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Answers to End-of-chapter questions 1 Kinematics – describing motion 1 a 120 × 2 = 4.0 km 60 b The car’s direction of motion keeps changing. Hence its velocity keeps changing. In the course of one lap, its displacement is zero so its average velocity is zero. Its average speed remains constant. c 1273 ≈ 1300 m 2 a 1000 m b 1000 m at an angle 53º W of N c 16.7 m s−1 at an angle 53º W of N 3 a b c d e

17.2 km 15 200 m at an angle 8º E of N 2000 s 8.6 m s−1 7.6 m s−1

4 2.6 m s−1 at an angle 23º E of N

AS and A Level Physics © Cambridge University Press

Chapter 1: Answers to end-of-chapter questions

Answers to Exam-style questions 1 Kinematics – describing motion 1 a Distance in a (particular) direction b When athlete returns to his original position/the start (direct) distance from original position is zero. 2

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boy

s / m 40 38 36 35

4 a Vector quantities have direction and scalar quantities do not. One example of a vector, e.g. velocity, acceleration, displacement, force One example of a scalar, e.g. speed, time, mass, pressure

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100 km h–1

girl

30 25 N

20 15 10

500 km h–1

resultant

5 0 0

1

2

3

4

5

6

7

8

9 10 11 12 t/s

a straight line from t = 0, s = 0 to t = 12, s = 36 b straight line from t = 0, s = 0 to t = 5, s = 10 straight line from t = 5, s = 10 to t = 12, s = 38 c 10 s 3 a Each second it travels a constant distance At least two examples: 108 − 84 = 24 cm, 84 − 60 = 24 cm, 60 − 36 = 24 cm b speed = d = 24 t 0.1 speed = 240 cm s−1 c 108 + 2 × 24 156 cm

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b Correct vectors drawn and labelled Scale stated and diagram of sufficient size Resultant velocity 510 (±5) km h−1 11° W of N or a bearing of 349° (±3°) c 0.25 × 510 = 128 ≈ 130 km 11° W of N 5 a

velocity of aircraft B

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7.5 m s–1

15 m s–1

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A

Correct vector diagram Velocity of aircraft in still air in easterly direction or calculation b tout = 5000 = 333 s and treturn = 5000 15 13.5 = 370 s

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total time = 703 or 704 s or 703.7 s average speed = 10000 = 14.2 m s−1 703.7

AS and A Level Physics © Cambridge University Press

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Chapter 1: Answers to exam-style questions

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Answers to End-of-chapter questions 2 Accelerated motion motio 1 100 m 2 Train slows to rest and covers a distance of 2500 m. 3 b 20.4 ≈ 20 m; −56.4 ≈ −56 m c 4.08 ≈ 4.1 s 4 a b c d

800 m 1.25 m s−2; 750 m 5 s (t = 25 s) 1000 m

5 a 2.8 m s−1; 0.57 s b 4.85 m s−1; 2.77 m

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Chapter 2: Answers to End-of-chapter questions

Answers to Exam-style questions 2 Accelerated motion 1 a 0.200 = 12 × 9.81 × t 2

[1]

0.202 ≈ 0.20 s [1] b i v2 = u2 + 2as leading to 2.912 = 1.922 + 2a × 0.25 [1] a = 9.56 m s −2 [1] ii Air resistance [1] Acts in the opposite direction to the velocity and so reduces the acceleration. [1] 2 a i ii iii iv

b i ii iii

Ball travels upwards (or reverses direction) on bouncing. [1] In both cases the ball is accelerating due to gravity only. [1] Initial height of the ball above the ground [1] Ball does not bounce as high as initial position [1] or (kinetic) energy is lost (as heat/internal energy) during the bounce. [1] v 2 = u 2 + 2as leading to v2 = 2 × 9.81 × 1.2 [1] v = 4.85 m s−1 [1] v2 = 2 × 9.81 × 0.8 [1] −1 v = 3.96 m s [1] v = u + at leading to 4.85 = −3.96 + a × 0.16 [1] a = 55.1 ≈ 55 m s − 2 [1] Upwards direction [1]

3 a Tangent drawn at t = 0.7 s and gradient of graph determined. a = 0.8 (±0.1) m s − 2 b Acceleration is constant from t = 0 to t = about 0.5 s Acceleration then decreases. Gradient constant at first then decreases.

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c Area under the graph used [1] Correct method, e.g. trapezium rule or squares counted [1] Distance = 0.202 (±0.002) m [1] d Random errors – the points are either side of the line [1] Systematic errors – the whole line is shifted up or down [1] 4 a v2= u2 + 2as leading to 0 = u2 − 2 × 2 × 140 [1] u = 23.7 or 24 m s−1 [1] 23.7 b t= v= = 0.39 s [1] s 60 T he reaction time is approximately 0.4 s so the driver was alert. [1] 100000 −1 −1 = 27.8 m s [1] c 100 km h = 60× 60 T he driver was not speeding as his speed (24 m s−1) is less. [1] 5 a Constant gradient [1] b i 1.55 (±0.05) s [1] ii Area under graph calculated between t = 0 and t = 1.55 s [1] 1.55 15 × = 11.6 ≈ 12 m [1] 2 iii Area between t = 1.55 s and t = 4.1 s [1] 31.8 ≈ 32 m; accept error carried forward from time in i [1] c i T he initial speed of the ball or the hot-air balloon is 15 m s−1 [1] ii T he acceleration is in the opposite direction to the initial speed of the ball, or the acceleration due to gravity is downwards and the ball initially rises. [1] 6 a v2 = u2 + 2as leading to 202 = 0 + 2 × 9.81 × s s = 20.4 or 20 m b v = u + at leading to 20 = 0 + 9.81 × t t = 2.04 or 2.0 s c distance = 80 × 2.04 = 163 or 160 m

Chapter 2: Answers to Exam-style questions

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Answers to End-of-chapter questions 3 Dynamics – explaining motio motion 1 a 6440 ≈ 6400 N b 656 ≈ 660 kg 2 a 112 N b 388 N c 5.54 m s−2 3 a At first the only force is the weight but as its speed increases viscous drag increases. When viscous drag equals weight the acceleration is zero and the speed is constant. b Put rubber bands around the cylinder the same vertical distance apart along the cylinder. Time the ball between the bands. When terminal velocity is reached the time taken between successive bands will be constant. 4 a 6 × 10−4 m s−1 b 2 × 10−3 m s−1 c 2 × 10−3 m s−1 5 a i The Earth, ii upwards, iii gravitational force b i The Earth or the ground under the man, ii downwards, iii contact force

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Chapter 3: Answers to End-of-chapter questions

Answers to Exam-style questions 3 Dynamics – explaining motion [1] 1 a i F = ma = 1100 × 1.5 = 1650 N ii 1650 + 600 = 2250 N (so that resultant force is still 1650 N) [1] 2 2 1 1 [1] b s = ut + 2 at = 2 × 1.5 × 10 s = 75 m

[1]

2 a 1.5 m s−1 [1] b Constant velocity reached when weight = upward force due to air resistance [1] Air resistance increases with speed. [1] Air resistance is less than weight of metal ball even at 2.5–3.0 m s−1 [1] c Initial acceleration is acceleration due to gravity or 9.81 m s−2. [1] Initially neither ball has any air resistance. [1] 3 a F = ma = 1200 × 8 2 F = 4800 N b i kg m s−2 ii kg m−1 iii 4800 = b × 502 b =1.92 (kg m−1 or N s2 m−2) iv Sketch graph showing increasing gradient and force values marked at speeds of 0 and 50 m s−1 Resistive force increases with speed so resultant force and acceleration decrease.

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4 a Mass is the amount of matter in a body. Weight is a force due to gravity acting on the body. b Body moves to Moon or rises high above Earth. Amount of matter constant but force due to gravity is less on Moon or at altitude. c mass: kg weight: kg m s−2

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5 a For a body of constant mass, the acceleration is directly proportional to the resultant or net force applied to it. [1] Direction of the acceleration and the resultant force are the same. [1] b i It increases the time. [1] ii If time increases then acceleration decreases. [1] Since F = ma, when acceleration is less the net force is less and there is less force between the ground and the legs. [1]

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Chapter 3: Answers to Exam-style questions

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Answers to End-of-chapter questions 4 Forces–vectors and moment moments 1 a

4000 N drag

4000 N

bbCorrect scale drawing giving a value of 6130 ≈ 6100 N 2 a

contact force friction

weight

b 5.03 N c 5.03 N d 13.8 N 3 a 0.50 N; these components cancel as there is no resultant horizontal force. b String 1: 0.87 N; string 2: 0.29 N c 1.16 N d 1.0 N weight

0.58 N

1.16 N 4 28.3 ≈ 28 N 5 9.83 ≈ 9.8 N

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Chapter 4: Answers to End-of-chapter questions

Answers to Exam-style questions 4 Forces – vectors and moments 1 a i

A couple is a pair of equal and opposite forces. [1] ii T he torque of a force about a point is the product of the force and distance. [1] T he distance is the perpendicular distance of the line of action of the force to the point. [1] b i Rotation shown clockwise [1] Force drawn forwards [1] axle ground

force exerted by road on wheels

ii

Zero [1] T he car moves at constant speed or the wheel turns at a constant rate. [1] torque 200 = 690 N [1] = iii force = radius 0.29 2 a The centre of gravity of an object is the point where all the weight of the object [1] may be considered to act. [1] b i Taking moments about the left-hand end of the flagpole: sum of clockwise moments = sum of anticlockwise moments (25 × 9.81) × 1.5 = Tx [1] where x = perpendicular distance of the line of action of the tension from the left-hand end of the flagpole, given by: x = 2.5 sin 30° = 1.25 m [1] (25 × 9.81) × 1.5 = T × 1.25 [1] [1] T = 25 × 9.81 × 1.5 = 294 ≈ 290 N 1.25 ii The net vertical force = 0 Vertical component of force at fixed end + vertical component of T = weight Vertical component of force at fixed end + 294 sin 30° = 25 × 9.81 [1] Vertical component of force ≈ 98 N [1]

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3 a No net/resultant force No net/resultant moment b i Torque of the couple about the centre = 30 × 90 = 2700 N cm ii Moment of force about the centre = (T × 24) N cm For equilibrium: 24T = 2700 T = 113 ≈ 110 N

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4 a The torque of a couple is the product of force, F, and distance, d, [1] where d is the perpendicular distance between the forces and F is one of the two equal and opposite forces. [1] b Tension in string B = vertical component of A = 8.0 sin 50° [1] TB = 6.13 ≈ 6.1 N [1] Tension in string C = horizontal component of A = 8.0 cos 50° [1] [1] TC = 5.14 ≈ 5.1 N 5 a Any two of the following: • No rotation • No movement in any direction • No net/resultant moment and no net/resultant force b Two tension forces in the cord Closed triangle of forces or parallelogram of forces including weight

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tension in cord weight of picture

tension in cord

c i Vertical component = 45 sin 50° = 34.5 ≈ 35 N [1] ii Weight = sum of vertical components = 2 × 34.5 = 69 N [1]

Chapter 4: Answers to Exam-style questions

1

Answers to End-of-chapter questions 5 Work, Work energy and power 1 a Loss of gravitational potential energy → gain in kinetic energy b Kinetic energy → thermal energy/heat (in the brakes) c Loss of gravitational potential energy → gain in kinetic energy 2 a i 1.39 × 103 ≈ 1.4 kJ ii 0 iii 0 b 86.9 ≈ 87 W 3 Truck: 9 MJ Dust particle: 14 MJ T he dust particle has greater kinetic energy than the truck. 4 a 4.66 × 105 J ≈ 4.7 × 105 J b 116.5 ≈ 120 s c 3.4 × 105 J 5 a Rate at which work is done; W b Kinetic energy = 21× mass × velocity2 c 7130 ≈ 7.1 kW

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Chapter 5: Answers to End-of-chapter questions

Answers to Exam-style questions 5 Work, energy and power 1 a i

Vertical distance = 40 sin 5° = 3.49 m p.e. lost = mgh = 90 × 9.81 × 3.49 p.e. lost = 3078 ≈ 3100 J ii k.e. increase = 21mv2 = 21 × 90 × 122

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k.e. increase = 6480 J [1] Energy produced by cyclist = 6480 − 3078 = 3402 J [1] energy 3402 useful power output = [1] = time 67 power = 50.8 ≈ 51 W [1] ii Energy is wasted [1] as work done against friction in the axle/chain or against air resistance. [1]

b i

2 a Work is the product of force and distance moved. [1] T he distance moved is in the direction of the force. [1] b i As he falls, his potential energy decreases, his kinetic energy increases and internal energy (thermal energy/heat) is produced as work is done against friction. [1] T he decrease in p.e. = increase in k.e. + internal energy produced. [1] ii Graph with axes labelled and Ep decreasing linearly from 1000 J to 0. [1] Ek increasing from 0 when h = 0 [1] Ek increases as a straight line to a value below 1000 J at h = 15 m [1]

Energy / J

1000

3 a k.e. = work done on body to increase speed from 0 to v = F × s Since F = ma, k.e.= ma × s [1] 2 From v2 = u2 + 2as with u = 0, s = v , 2a leading to 2 k.e. = ma × v = 1 mv 2 [1] 2a 2 b i k.e. = 21mv2 = 21 × 800 × 202 = 160 000 J [1] energy 160 000 = time 6 = 2.67 × 104 ≈ 2.7 × 104 W [1] Air resistance increases (with speed). [1] Net driving force less or more energy (per second) wasted so less available to increase k.e. [1] power =

ii

4 a i

The potential energy of a body is the energy stored in the body by reason of its position or shape. [1] ii Gravitational p.e. is energy due to position in a gravitational field. [1] Elastic p.e. is energy contained in a stretched or squashed object. [1] or When an object is raised above the Earth’s surface its gravitational p.e. increases. [1] When a spring or wire is stretched its elastic p.e. increases. [1]

b i Ep

Ek

15

h/m

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mass = density × volume = 1030 × 1.4 × 106 × 10.0 = 1.442 × 1010 ≈ 1.4 × 1010 J [1] ii Water falls an average distance of 5.0 m when released through turbines. p.e. lost = mgh = 1.442 × 1010 × 9.81 × 5.0 (accept also use of h = 10 m) [1] 11 11 p.e. lost = 7.07 × 10 ≈ 7.1 × 10 J (or 1.4 × 1012 J if h = 10 used) [1]

Chapter 5: Answers to Exam-style questions

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iii Energy output = 0.5 × value in ii = 3.5 × 1011 J (or 7.1 × 1011 J if h = 10 used) [1] 11

power = energy = 3.5×10 J 6 × 60× 60 time 11

7.1×10 J if h = 10 used) 6 × 60× 60 power = 1.64 × 107 ≈ 1.6 × 107 W (or 3.3 × 107 W if h = 10 used) (or

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5 a Mention of kinetic energy and potential energy of molecules [1] Sum of potential energy and random kinetic energy of the molecules in a body [1] b i Internal energy increases. [1] Molecules gain random kinetic energy as they move faster at higher temperature. [1] ii Internal energy increases. [1] Molecules gain potential energy as they move away from each other or as work is done against the intermolecular force. [1]

2

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Chapter 5: Answers to Exam-style questions

Answers to End-of-chapter questions 6 Momentum 1 T he law of conservation of momentum applies if the Earth is considered to rise as the ball falls. The momentum of the Earth upwards equals the momentum of the ball downwards. The weight of the ball has an equal and upwards force on the Earth due to Newton’s third law. 2 a Change in momentum of ball is 12 kg m s−1 away from the wall. b No change in kinetic energy. 3 a b c d

linear momentum = mass × velocity kg m s−1 1.5 × 104 kg m s−1 The objects move to the left with a combined speed of 0.50 m s−1.

4 a i In an elastic collision both momentum and kinetic energy are conserved. ii In an inelastic collision momentum is conserved but not kinetic energy. b 1.855 ≈ 1.9 kg m s−1 c When the table and Earth are also considered then the initial momentum of the ball is equal to the final momentum of the ball added to the momentum of the snooker table and Earth, and so momentum is conserved. 5 a −26 400 ≈ −26 000 kg m s−1 b 1320 ≈ 1300 N c 240 m

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Chapter 6: Answers to End-of-chapter questions

Answers to Exam-style questions 6 Momentum 1 a Initial momentum of ball = 0.16 × 25 = 4.0 kg m s−1 [1] Change in momentum = 4.0 − (−4.0) = 8.0 kg m s−1 [1] change in momentum 8 [1] = b force = time 0.003 force = 2667 ≈ 2700 N [1] c The bat slows down. [1] T he law of conservation of momentum requires that the change in momentum of the ball and of the bat are equal but in opposite directions. [1] Energy is neither created nor destroyed but internal energy (heat) and sound are created from the drop in k.e. (of the bat). [1] T he impact is non-elastic. [1] 2 a The total momentum before an interaction is equal to the total momentum after the interaction. [1] T he system is closed or there are no external forces acting. [1] b i Final momentum = initial momentum 0.35v = 0.25 × 30 [1] v = 21.4 ≈ 21 m s−1 [1] ii Change in momentum = 0.25 × 30 − 0.25 × 21.4 [1] change in momentum = 2.14 ≈ 2.1 kg m s−1 or 2.15 ≈ 2.2 kg m s−1 [1] iii Change in total kinetic energy [1] = 12 × 0.25 × 302 − 21× 0.35 × 21.42 change in total k.e. = 32.4 ≈ 32 J [1] iv T h e arrow stops and the ball moves off with a speed of 30 m s−1 [1] Relative speed remains unaltered in an elastic collision, 30 m s−1 [1] 3 a i T he total kinetic energy before the collision is equal to the total kinetic energy after the collision.

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ii In a completely inelastic collision the maximum amount of kinetic energy is lost (subject to the law of conservation of momentum, which must be obeyed). [1] b i Momentum is conserved as there are no external forces/the system is closed. [1] Momentum of α-particle in one direction must equal that of uranium nucleus in the exactly opposite direction for the change to be zero. [1] −27 −25 ii 6.65 × 10 × vα + 3.89 × 10 × vX = 0 [1] v iii α = −58.5 ≈ −58 or −59 [1] vX 4 a Momentum and kinetic energy [1] b i Momentum = 0.014 × 640 = 8.96 ≈ 9.0 kg m s−1 [1] ii Bullets leave with momentum forward and gun has equal momentum backwards. [1] To stop motion/momentum of the gun the soldier must provide a force. [1] ∆p iii F = ; 140 = n × 8.96 [1] ∆t Number of bullets per second = 15.6 or 15 or 16 [1] 5 a

Change in Initial momentum / kinetic energy / J kg m s–1

Final kinetic energy / J

truck X

6.0 × 104

2.5 × 105

4.0 × 104

truck Y

6.0 × 104

1.5 × 104

1.35 × 105

One mark for each change in momentum. [2] One mark for kinetic energy values for X. [1] One mark for kinetic energy values for Y. [1] b Total initial k.e. = 2.65 × 105 J and total final k.e. = 1.75 × 105 J [1] Collision is not elastic as the total k.e. has decreased in the collision. [1] 4 ∆p = 6.0× 10 1.6 ∆t 4 = 3.75 × 10 ≈ 3.7 or 3.8 × 104 N

[1]

c force =

Chapter 6: Answers to Exam-style questions

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Answers to End-of-chapter questions 7 Matter 1 a The density of ice is roughly the same as the density of water. b The force required to break a solid, e.g. a metal, is large. 2 a Many molecules of gas hit the tyre each second. Each impact involves a change in momentum of a molecule as it rebounds. This change in momentum is caused by a contact force on the molecule. There is an equal and opposite force on the tyre wall. Many random impacts per second on unit area cause a steady force and pressure. b With more molecules there is a greater rate of change of momentum on the tyre walls. 3 a Average spe...