = 0

WD = Fd =

2

−5 ± 5 − 4(7)(−15 050) = 46.0 m s–1 14

ms–1

faster than V

Change in KE + Change in PE + Work done = 0 1 3 1 Mv2 – Mgh + Mgh = 0 2 2 2

Since d1 : d2 = 3 : 1, d1 = 37.5 m and d2 = 12.5 m

v2 = 2gh

PE = mgh = 6.8(10)(15 sin a) = 480

v = 2gh

480 8 sin a = = 1020 17

Mathematics in life and work

b PE = mgh = 6.8(10)(12.5 sin b)

( )

13 = 130 J PE = 6.8(10)(12.5) 85

c Change in PE = 130 – 480 = –350 J

Change in KE = 0

WD = Fd = 65F

Using the conservation of energy:

Change in KE + Change in PE + Work done = 0

–350 + 65F = 0

F =

1 Mgh 2

Using the conservation of energy:

18 a d1 + d2 = 65 – 15 = 50 m

1 Mv2 2

70 = 5.4 N 13

d Change in PE = 0 – 480 = –480 J

1 288 ÷ 3.6 = 80 m s–1 1 1 Change in KE = m(v2 – u2) = × 720 × (v2 – 802) 2 2 = (360v2 – 2304 000)  J.

Decrease in PE = mgh = 720 × 10 × 20 = 144 000 J.

So change in PE = –144 000 J.

The work done against the resistive force, WD = Fd = 2400 × 140 = 336 000 J.

Using the conservation of energy:

Change in KE + Change in PE + Work done = 0

360v2 – 2304 000 – 144 000 + 336 000 = 0

360v2 = 2112 000 v = 76.6 m s–1

1 1 Change in KE = m(v2 – u2) = (6.8)v2 = 3.4v2 2 2

WD = Fd =

Using the conservation of energy:

Change in KE + Change in PE + Work done = 0

3.4v2 – 480 +

D – F – 7200 sin 3° = ma = 720 × 0.4 P – 0.2 × 7200 cos 3° – 7200 sin 3° = 288 76.6 P = 161 kW

3675 =0 13

v2 = 58.0 v = 7.6 ms–1 19 The roof is the hypotenuse of a right-angled triangle. If the angle to the horizontal is q, then sin q =

1 3 and cos q = 2 2

R – 720g cos 3° = 0 R = 7200 cos 3° R( ):

70 3675 (15 + 37.5) = 13 13

2 R( ):

3 161 078 – 0.2 × 7200 cos 3° – 7200 sin 3° = 0 v v =

161 078 = 88.8 m s–1 0.2 × 7200 cos3° + 7200sin 3°

Let the mass of the tile be M kg

R( ): R = Mg cos q = F = mR =

3 Mg 2

3 3  1 Mg = Mg 3  2  2

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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

2 For A:

Distance = area under graph = area of trapezium. Distance = 12 (60 + 360) = 2520 metres. 2

2 Sine rule 8 = 6 sin 30° sin D

16 =

30

30

W

6 sin D

sin D = 3 8

D = 22.02°

Angles in a triangle = 180o, so C = 127.98o.

Therefore, the bearing of D from C is 360 – 90 – 127.98 = 142.0o (1 d.p.).

3

18 N

R

1 Remember that units must be in seconds.

Resolving parallel to plane:

W sin 30° = 18

For B:

W = 36 N R a

W

6 cm a

8 cm

B

Resolving parallel to plane:

W sin 30° = 18 cos 30°

W = 18 3 2 2

W = 31.2 N

3

i tan a = 6 8

a = 36.869… ˆ C = 53.1°. Therefore AB

∴ Area of kite =

61 11

a

ii Area of triangle DCB = 20 × 6 = 60 cm2. 2

30

30

C

X

18 N

120 cm2.

A Level questions

60

tan a = 11 60

cos a = 60 61

sin a = 11 61

1 P = Fv

1 330 000 = 28 000 V

V = 47.5

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Summary REVIEW

R

a

F a

a 6.1 N

When s = 0, t = 0 ⇒ c = 0.

Therefore, when t = 5,

s = 0.3 × 25 − 0.01 × 125 = 6.25 m.

By differentiation: a = 0.6 − 0.06t.

When t = 5,

a = 0.6 − 0.06 × 5 = 0.3 m s−2.

ii a = dv = 0.6 − 0.06t dt At vmax, dv = 0 dt

Resolving perpendicular to plane:

R = 6.1 × 60 = 6 N 61

0.06t = 0.6

F = μR

= 1 × 6 = 1.5 N 4

2 Check: d v2 = −0.06 ⇒ maximum dt

Newton’s second law:

6.1 sin a − F = 0.61a

Therefore vmax = 6 – 3 = 3 m s−1.

At half of vmax, v = 1.5 m s−1.

1.5 = 0.6t − 0.03t2

6.1 × 11 − 1.5 = 0.61a 61 −0.4 = 0.61a

150 = 60t − 3t2

3t2 − 60t + 150 = 0

u = 2 m s−1  a = −0.656 m s−2  v = 0 m s−1  s = ?

v2

t=

0 = 4 + 2(−0.656) s

s = 2.69 m

= 60 ± 3600 − 1800 6

= 60 ± 1800 6

= 60 ± 30 2 6

= 10 ± 5 2

t = 2.93 s   or   t = 17.1 s

6.1 cos a = R

a = − 0.656 m s−2 =

u2

+ 2as

4 i 400 m R O

A 1400g

24 m s−1

q

300 m

B

Gain in KE from O to B = 1 × 14 000 × 242 2 = 4 032 000 J

Loss in PE = 14 000 g (300 sin q ) = 42 000 000 sin q J

ii Conservation of energy

(work done by driving force) + (loss of potential energy) = (work done against resistance) + (gain in kinetic energy) 5 000 000 + 42 000 000 sin q  = 4 032 000 + 4 800 × 700 2 392 000 sin  q= = 0.0569…   42 000 000 q  = 3.3°

5 i v = 0.6t −

s=

0.3t2

0.03t2 0.01t3

+c

t = 10 s

60 ± 60 2 − 4 ( 3)(150 ) 6

6 i a

T

T A

a

B

0.25g 0.75g

Newton’s second law:

T − 0.25g = 0.25a

1

0.75g − T = 0.75a

2

①+②

0.5g = a

a = 5 m s−2

u = 0 m s−1  a = 5 m s−2  t = 0.6 s   s = ?

s = ut + 1 at2 2

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WORKED Solutions

s = 1 × 5 × 0.62 2

s = 0.9 m ii v = u + at

2

7 − T = 0.7a 1 + 2 a = 6.4 m s−2

u = 0 m s−1  v = ?  a = 6.4 m s−2  t = 0.25 s

v = u + at

V = 5 × 0.6

=3

From 0.6 to T :  u = 3 m s−1  v = 0 m s−1  a = −10 m s−2   t = ?

= 6.4 × 0.25

= 1.6 m s−1

v = u + at

0 = 3 − 10t

t = 0.3 s

s = ut + 1 at2 2 = 1 × 6.4 × 0.252 2

Therefore: T = 0.6 + 0.3 = 0.9

= 0.2 m

After the string breaks:

iii From T to 1.6:  u = 0 m s−1  v = ?  a = 10 m s−2    t = 0.7 s

v = u + at

= 10 × 0.7

= 7 m s−1

u = 1.6 m s−1  s = 0.3 m  a = 10 m s−2  v = ?

0.9 × 3down Distance up = distance + h = 0.7 × 7 = 2.45 m, so 2 2 0.9 × 3 + h = 2.45 0.7 × 7 2 2

1.35 + h = 2.45

7 i

v2 = u2 + 2as

= 1.62 + 2 × 10 × 0.3

= 8.56

v = 2.93 m s−1

ii Before the string breaks, A has travelled 0.2 m.

After the string breaks: a

h = 1.1

R

a m = 0.2 R F

T

F A

T 0.3g

B a 0.5 m 0.7g

0.3g

Newton’s second law:

−F = 0.3a

−0.2R = 0.3a

−0.6 = 0.3a

a = −2 m s−2

u = 1.6 m s−1  s = ?  a = −2 m s−2  v = 0 m s−1

Before the string breaks:

v2 = u2 + 2as

Resolving vertically (A):

0 = 1.62 − 4s

R = 0.3g = 3 N

4s = 1.62

Newton’s second law (A):

s = 0.64 m Therefore, total distance travelled = 0.84 m.

T − 0.2R = 0.3a

T − 0.6 = 0.3a

Newton’s second law (B):

0.7g − T = 0.7a

P = 1166.7 = 1170 W

T − F = 0.3a

8 i Work done = Fd = 200g × 0.7 = 1400 J 1

ii E = Pt 1400 = P × 1.2

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Summary REVIEW

9 i

11 i

R

25

X 3200 N

7

a 3˚

24

24 000g

Newton’s second law:

X – 3200 – 24 000g sin 3o = 24 000a

When a = 0.2 m s–2 ⇒ X = 20 561 N

P = Fv = 20 561 × 25 = 514 016 W = 514 kW

7 24 7 sin α = 25 tan α =

cosα =

24 25 R 7.2 N F a

a

ii Resolving parallel to plane: X = 3200 + 24 000g sin 3o = 15 761 N

P = Fv

500 000 = 15 761v

v = 31.7 m s−1

10 Vertical component of resultant force   = 7 sin 60° − 3 sin 30° − 4

= 7 3 −3−4 2 2

=

Horizontal component of resultant force   = 5 − 3 cos 30° − 7 cos 60°

7 3 − 11 = 0.562... 2

=5 − 3 3 − 7 2 2

=

3−3 3 = −1.098 ... 2 F

0.562

a 1.098

By Pythagoras’ theorem:

F 2

F = 1.23 N(magnitude of resultant force).

0.562 tan α = = 0.5119… 1.098

Therefore the resultant force is at an angle of 152.9o from the positive x-axis (anti-clockwise).

=

0.5622

+

7.5 N

Resolving perpendicular to plane:

R = 7.5 cos a

= 7.5 × 24 25

= 7.2 N

Resolving parallel to plane:

7.2 = 7.5 sin a + F

F = 7.2 − 7.5 × 7 25

The block is at rest (in equilibrium), so F  µR

F = µR

5.1  µ × 7.2

µ

51 72

17 24

= 5.1 N

ii

R F

1.0982 7.2 N

a

a

a = 27.1°

7.5 N

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WORKED SOLUTIONS

Resolving perpendicular to plane:

Velocity (m s –1)

A – particle enters liquid B – particle hits bottom of container

R = 7.5 cos a = 7.2 N Given that the block is sliding (accelerating), the forces pulling the block down the plane must be greater than the resistive forces (pulling the block up the plane) and friction must be at its maximum value. Therefore:

12

B

10

A

F = μR = 7.2μ 0

and 7.2 + 7.5 sin a > F

1

13 i

7.2 + 7.5 × 7 > 7.2μ 25

1.36

13

Time (s)

5

a

μ < 31 24

12 sin a = 5 cos a = 12 tan a = 5 12 13 13 a R

12 i In the liquid: F

µ = 0.2

T F

a

Resolving perpendicular to the plane:

F = 30 − 3 × 5.5 = 13.5 N

R = 0.26g cos a

ii In the air: u = 0 m s−1 a = 10 m s−2 s = 5 m

t=?

v=?

v2 = u2 + 2as

= 0.48 N

v = 10 m s−1

Newton’s second law (A):

v = u + at

T = F − 0.26g sin a = 0.26a T − 0.48 − 2.6 × 5 = 0.26a 13

10 = 10t t = 1s

T − 1.48 = 0.26a

In the liquid:

v = u + at

= 2.6 × 12 = 2.4 N 13 F = μR = 0.2 × 2.4

= 2 × 10 × 5 = 100

v = 12 m s−1

a

0.26 g

3g − F = 3a

= 100 + 2 × 5.5 × 4 = 144

T B 0.54 g

Newton’s second law:

v2 = u2 + 2as

A

a

3g

u = 10 m s−1 a = 5.5 m s−2

a

s = 4m

t=?

v=?

1

Newton’s second law (B): 0.54g − T = 0.5a 5.4 − T = 0.54a

2

1 + 2 3.92 = 0.8a a = 4.9 m s −2

12 = 10 + 5.5t t = 0.36 s

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Summary REVIEW

In 1

T = 0.26 × 4.9 + 1.48 = 2.754 N = 2.75 N (3 sf.)

ii u = 0 m s−1  a = 4.9 m s−2  s = ?  t = 0.4 s = ut + 1 at2 2 1 = × 4.9 × 0.42 = 0.392 m 2

(WD by pulling force) + (loss in KE) = (gain in KE) + (WD by resistive force)

(50 cos a)250 + 1 × 60(8.52 − 3.52) 2 = 60g × 17.5 + 6 × 250

12 500 cos a + 1800 = 10 500 + 1500

cos a = 10 200 = 0.816 12 500

2=

1 a × 1.52 2

a=

16 ms−2 9

ii Speed of A at time of collision: v = u + at 16 3 8 v= × = 9 2 3

14 Conservation of energy:

Use the principle of conservation of momentum: Momentum before collision = momentum after collision mA uA + mB uB = mA vA + mB vB

3m × 8 = 3m × 4 + m × v B 3 3

8 = 4 + vB

vB = 4 m s–1

a = 35.3°

17 m s–1,

15 i Particle P: u = 2 a = 0.05 v = 5 m s–1, t = ?, s = ?

1

m s–1,

0.96

v = u + at

sin α = 0.28

cos α = 0.96

5 = 2 + 0.05t

0.28

a

t = 60 s v2

=

u2

R

0.6 N

+ 2as

52 = 22 + 2 × 0.05 × s

s = 210 m ii Particle Q: u = 0 m s–1, a = 0.05 m s–1, v = kt 3 m s–1 at time t, t = 60 s when s = 210 m ds = kt 3 v= dt

s = ∫kt 3dt

s=k

s=k

60 4 7 , ∴k = 4 108000

At B: speed = kt 3 =

7 × 603 = 14 ms−1 108000

16 i Particle A: u = 0 m s–1,  s = 2 m,  t = 1.5 s,  a = ?

1 s = ut + at 2 2

i Resolving perpendicular to the plane: 5 × 0.96 = R + 0.6 × 0.28

R = 4.63 N

t4 4

210 = k

a 0.5g

t4 +c 4

When t = 60, s = 210

a

0.5g cos α = R + 0.6 sin α

When t = 0 s = 0, ∴ c = 0

a

P

25 = 4 + 0.1s

ii The 0.6 N force is just enough to prevent P sliding down the plane, so friction acts up the plane.

Resolving parallel to the plane:

0.6 cos α + F = 0.5g sin α

0.576 + F = 1.4

F = 0.824 N iii For limiting friction, Fmax = µR

0.824 = µ × 4.63

µ = 0.1779 … µ = 0.178

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WORKED Solutions

18

For Q: u = 3 m s–1, a = 2.5 m s–2, t = T, s = x, v = ?

+

Use s = ut + Before: 5ms

−1

After: 4ms

vP m s−1

10 m s−1

Q 3m

P 7m

Q 3m

−1

P 7m

x = 3T −

1 2 at 2

1 2.5 T 2 2

x = 3T – 1.25 T 2

Use the principle of conservation of momentum: Momentum before collision = momentum after collision

Substitute this into the equation for P:

5 + 3T – 1.25T 2 = 8T – T 2

0 = 0.25T 2 + 5T – 5

mA uA + mB uB = mA vA + mB vB

−5 ± 25 + 5 = 0.9545 0.5 Note: the positive value only as T > 0.

7m × 5 – 3m × 4 = 7m × vP + 3m × 10

Now use v = u + at for both P and Q.

35 – 12 – 30 = 7vP

For P:

vP = –1 m s–1

v = 8 – 2(0.9545) = 6.091 m s–1 = 6.09 m s–1 m s–1

P has a speed of 1 direction of motion. P µ = 0.2 19 i

T =

and has reversed its

For Q: v = 3 – 2.5(0.9545) = 0.6138 m s–1 = 0.614 m s–1

Q µ = 0.25

20 i u = ?, a = –g, v = 0, s = 45

Resolving vertically (P):

v2 = u2 + 2as

RP = mP g

As P is moving, friction is maximum, so

u2 = 900

Fmax = µRP

u = 30 m s–1

Fmax = 0.2mP g = 2mP

ii Find the time, t, when the signal is 40 m above sea level.

Newton’s second law: F = mP a

s = 40, u = 30, a = –g, t = ?

2mP = mP a a = 2

m s–2

Resolving vertically (Q):

RQ = mQg As Q is moving, friction is maximum, so Fmax = µRQ

2.5 mQ = mQa a = 2.5 m s–2 8 m s−1

Q

s = ut +

40 = 30t – 5t2

t2 – 6t + 8 = 0 (t – 2)(t – 4) = 0 The signal passes the top of the cliff after 2 seconds, reaches the top of its flight then returns and passes the top of the cliff again 4 seconds after launch.

Newton’s second law: F = mQa

ii P

1 2 at 2

t = 2, t = 4

Fmax = 0.25mQg = 2.5mQ

0 = u2 – 2 × 10 × 45

3 m s−1

Total time above the cliff = 4 – 2 = 2 seconds.

iii Time above cliff = 17 seconds, so time A

B

Let time at impact be T, and the distance from B to the point of collision x.

17 = 2 time to return from highest point to top of cliff when falling.

For P: u = 8 m s–1, a = 2 m s–2, t = T, s = 5 + x, v = ?

When falling from highest point to top of cliff:

5m

1 Use s = ut + at 2 2

1 5 + x = 8T – 2T 2 2

5 + x = 8T – T 2

x

Collision

from top of cliff to highest point =

u = 0, t =

17 ,a=g 2

Use s = ut +

1 2 at 2

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Summary REVIEW

2

 17  s =0+5× = 21.25m  2 

From launch to highest point:

s=

v2 = u2 + 2as

23

12

∫0

12

12 v dt = ∫ 3 t 2 − 1 t 3 dt =  1 t 3 − 1 t 4   4 64 0 16 0 4

= ( 432 − 324 ) − ( 0 ) = 108 m

u = ?, v = 0, a = –g, s = 40 + 21.25 = 61.25

Distance travelled before P changes direction:

4

0 = u2 – 2 × 10 × 61.25

600 m

u2 = 1225

30 m s−1

u = 35 m s–1 2.5°

21

a

T

T

A

B

2.5°

a

1250 g

Work done against the resistive force = 400 × 600 = 240 000 J 0.65g 0.35g Use Newton’s second law: F = ma

Work done by the driving force = 450 000 J Increase in PE = mgh = 1250 × 10 × 600 sin 2.5 = 327 145.4 ...

For A: 0.65g – T = 0.65a

Change in KE = KE at bottom – KE at top 1 1 2 2 = × 1250 × 30 2 − × 1250 × v top = 562500 − 625v top 2 2 Work done by the driving force

6.5 – T = 0.65a For B: T – 0.35g = 0.35a

= Increase in PE + work done against the resistive force – loss in KE

T – 3.5 = 0.35a a=

T − 3.5 0.35

2 450 000 = 327 145.4 + 240000 − (562500 − 625v top ) 2 562 500 – 625v top = 327 145.4 + 240 000 – 450 000

Substitute into the equation for particle A. 6.5 − T = 0.65 ×

(T

− 3.5) 0.35

2 625v top = 562 500 – 327 145.4 – 240 000 + 450 000

= 445 355

7(6.5 – T ) = 13(T – 3.5)

2 v top = 712.567

45.5 – 7T = 13T – 45.5

vtop= 26.69 … = 26.7 m s–1

91 = 20T

24 i

T = 4.55 N

V

Total downward force on the pulley = 2 × T = 9.1 N

a=

F

960

22 i v = 0.75t2 – 0.0625t3 dv = 1.5t − 0.1875t 2 dt

1200 g

a = t(1.5 – 0.1875t)

Rate of work = power = 17 280 W

a = 0 when t = 0, t = 8

Power = Fv = 17 280

F =

When a = 0 the positive value of t is 8. ii The particle changes its direction of motion when v = 0

(

F = ma

)

2 v = 0.75t 2 − 0.0625t 3 = 3 t 2 − 1 t 3 = t 3 − 1 t , 4 4 16 4

v = 0 when t = 0 or t = 12

17 280 = 1440 12

1440 – 960 = 1200a

480 = a = 0.4 ms−2 1200

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WORKED Solutions

ii Speed is constant, V, so acceleration = 0

i Use F = ma for both particles

F = ma = 0 17 280 As above, power = Fv = 17 280, so F = V 17 280 ∴ − 960 = 0 V

For A:

V =

17 280 = 18 as required 960

T – 0.3g = 0.3a T – 3 = 0.3a T −3 a= 0.3 For B:

iii After passing point B the only force is the resistive force

0.7g – T = 0.7a

7 – T = 0.7a

Substitute for a in this equation

a = –0.8 m s–2

7 − T = 0.7 ×

u = 18 m s–1, v = 0 m s–1, s = ?, a = –0.8 m s–2

3(7 – T ) = 7(T – 3)

v2

+ 2as

21 – 3T = 7T – 21

0 = 324 – 2 × 0.8 × s

42 = 10T

F = ma

–960 = 1200a

=

u2

T −3 0.3

s = 202.5 = 203 m

T = 4.2 N

For AB, time taken = 52.5 – 22.5 = 30 seconds

Distance at constant speed of 18 m s–1 = 18 × 30 = 540 m

ii Substitute T = 4.2 N into T − 3 1.2 a= = = 4 ms−2 0.3 0.3

So distance AC = 540 + 203 = 743 m

25 Before:

u = 0, a = 4, v = 1.6, s = ?

After: ku m s–1

4u m s–1

2u m s–1

For initial motion:

ku/2 m s–1

2 = u2 + 2as v

1.62 = 8s s = 0.32 m

mP = 2 m kg

mQ = 5 m kg

mP = 2 m kg

mQ = 5 m kg

Use the principle of conservation of momentum: Momentum before collision = momentum after collision mP uP + mQ uQ = mP vP + mQ vQ

ku 2m × 4u – 5m × ku = –2m × 2u + 5m × 2 5k 8mu – 5kmu = –4mu + mu 2 5k 8mu + 4mu = 5kmu + mu 2 12 = 7.5k k = 1.6

At the point that the string breaks, for particle A:

s = 0.32 + 0.52 = 0.84, u = –1.6, a = g, t = ? 1 s = ut + at 2 2 0.84 = –1.6t + 5t2 5t2 – 1.6t – 0.84 = 0

( −1.6 )2 + 4 × 5 × 0.84

1.6 ±

t=

1.6 ± 4.4 t= 10

10

6 = 0.6 s, as only the positive value is 10 possible.

t= 

3

27 i v = 0.16t 2 − 0.016t 2

26 T

T

A

B

0.3g

0.7g

3

v = 0.16 (100 ) 2 − 0.016 (100 ) = 160 − 160 = 0

At rest at point A when t = 100.

ii For max 0.52 m

2

dv =0 dt

1 dv 3 = × 0.16t 2 + 2 × 0.016 × t = 0 dt 2

0.24t 2 + 0.032t = 0

1

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5 .16t 2

Summary REVIEW

1 1 t 2  7.5 − t 2  = 0

t = 7.52 = 56.25 seconds This is the time at which the max velocity is reached. 3 Max velocity = 0.16(56.25)2 – 0.016 (56.25)2 m s–1

= 67.5 – 50.625 = 16.9 iii Distance OA = distance travelled in the first 100 seconds. 100

100

5 ⌠ 2  1 s =  v dt =  × 0.16t 2 − × 0.016t 3  5 3  0 ⌡0

100 1  − × 0.016t 3     = ( 6400 − 5333) − 0 = 1066.6 = 1070 m 3 

4 3 T + T 5 1 5 2

8=

40 = 4T1 + 3T2

Resolving vertically at B:

T1 cos α = T2 cos θ

3 4 T = T 5 1 5 2

3T1 = 4T2 T1 =

4 T 3 2

) −Substitute intom the equation above. = ( 6400 − 5333 0 = 1066.6this = 1070

4 40 = 4 × T2 + 3T2 3

120 = 25T2 5 1 2 3 iv s = × 0.16t 2 − × 0.016t + c, when t = 0, s = 0 so c = 0 T2 = 4.8 N 5 3 4 4 3 Substitute into T1 = T2 = × 4.8 = 6.4N 016t + c, when t = 0, s = 0 so c = 0 3 3 5 2 1 Tension in AB is 6.4N and tension in BC is 4.8 N. s = × 0.16t 2 − × 0.016t 3 5 3 ii Horizontal force at When P passes through O, s = 0 4 A = T1 sin α = 6.4 × = 5.12 5 2 1 5 0 = × 0.16t 2 − × 0.016t 3 5 3 Resolving vertically at A: 5 1 3 2 T1 cos α = Fmax + 0.2g 0 = 0.064t − × 0.016t 3 3 1 6.4 × − 2 = Fmax = 3.84 − 2 = 1.84  1 52  5 2 0 = t  0.192 − 0.016t  3   Fmax = µR 1 1 0.192 1.84 = µ × 5.12 = 12 , As t ≠ 0, 0.192 − 0.016t 2 = 0, ∴ t 2 = 0.016 1.84 so t = 144 µ= = 0.359 5.12 28 C 3 29 Use tan α = and Pythagoras’ rule to T2 4 q find sin α, cos α 2 T2 2.5 q 5 3 a B 8N T a 1.5 1 a A T1 4 0.2 g N sin α = 0.6, cos α = 0.8 Use Pythagoras’ rule to find AC = 2.5 m i 12 N R1 Let T1 and T2 be the tension in AB and BC respectively. a Angle BCA = θ, and angle BAC = α as shown in the diagram, 2g 3 4 4 3 then sin θ = ,cos θ = , sin α = , cos α = 5 5 5 5 Resolving forces vertically: 0

i Resolving horizontally at B:

8 = T1 sin α + T2 sin θ

R1 = 2g + 12 sin α = 20 + 7.2 = 27.2 F  µR

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WORKED Solutions

12 cos α  µ × 27.2

9.6 µ 27.2

µ  6 as required 17

s=−

1 3 1000 t + 50t − 15 3

When t = 250 ,

12 N s = − 1 15

ii R2

s=−

(

250

)

3

+ 50

(

1 15

(

)

3

+ 50 1000 250 − = 193.7… = 194 m 3

)

250

(

)

250 −

1000 = 193.7… = 194 m 3

Extension questions 1 For the white snooker ball: a

2g R2 = 2g – 12 sin α = 20 – 7.2 = 12.8

As the block is moving F > µR

12 cos α > µ × 12.8

9.6 > µ 12.8

µ X. Here is a sample space diagram to show how this can occur.

For ball B:  a = −10 m s−2  u = u  s = 10 m  t=T–1=1s 1 Using s = ut + at 2 , 2 10 = u – 5 ⇒ u = 15 m s−1

30 = 5t2 ⇒ t = 6  (since t > 0)

Ball B hits the ground when: a = −10 m s−2  u = 15 m s−1  s = 0 m  t = ? 1 Using s = ut + at 2, 2

Second roll of dice (magnitude of Y)

0 = 15t – 5t2 ⇒ 0 = 5t(3 – t) ⇒ t = 3 (since t > 0)

1

Therefore the time between the balls hitting the ground for the first time is 3 − 6 + 1 (note the extra second because the motion of ball B began 1 second after the motion of ball A.

Therefore, time between balls hitting the ground is 4 − 6 seconds.

2

3

4

5

6

1 2 3

7 For each particle, a = 4 m s–2 and u = 0 m s–1. 1 Therefore you can use s = ut + at 2 for 2 different values of t to calculate the distance travelled by each particle.

4 5

6 P ( P accelerates to the right ) =

15 5 = 36 12

ii For P to move with a constant velocity, the resultant force must be zero. However, this does not guarantee that P is moving, since the constant velocity could be equal to zero. Therefore the maximum probability is found by assuming that when the dice rolls are equal, P has a constant velocity that is not equal to zero. P ( P moves with a constant velocity )

6 1 = 36 6

6 i Let T be the number of seconds after which ball A reaches the required height. Therefore ball B reaches the required height after T – 1 seconds.

20 = 5T 2 ⇒ T = 2 s   (since T > 0)

ii Ball A hits the ground when: a = 10 m s−2  u = 0 m s−1  s = 30 m  t = ? 1 Using s = ut + at 2, 2

5 This question combines statistics and mechanics.

Given the components of F are 20 N west and 20 N south, then F must be acting south-west. i

Using s = ut +

F = 20 2 + 20 2 = 28.3 N

1 2 at , 2

For ball A:  a = 10 m s−2  u = 0 m s−1  s = 20 m   t = T

Particle 1 is in motion for 19 seconds, 1 so s = × 4 × 192. 2 Particle 2 is in motion for 18 seconds, 1 so s = × 4 × 182. 2 Particle 3 is in motion for 17 seconds, 1 so s = × 4 × 172. 2 … Particle 18 is in motion for 2 seconds, 1 so s = × 4 × 22. 2 Particle 19 is in motion for 1 seconds, 1 so s = × 4 × 12. 2 So the cumulative distance travelled by the particles is 2(12 + 22 + 32 + … + 182 + 192) Using the result provided, we can sum the squares, so the result is 19 Cumulative distance travelled = × 20 × 39 3 = 4940 metres.

81

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Summary REVIEW

8 Integrating:

m = t3 − 2 t + d When t = 0, m = 2 ⇒ d = 2. The initial momentum (when t = 0) is 1 × 2 = 2 kg m s−1. The final momentum (when t = 3) is (18 – 6 + 1) (27 – 6 + 2) = 299 kg m s−1. Therefore the change in momentum is 297 kg m s−1.

11 i Using Newton’s second law for P: −2 = 2a ⇒ a = −1 m s−2 So for P:  a = −1 m s−2  u = u m s−1  s = 5u m  t = T

9 Assume the particles collide after T seconds.

For particle A, Newton’s second law can be used to calculate the acceleration.

mgsin 30° = ma ⇒ a = 5 m s−2

So:  a = 5 m s−2  u = 0 m s−1  s = (10 – h) m  t = T. 1 2 at , 2

Using s = ut +

1 1 10 − h = × 5T 2 ⇒ h = 10 − × 5T 2. 2 2

For particle B, Newton’s second law can be used to calculate the acceleration.

1

−mgsin 30° = ma ⇒ a = −5 m s−2

So:  a = −5 m s−2  u = 8 m s−1  s = h m  t = T. 1 Using s = ut + at 2 , 2 1 h = 8T − × 5T 2. 2

Equating 1 and 2

10 −

2

()

2

1 1 u mu 2 = m 2 2 2

1 2 1 2 10u u = u + 2 8 16

8u2

=

3u2

– 5u = 0

2u2

( )

+ 10u

5 3 5 Since u > 0, you can say that u = ms−1. 3

u = 0 or u =

J. ( ) ( ) − ( 65 )  = 125 24

5  5 1 3×  2 3  3

Loss in KE is

v −u Using v = u + at, then t = a

82

1 5u = uT − T 2 ⇒ T 2 – 2uT + 10u = 0. 2 By the quadratic formula:

T =

2u ± 4u 2 − 4(1)(10u) 2

T =

2u ± 4u 2 − 40u 2

T =

2u ± 2 u 2 − 10u 2

T = u + u 2 − 10u (this is the time taken for P to reach the initial position of Q).

For Q:  a = −10 m s−2  u = v m s−1  v = −v m s−1  t = T1. Using v = u + at,

v (this is the time 5  taken for Q to return to the ground).

You require T1 > T

–v = v – 10T1 ⇒ T1 =

v > u + u 2 − 10u ⇒ v > 5u + 5 u 2 − 10u . 5 ii The discriminant must be positive ⇒ u2 – 10u > 0 ⇒ u > 10. If u is not greater than 10, P will not pass the initial position of Q. 12 Let the speed of the particle at A be u m s−1 and the speed of the particle at B be v m s−1.

u + mg 16

2

Using s = ut +

From change in KE: 1 m m v 2 − u 2 = 25 ⇒ (v + u )(v − u ) = 25. 2 2

So: m(v – u) (v + u) = 50.

1

From change in momentum: mv – mu = 10 ⇒ m (v – u) = 10.

2

u(3u – 5) = 0

1 2 at , 2

1 1 × 5T 2 = 8T − × 5T 2 ⇒ 10 = 8T ⇒ 2 2 T = 1.25 s 1 Therefore, h = 10 − × 5 × 1.252 = 6.09 m 2

10 By conservation of energy,

5 u 5 , v = = , a = −10 3 2 6 1 1 then t = , or T = 12 12 where u =

2 v = t 3 − 2t + c  When t = 0,  v = 1 ⇒ c = 1. 3

(

)

Substituting 2 into 1 gives

10 (v + u) = 50 ⇒ u + v = 5.

To find the distance travelled by the particle, u+v use s = t. 2

You know that u + v = 5 and that t = 5.

s=

2

( )

( 52 )5 = 12.5 metres

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