Chapter 1 Motion

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Heinemann Physics 12(3e) Copyright © Pearson Education Australia 2008 Teacher’s Resource and Assessment Disk (a division of Pearson Australia Group Pty Ltd) ISBN 9781442501263 1 Worked solutions Chapter 1 Motion 1.1 Mechanics review 1 S, V, S, V, S, V, V, S 2 C is the correct answer. The units associated with this area = (m s –1 )(s) = m, which are the units of displacement. 3 a i x = area under graph from t = 0 to t = 4 s = 40 m ii x = area under graph from t = 4 s to t = 12 s = 160 m iii x = area under graph from t = 12 s to t = 28 s = 160 m b The displacement after 28 s = 40 m + 160 m + 80 m – 80 m = 200 m c The total distance travelled after 28 s = 40 m + 160 m + 80 m + 80 m = 360 m d The average speed = total distance travelled/time = 360 m/28 s = 13 m s –1 e Acceleration = rate of change of velocity = 0 m s –2 4 80 km h –1 = 22.2 m s –1 and 100 km h –1 = 27.8 m s –1 Let the required time = t. Then at time t the distance travelled by the car: d = v × t = 22.2t. The distance travelled by the policeman in the first 10.0 s d = 0.5(u + v)t = 0.5 × 22.2 × 10 = 111 m, and in the next 5.0 s he travels a distance of: 0.5 × (22.2 + 27.8) × 5.0 = 125 m. Then at time t the policeman has travelled a distance of 236 + 27.8(t – 15). When the policeman catches up with the car, it is: 22.2t = 236 + 27.8)(t – 15) and so t = 32.5 s. 5 a v = u + at = 0 = 9.8 × 1.00 = 9.8 m s –1 down b x = ut + 0.5at 2 = 0 + 0.5 × 9.8 × 2.00) 2 = 19.6 m c v 2 + u 2 + 2ax = 0 + 2(9.8 × 10.0) = 14.0 m s –1 d x = ut + 0.5at 2` 78 = 0 + 0.5 × 9.8t 2 and t = 4.0 s Answer: after 4.0 s e i 9.8 m s –2 down ii 9.8 m s –2 down 6 a F h = F cos θ = 120 × cos 35° = 120 × 0.819 = 98 N

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Transcript of Chapter 1 Motion

Page 1: Chapter 1 Motion

Heinemann Physics 12(3e) Copyright © Pearson Education Australia 2008 Teacher’s Resource and Assessment Disk (a division of Pearson Australia Group Pty Ltd) ISBN 9781442501263

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Worked solutions

Chapter 1 Motion

1.1 Mechanics review 1 S, V, S, V, S, V, V, S 2 C is the correct answer. The units associated with this area = (m s–1)(s) = m, which are the units

of displacement. 3 a i x = area under graph from t = 0 to t = 4 s = 40 m

ii x = area under graph from t = 4 s to t = 12 s = 160 m iii x = area under graph from t = 12 s to t = 28 s = 160 m

b The displacement after 28 s = 40 m + 160 m + 80 m – 80 m = 200 m c The total distance travelled after 28 s = 40 m + 160 m + 80 m + 80 m = 360 m d The average speed = total distance travelled/time = 360 m/28 s = 13 m s–1 e Acceleration = rate of change of velocity = 0 m s–2

4 80 km h–1 = 22.2 m s–1

and 100 km h–1 = 27.8 m s–1 Let the required time = t. Then at time t the distance travelled by the car: d = v × t = 22.2t. The distance travelled by the policeman in the first 10.0 s d = 0.5(u + v)t = 0.5 × 22.2 × 10 = 111 m, and in the next 5.0 s he travels a distance of: 0.5 × (22.2 + 27.8) × 5.0 = 125 m. Then at time t the policeman has travelled a distance of 236 + 27.8(t – 15). When the policeman catches up with the car, it is: 22.2t = 236 + 27.8)(t – 15) and so t = 32.5 s.

5 a v = u + at = 0 = 9.8 × 1.00 = 9.8 m s–1 down

b x = ut + 0.5at2 = 0 + 0.5 × 9.8 × 2.00)2 = 19.6 m c v2 + u2 + 2ax = 0 + 2(9.8 × 10.0) = 14.0 m s–1 d x = ut + 0.5at2` 78 = 0 + 0.5 × 9.8t2 and t = 4.0 s Answer: after 4.0 s e i 9.8 m s–2 down ii 9.8 m s–2 down

6 a Fh = F cos θ = 120 × cos 35° = 120 × 0.819 = 98 N

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Worked solutions Chapter 1 Motion

Heinemann Physics 12(3e) Copyright © Pearson Education Australia 2008 Teacher’s Resource and Assessment Disk (a division of Pearson Australia Group Pty Ltd) ISBN 9781442501263

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b Fv = F sin θ = 120 × sin 35° = 120 × 0.574 = 69 N 7 a vav = d/t = 3.2/2.4 = 1.3 m s–1

b u = 0, x = 3.2 m, t = 2.4 s, a = ? x = ut + ½at2 3.2 = 0 + 0.5 × a × 2.42 a = 6.4/2.42 = 1.1 m s–2 c v = u + at = 0 + 1.11 × 2.4 = 2.7 m s–1 d u = 0, a = 1.1 m s–2, x = 1.6 m, v = ? v2 = u2 + 2ax = 0 + 2 × 1.1 × 1.6 v = 1.9 m s–1

8 a Taking down as positive:

u = 0, a = 9.8 m s–2, x = 2.00 m, v = ? v2 = u2 + 2ax = 0 + 2 × 9.8 × 2.00 = 39.2 v = 6.3 m s–1 b Same as for golf ball: v = 6.3 m s–1 c a = 9.8 m s–2, x = –1.5 m, v = 0, u = ? v2 = u2 + 2ax 0 = u2 + 2 × 9.8 × –1.5 u = 5.4 m s–1 d The tomato had the greater speed change: Δv = v – u = 0 – 6.3 = –6.3 m s–1. e The golf ball experienced the greater velocity change: Δv = v – u = –5.4 – 6.3 = –11.7 m s–1 or 11.7 m s–1 up

9 a u = 8.0 m s–1 , a = –2.0 m s–2 , v = 0, x = ?

v2 = u2 + 2ax 0 = 8.02 – 2 × 2.0 × x x = 16 m b v = u + at 0 = 8.0 – 2.0t t = 4.0 s c u = 8.0 m s–1 , a = –2.0 m s–2 , t = 5.0 s, v = ? v = u + at = 8.0 – 2.0 × 5.0 = –2.0 m s–1, i.e.2.0 m s–1 down ramp

10 a The velocity of the plane is found from the vector addition of 100 m s–1 south +25 m s–1 west = 103 m s–1 south 14° west.

b The pilot should steer the plane in a direction south 14° east to compensate for the effect of the crosswind.

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Worked solutions Chapter 1 Motion

Heinemann Physics 12(3e) Copyright © Pearson Education Australia 2008 Teacher’s Resource and Assessment Disk (a division of Pearson Australia Group Pty Ltd) ISBN 9781442501263

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1.2 Newton’s laws of motion 1 When the car is at rest, its tendency as described by Newton’s first law is to remain at rest. A

large force is needed to overcome its inertia and start the car rolling. Once the car is rolling, however, its tendency is to continue rolling. Only a small pushing force, to overcome resistance forces, is needed to maintain the car’s motion.

2 No. Phil stayed where he was as the tram moved forwards. This is an example of the law of

inertia. Objects will remain at rest unless forces act to change the motion. 3 Forces are balanced, so air resistance force is equal in magnitude to the weight of the ball.

Fa = Fg = mg = 0.010 × 9.8 = 0.098 N upwards 4 C is the correct answer. This is an example of Newton’s third law. 5 a Constant speed, so forces are balanced: Fd = 45 N.

b ΣF = ma = 80 × 1.5 = 120 N Fd – 45 = 120 Fd = 165 N

6 a a = 0, v = 7.5 m s–2, t = 5.0, a = ?

v = u + at 7.5 = 0 + 5.0a a = 1.5 m s–2 b ΣF = ma = 80 × 1.5 = 120 N c Constant speed, so forces are balanced, i.e. ΣF = 0. The frictional force will equal 60 N.

7 a Constant speed, so net force is zero.

b The horizontal component of the pulling (tension) force is in balance with the frictional force of 60 N.

Th = T cos 25° = 60 T = 60/0.91 = 66 N c The rope is exerting a force of 66 N on Matt.

8 a According to Newton’s third law, the forces are equal in magnitude.

b According to Newton’s third law, the forces are opposite in direction. 9 The trolley will move. The horizontal forces acting on the trolley (pushing force and friction)

are unbalanced. The pushing force is greater than friction, so the trolley will accelerate forwards.

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Worked solutions Chapter 1 Motion

Heinemann Physics 12(3e) Copyright © Pearson Education Australia 2008 Teacher’s Resource and Assessment Disk (a division of Pearson Australia Group Pty Ltd) ISBN 9781442501263

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a The reaction force is the force that the ball exerts on the racquet.

b The reaction force is the downward force that the pig exerts on the ground. c The reaction force is the forward force that the wardrobe exerts on the ground. d The reaction force is the gravitational force of attraction that the whale exerts on the

Earth.

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Worked solutions Chapter 1 Motion

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1.3 The normal force and inclined planes 1 A, the frictional force is opposite to the velocity. 2 C, perpendicular to the slope. 3 ΣF = 0 so Ff = Fgsinθ = 100 × 9.8 × sin30° = 490 N up the hill 4 Acceleration a = g sin θ = 9.8 sin 30° = 4.9 m s–2 5 Acceleration is not affected by mass if there is no friction. 6 a FN = mg cos θ = 60 × 9.8 × cos 65° = 250 N

b The resultant force parallel to ramp = mg sin θ = 60 × 9.8 sin 65° = 530 N c a = ΣF/m = 530/60 = 8.9 m s–2 down the ramp

7 a u = 0, x = 5.0 m, a = 8.9 m s–2, v = ?

v2 = u2 + 2ax = 0 + 2 × 8.9 × 5.0 = 89 v = 9.4 m s–1 b

8 ΣF = 0 so forces parallel to incline are balanced.

Ff = mg sinθ = 530 N up the ramp 9 a FN = mg cosθ = 0.050 × 9.8 cos15° = 0.47 N

b a = gsinθ = 9.8 × sin 15° = 2.5 m s–2 a = –2.5 m s–2, u = 4.8 m s–1, x = 1.75 m, v = ?

v2 = u2 + 2ax = 4.82 – 2 × 2.5 × 1.75 = 14.3

v = 3.8 m s–1 10 a When the ball is falling toward the floor, it has not yet made contact with the floor and

FN = 0. Therefore Fg > FN. b When the ball has just made contact with the floor, it will start to slow down. This means

that the upward force that the floor exerts on the ball (FN) must be greater than the downward force of gravity (Fg).

c At its point of maximum compression, the ball is stationary but has an upward acceleration. This indicates that the net force is upwards and therefore FN > Fg.

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Worked solutions Chapter 1 Motion

Heinemann Physics 12(3e) Copyright © Pearson Education Australia 2008 Teacher’s Resource and Assessment Disk (a division of Pearson Australia Group Pty Ltd) ISBN 9781442501263

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1.4 Projectile motion 1 (Taking down as positive)

a x = ut + 0.5at2 then 4.9 = 0 + 0.5 × 9.8t2 and t = 1.0 s b x = v × t = 20 × 1.0 = 20 m c The acceleration of the ball is constant at any time during its flight, and is equal to the

acceleration due to gravity = 9.8 m s–2 down d After 0.80 s, the ball has two components of velocity: vx = 20 m s–1 and vy = u + at = 0 + 9.8 × 0.80 = 7.84 m s–1 The speed of the ball at 0.80 s is given by: [(20)2 + (7.84)2]½ = 21.5 m s–1 e The ball will strike the ground 1.0 s after it is struck. Then vx = 20 m s–1 and vy = u + at = 0 + 9.8 × 1.0 = 9.8 m s–1 The speed of the ball at 1.0 s is given by: [(20)2 + (9.8)2] ½ = 22.3 m s–1

2 (Taking down as positive)

a The horizontal velocity of the ball remains constant and vx = 10 m s–1. b v2 = u2 + 2ax and vy

2 = 02 + 2 × 9.8 × 1.0 and vy = 4.4 m s–1 down c v = [(10)2 + (4.43)2] ½ = 10.9 m s–1 at 24° to the horizontal, where the angle is determined from tan θ = 4.43/10 = 0.44 and θ = 24° d x = ut + 0.5at2 and 1.0 = 0 + 0.5 × 9.8 × t2 so t = 0.45 s e Horizontal distance = vh × t = 10 × 0.45 = 4.5 m f

3 a vx = 28 cos 30° = 24 m s–1 and remains constant throughout the flight. b 24 m s–1 c 24 m s–1

4 (taking up as positive)

a vy = 28 sin 30° = 14 m s–1 up b vy = u + at = 14 – 9.8 × 1.0 = 4.2 m s–1 up

c The time for the ball to reach its maximum height is determined from v = u + at. Then at maximum height, the vertical velocity of the ball = 0

and v = u + at

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Worked solutions Chapter 1 Motion

Heinemann Physics 12(3e) Copyright © Pearson Education Australia 2008 Teacher’s Resource and Assessment Disk (a division of Pearson Australia Group Pty Ltd) ISBN 9781442501263

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0 = 14 – 9.8t and t = 1.43 s

Therefore at t = 2.0 s the ball is 0.57 s into its downward flight. vy = u + at = 0 + 9.8 × 0.57 = 5.6 m s–1 down

5 a The time for the ball to reach its maximum height is determined from v = u + at.

Then at maximum height, the vertical velocity of the ball = 0 and 0 = 14 – 9.8 × t

and t = 1.43 s b v2 = u2 + 2ax then 0 = 142 – 2 × 9.8 × x and x = 10 m c The acceleration of the ball is constant at any time during its flight, and is equal to the

acceleration due to gravity = 9.8 m s–2 down. 6 a The minimum speed will occur when the vertical components of the ball’s velocity = 0,

i.e. at the maximum height. b The minimum speed of the ball during its flight occurs at the maximum height, and is

equal to the horizontal component of the ball’s velocity = 24.2 m s–1. c The minimum speed of the ball during its flight occurs at the maximum height at

t = 1.43 s. d

7 a The flight of the ball is symmetrical. Therefore the time for it to reach the ground after launching =2 × 1.43 = 2.86 s.

b The flight of the ball is symmetrical. Therefore the ball will strike the ground at the same velocity as that when it was launched: 28 m s–1 at an angle of 30° to the horizontal.

c Horizontal range: xh = vh × t = 24.2 × 2.86 = 69.2 m

8 C is the correct answer. Air resistance is a force that would be acting in the opposite direction to

the horizontal velocity of the ball, thereby producing a horizontal deceleration of the ball during its flight.

9 a The initial Ek of the ball = ½mv2 = 0.5 × 0.250 × 162 = 32 J

At its maximum height Ek = 16 J, then loss in Ek = 32 J – 16 J = 16 J = 0.250 × 9.8 × h and maximum height h = 6.53 m b v2 = u2 + 2ax in vertical component and 0 = u2 – 2 × 9.8 × 6.53 then the initial vertical velocity u = 11.3 m s–1 up c At its maximum height the velocity of the ball is equal to its horizontal component vx and

Ek = 16 = 0.5 × 0.250vx2

and vx = 11.3 = 16 cos θ then cos θ = 0.706 and θ = 45° d vx = 16 cos 45° = 11.3 m s–1

Page 8: Chapter 1 Motion

Worked solutions Chapter 1 Motion

Heinemann Physics 12(3e) Copyright © Pearson Education Australia 2008 Teacher’s Resource and Assessment Disk (a division of Pearson Australia Group Pty Ltd) ISBN 9781442501263

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vy = 11.3 – 9.8 × 1.0 = 1.51 m s–1 then the speed of the ball at t = 1.0 s = [(11.3)2 + (1.51)2]½ = 11.4 m s–1 e The horizontal displacement of the ball at t = 1.0 s = 11.3 × 1.0 = 11.3 m, while the vertical displacement at t = 1.0 s using x = ut + 0.5at2 = 11.3 × 1.0 – 0.5 × 9.8 × 1.02 = 6.41 m The resultant displacement after 1.0 s = [(11.3)2 + (6.41)2] ½ = 13.0 m The angle θ of the displacement from the horizontal is given by: tan θ = 6.41 m/11.31 m = 0.5668 and θ = 30° f v = u + at then at maximum vertical height vy = 0 therefore 0 = 11.3 – 9.8t and t = 1.15 s Since flight is symmetrical, time of flight T = 2 × 1.15 = 2.30 s. g Horizontal distance = (horizontal speed)(time) = 11.3 × 2.31= 26.1 m

10 a Taking down as positive and the top of the ramp as the zero position. Using the vertical

component and finding initial velocity by analysing motion to maximum height: a = 9.80 m s–2, t = 1.5 s, v = 0, u = ? v = u + at 0 = u + 9.80 × 1.50 uv = –14.7 m s–1; i.e. 1.47 m s–1 up Trigonometry can be used to determine initial speed: v = uv/sin 40° = 14.7/0.643 = 22.9 m s–1 b Using vertical component: a = 9.80 m s–2, t = 1.50 s, u = –14.7 m s–1, v = 0, x = ? x = ½(u + v)t = 0.5 × –14.7 × 1.50 = 11.0 m c Using vertical component: u = –14.7 m s–1, a = 9.8 m s–2, x = 10 m, t = ? First find final vertical velocity (to avoid quadratic equation): v = ? v2 = u2 + 2ax = (–14.7)2 + 2 × 9.80 × 10 = 412 v = 20.3 m s–1 Now find total flight time: t = ? v = u + at 20.3 = –14.7 + 9.80 × t t = 3.57 s

Page 9: Chapter 1 Motion

Worked solutions Chapter 1 Motion

Heinemann Physics 12(3e) Copyright © Pearson Education Australia 2008 Teacher’s Resource and Assessment Disk (a division of Pearson Australia Group Pty Ltd) ISBN 9781442501263

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Chapter review 1 a Δv= v – u = 3.0 – 3.5 = -0.5 m s–1

b Δv= v – u = 3.0 m s–1 up – 3.5 m s–1 down = 6.5 m s–1 up 2 a The force that the trampoline exerts on David is initially very small, but increases in

strength until the point of maximum extension. The reverse process then occurs. This force is usually called the normal force.

b D, this is an example of Newton’s third law. 3 a When the arrow reaches its maximum height, its vertical velocity = 0.

Then v = u + at and 0 = 100 – 9.8t hence t = 10.2 s b x = ut + 0.5at2 = 0 + 0.5 × 9.8 × 10.22 = 510 m c The acceleration of the arrow is constant during its flight and is equal to 9.8 m s–2 down.

4 a u= 60/3.6 = 16.7 m s–1 , v = 0, x = 15 m, a = ?

v2 = u2 + 2ax 0 = 16.72 + 2 × a × 15 a = -9.3 m s–2 Magnitude of a = 9.3 m s–2 b v = u + at = 16.7 – 9.3 × 1.5 = 2.8 m s–1 c ΣF = FB = ma = 120 × –9.3 = 1120 = 1.1 kN

5 a C, the ground speed of the plane is the sum of its air speed and the wind speed.

vg = vp + vw = 500 km h–1 north + 100 km h–1 west = √(5002 + 1002) = 510 km h–1 at 349° T b vg = vp + vw vp = vg – vw = 500 km h–1 north – 100 km h–1 west = 500 km h–1 north + 100 km h–1 east Using vectors and trigonometry, this gives vp = 510 km h–1 at 11° T so D is correct.

6 a Using the formula x = ut + 0.5at2,

then 2.0 = 0 + 0.5 × 9.8t2 and t = 0.64 s for both b For ball X: horizontal velocity = 5.0 m s–1, vertical velocity on striking ground v = u + at = 0 + 9.8 × 0.639= 6.26 m s–1 then the speed of ball X just before it strikes the ground = [(5.0)2 + (6.26)2]½ = 8.0 m s–1 c For ball Y: horizontal velocity = 7.5 m s–1, vertical velocity on striking ground v = u + at = 0 + 9.8 × 0.639 = 6.26 m s–1 then the speed of ball Y just before it strikes the ground = [(7.5)2 + (6.26)2] ½ = 9.8 m s–1 d Horizontal range x = vt = 7.5 × 0.639 = 4.8 m for ball Y Horizontal range x = vt = 5.0 × 0.639 = 3.2 m for ball X Difference in range = 4.8 – 3.2 = 1.6 m

7 At point Q, the vertical velocity = 0

Then if u = initial vertical velocity of Vortex,

Page 10: Chapter 1 Motion

Worked solutions Chapter 1 Motion

Heinemann Physics 12(3e) Copyright © Pearson Education Australia 2008 Teacher’s Resource and Assessment Disk (a division of Pearson Australia Group Pty Ltd) ISBN 9781442501263

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0 = u2 – 2 × 9.8 × 4.0 and u = 8.85 m s–1 Since u = 10 sin θ, then θ = 62.3°

8 At point Q the speed of the Vortex is equal to the horizontal component of velocity

= 10 cos 62.3° = 4.6 m s–1 9 D, the acceleration of the Vortex at point Q = acceleration due to gravity = 9.8 m s–2 down 10 Horizontal distance X = v × t = 4.6T,

where T = time of flight From v = u + at, 0 = 8.85 – 9.8t and t = 0.90 s and T = 2 × 0.90 = 1.80 s Then X = v × t = 4.65 × 1.80 = 8.4 m

11 a vx = 8.0 cos 60° = 4.0 m s–1

b vy = 8.0 sin 60° = 6.9 m s–1 c v = u + at then 0 = 6.93 – 9.8t and t = 0.71 s d v2 = u2 + 2ax then 0 = 6.932 – 2 × 9.8x and x = 3.9 m e Using the vertical component: u = 6.9 m s–1, a = –9.80 m s–2, x = –1.5 m, t = ? x = ut + ½at2 –1.5 = 6.9t – 4.9t2 4.9t2 – 6.9t – 1.5 = 0 Taking positive solution gives t = 1.6 s. f Using horizontal component: vy = d/t 6.9 = d/1.6 d = 6.4 m

12 At its maximum height the speed of the shot = vx = 4.0 m s–1 since vy = 0 13 Minimum Ek = 0.5 × 2.0 × 4.02 = 16 J 14 At maximum height the acceleration of the shot = acceleration due to gravity

= 9.8 m s–2 down 15 C is the correct answer.

45˚ is the optimum angle for achieving the maximum range. 16 Ek = ½ mv2 = 0.5 × 0.140 × 122 = 10 J so A is correct. 17 Ug = mgh = 0.140 × 9.8 × 2.0 = 2.7 J 18 Mechanical energy = 10 + 2.7 = 12.7 J

At maximum height: Ek + Ug = 12.7 Ek + (0.140 × 9.8 × 4.5) = 12.7 Ek = 12.7 – 6.2 = 6.5 J

Page 11: Chapter 1 Motion

Worked solutions Chapter 1 Motion

Heinemann Physics 12(3e) Copyright © Pearson Education Australia 2008 Teacher’s Resource and Assessment Disk (a division of Pearson Australia Group Pty Ltd) ISBN 9781442501263

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½ mv2 = 6.5 v2 = 92.9 v = 9.6 m s–1 so B is correct.

19 At maximum height v = 7.5 cos 36° = 6.1 m s–1

and R = 3.78 × 10–5 × 6.12 = 1.4 × 10–3 N The shot is launched at 7.5 m s–1, so Fa = 3.78 × 10–5 × (7.5)2 = 2.1 × 10–3 N.

20 Fg/R(max) = 19.6/2.1 × 10–3 = 9.2 × 103 : 1

This indicates that the gravitational force is 9200 greater than the drag force. It is evident that air resistance could validly be ignored when analysing the motion of the shot-put.