Equations of motion — Paper 2 · A Level Physics 9702

5 An electron is travelling in a straight line through a vacuum with a constant speed of 1.5 × 107 m s–1. The electron enters a uniform electric field at point A, as shown in Fig. 5.1. uniform electric field 2.0 cm electron speed A B 1.5 × 107 m s–1 Fig. 5.1 The electron continues to move in the same direction until it is brought to rest by the electric field at point B. Distance AB is 2.0 cm. (a) State the direction of the electric field. … [1] (b) Calculate the magnitude of the deceleration of the electron in the field. deceleration = … m s–2 [2] (c) Calculate the electric field strength. electric field strength = … V m–1 [3] (d) The electron is at point A at time t = 0. On Fig. 5.2, sketch the variation with time t of the velocity v of the electron until it reaches point B. Numerical values of v and t do not need to be shown. v 0 0 t Fig. 5.2 [1] [Total: 7]

7 marks

Mark scheme: 5(a) to the right / from the left / from A to B / in the same direction as electron velocity B1 5(b) v 2 = u 2 + 2as a = (1.5 × 107)2 / (2 × 2.0 × 10–2) Other alternative calculations for the C1 mark: e.g. a = 1.5×107 / 2.67×10–9 e.g. a = [(1.5×107 × 2.67×10–9) – 2.0×10–2] × [2 / (2.67×10–9)2] e.g. a = (2.0×10–2 × 2) / (2.67×10–9)2 C1 = 5.6 × 1015 m s–2 A1 5(c) E = F / Q C1 = (9.1 × 10–31 × 5.6 × 1015) / 1.6 × 10–19 C1 = 3.2 × 104 V m–1 A1 5(d) straight line with negative gradient starting at an intercept on the v-axis and ending at an intercept on the t-axis. B1

This question in 9702/22 Feb/March 2017

1 (a) Complete Fig. 1.1 to indicate whether each of the quantities is a vector or a scalar. quantity vector or scalar acceleration speed power Fig. 1.1 [2] (b) A ball is projected with a horizontal velocity of 1.1 m s–1 from point A at the edge of a table, as shown in Fig. 1.2. table ball 1.1 m s–1 A path of ball B horizontal ground 0.43 m Fig. 1.2 The ball lands on horizontal ground at point B which is a distance of 0.43 m from the base of the table. Air resistance is negligible. (i) Calculate the time taken for the ball to fall from A to B. time = … s [1] (ii) Use your answer in (b)(i) to determine the height of the table. height = … m [2] (iii) The ball leaves the table at time t = 0. For the motion of the ball between A and B, sketch graphs on Fig. 1.3 to show the variation with time t of 1. the acceleration a of the ball, 2. the vertical component sv of the displacement of the ball from A. Numerical values are not required. a sv 0 0 0 t 0 t Fig. 1.3 [2] (c) A ball of greater mass is projected from the table with the same velocity as the ball in (b). Air resistance is still negligible. State and explain the effect, if any, of the increased mass on the time taken for the ball to fall to the ground. … … [1] [Total: 8]

8 marks

Mark scheme: 1(a) acceleration: vector speed: scalar power: scalar All three correct scores 2 marks. Only two correct scores 1 mark. B2 1(b)(i) time = 0.43 / 1.1 = 0.39 s A1 1(b)(ii) s = ut + ½at 2 = ½ × 9.81 × 0.392 C1 = 0.75 m A1 1(b)(iii) 1 horizontal line at a non-zero value of a. B1 2 curved line from origin with increasing gradient. B1 1(c) acceleration (of free fall) is unchanged / not dependent on mass and so no effect (on time taken). A1

This question in 9702/22 Feb/March 2018

3 A child on a sledge slides down a steep hill and then travels in a straight line up an ice-covered slope, as illustrated in Fig. 3.1. ice-covered slope child and sledge total mass 70 kg B 18 m s–1 A Fig. 3.1 (not to scale) The sledge passes point A with speed 18 m s–1 at time t = 0 and then comes to rest at point B. The child applies a brake to the sledge at point B. The brake does not keep the sledge stationary and it immediately slides back down the slope towards A. The variation with time t of the velocity v of the sledge from t = 0 to t = 24 s is shown in Fig. 3.2. 20 v / m s–1 10 0 0 4 8 12 16 20 24 t / s –10 Fig. 3.2 (a) State the time taken for the sledge to travel from A to B. time = … s [1] (b) Determine the displacement of the sledge up the slope from point A at time t = 24 s. displacement = … m [3] (c) Show that the acceleration of the sledge as it moves from B back towards A is 0.50 m s–2. [2] (d) The child and sledge have a total mass of 70 kg. The component of the total weight of the child and sledge that acts down the slope is 80 N. Determine (i) the frictional force on the sledge as it moves from B towards A, frictional force = … N [2] (ii) the angle θ of the slope to the horizontal. θ = … ° [2] (e) The child on the sledge blows a whistle between t = 4.0 s and t = 8.0 s. The whistle emits sound of frequency 900 Hz. The speed of the sound in the air is 340 m s–1. A man standing at point A hears the sound. Use Fig. 3.2 to (i) determine the initial frequency of the sound heard by the man, initial frequency = … Hz [2] (ii) describe and explain qualitatively the variation, if any, in the frequency of the sound heard by the man. … … [1] [Total: 13]

13 marks

Mark scheme: 3(a) time = 12 s A1 3(b) distance (up slope) = ½ × 12 × 18 (= 108) C1 distance (down slope) = ½ × 12 × 6 (= 36) C1 displacement from A = 108 – 36 = 72 m A1 3(c) v = u + at or a = gradient or a = ∆v / (∆)t C1 a = 6 / 12 = 0.50 (m s–2) (other points from the line may be used) A1 or v2 = u2 + 2as and u = 0 or v2 = 2as (C1) a = 6.02 / (2 × 36) = 0.50 (m s–2) (A1) or s = ut + ½at2 and u = 0 or s = ½at2 (C1) a = 2 × 36 / 122 = 0.50 (m s–2) (A1) or s = vt – ½at2 (C1) a = 2 × (6 × 12 – 36) / 122 = 0.50 (m s–2) (A1) Question Answer Marks 3(d)(i) F = 70 × 0.50 (= 35) C1 frictional force = 80 – 35 = 45 N A1 3(d)(ii) sin θ = 80 / (70 × 9.81) C1 θ = 6.7° A1 3(e)(i) f0 = (900 × 340) / (340 + 12) C1 = 870 Hz A1 3(e)(ii) speed/velocity (of sledge) decreases and (so) frequency increases B1

This question in 9702/22 May/June 2018

3 A ball is thrown vertically upwards towards a ceiling and then rebounds, as illustrated in Fig. 3.1. ceiling ball leaving speed 3.8 m s–1 ceiling ball thrown speed 9.6 m s–1 upwards Fig. 3.1 The ball is thrown with speed 9.6 m s–1 and takes a time of 0.37 s to reach the ceiling. The ball is then in contact with the ceiling for a further time of 0.085 s until leaving it with a speed of 3.8 m s–1. The mass of the ball is 0.056 kg. Assume that air resistance is negligible. (a) Show that the ball reaches the ceiling with a speed of 6.0 m s–1. [1] (b) Calculate the height of the ceiling above the point from which the ball was thrown. height = … m [2] (c) Calculate (i) the increase in gravitational potential energy of the ball for its movement from its initial position to the ceiling, increase in gravitational potential energy = … J [2] (ii) the decrease in kinetic energy of the ball while it is in contact with the ceiling. decrease in kinetic energy = … J [2] (d) State how Newton’s third law applies to the collision between the ball and the ceiling. … … … … [2] (e) Calculate the change in momentum of the ball during the collision. change in momentum = … N s [2] (f) Determine the magnitude of the average force exerted by the ceiling on the ball during the collision. average force = … N [2] [Total: 13]

13 marks

Mark scheme: 3(a) v = u + at v = 9.6 – (9.81 × 0.37) = 6.0 m s–1 A1 3(b) s = ½ × (9.6 + 6.0) × 0.37 or 6.02 = 9.62 – (2 × 9.81 × s) or s = (9.6 × 0.37) – (½ × 9.81 × 0.372) or s = (6.0 × 0.37) + (½ × 9.81 × 0.372) C1 s = 2.9 m A1 3(c)(i) (∆)E = mg(∆)h C1 ∆E = 0.056 × 9.81 × 2.9 = 1.6 J A1 3(c)(ii) E = ½mv 2 C1 ∆E = ½ × 0.056 × (6.02 – 3.82) = 0.60 J A1 3(d) force on ball (by ceiling) equal to force on ceiling (by ball) M1 and opposite (in direction) A1 3(e) (p =) mv or 0.056 × 6.0 or 0.056 × 3.8 C1 change in momentum = 0.056 × (6.0 + 3.8) = 0.55 N s A1 Question Answer Mark 3(f) resultant force = 0.55 / 0.085 (= 6.47 N) C1 force by ceiling = 6.47 – (0.056 × 9.81) = 5.9 N A1

This question in 9702/23 May/June 2018

1 (a) Define (i) displacement, … … [1] (ii) acceleration. … … [1] (b) A remote-controlled toy car moves up a ramp and travels across a gap to land on another ramp, as illustrated in Fig. 1.1. path of car 5.5 m s–1 car ramp P ramp Q d ground θ Fig. 1.1 The car leaves ramp P with a velocity of 5.5 m s–1 at an angle θ to the horizontal. The horizontal component of the car’s velocity as it leaves the ramp is 4.6 m s–1. The car lands at the top of ramp Q. The tops of both ramps are at the same height and are distance d apart. Air resistance is negligible. (i) Show that the car leaves ramp P with a vertical component of velocity of 3.0 m s–1. [1] (ii) Determine the time taken for the car to travel between the ramps. time taken = … s [2] (iii) Calculate the horizontal distance d between the tops of the ramps. d = … m [1] (iv) Calculate the ratio kinetic energy of the car at its maximum height . kinetic energy of the car as it leaves ramp P ratio = … [3] (c) Ramp Q is removed. The car again leaves ramp P as in (b) and now lands directly on the ground. The car leaves ramp P at time t = 0 and lands on the ground at time t = T. On Fig. 1.2, sketch the variation with time t of the vertical component vy of the car’s velocity from t = 0 to t = T. Numerical values of vy and t are not required. vy 0 0 TT t t Fig. 1.2 [2] [Total: 11]

11 marks

Mark scheme: 1(a)(i) distance in a specified direction (from a point) B1 1(a)(ii) change in velocity / time (taken) B1 1(b)(i) vertical component of velocity = (5.52 – 4.62)1/2 = 3.0 (m s–1) or 5.5 cos θ = 4.6 (so θ = 33.2°) and 5.5 sin 33.2° = 3.0 (m s–1) A1 1(b)(ii) s = ut + ½at 2 0 = (3.0 × t) – (½ × 9.81 × t 2) or v = u + at –3.0 = 3.0 – 9.81t C1 t = 0.61 s A1 1(b)(iii) d = 4.6 × 0.61 = 2.8 m A1 1(b)(iv) E = ½mv2 C1 ratio = (½ × m × 4.62) / (½ × m × 5.52) or ratio = (½ × m × 5.52 – m × 9.81 × 0.459) / (½ × m × 5.52) C1 ratio = 0.70 A1 1(c) straight line from positive value of vy at t = 0 to negative value of vy M1 straight line ends at t = T and final magnitude of vy greater than initial magnitude of vy A1

This question in 9702/21 Oct/Nov 2018

1 A golfer strikes a ball so that it leaves horizontal ground with a velocity of 6.0 m s–1 at an angle θ to the horizontal, as illustrated in Fig. 1.1. vY 6.0 m s–1 4.8 m s–1 ball θ ground vX Fig. 1.1 (not to scale) The magnitude of the initial vertical component vY of the velocity is 4.8 m s–1. Assume that air resistance is negligible. (a) Show that the magnitude of the initial horizontal component vX of the velocity is 3.6 m s–1. [1] (b) The ball leaves the ground at time t = 0 and reaches its maximum height at t = 0.49 s. On Fig. 1.2, sketch separate lines to show the variation with time t, until the ball returns to the ground, of (i) the vertical component vY of the velocity (label this line Y), [2] (ii) the horizontal component vX of the velocity (label this line X). [2] 5.0 4.0 velocity / m s–1 3.0 2.0 1.0 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 t / s –1.0 –2.0 –3.0 –4.0 –5.0 Fig. 1.2 (c) Calculate the maximum height reached by the ball. maximum height = … m [2] (d) For the movement of the ball from the ground to its maximum height, determine the ratio kinetic energy at maximum height . change in gravitational potential energy ratio = … [4] (e) In practice, significant air resistance acts on the ball. Explain why the actual time taken for the ball to reach maximum height is less than the time calculated when air resistance is assumed to be negligible. … … … [1] [Total: 12]

12 marks

Mark scheme: 1(a) or 6.0 sinθ = 4.8 (so θ = 53.1°) and vx = 6.0 cos 53.1° = 3.6 (m s–1) A1 1(b)(i) straight line from (0, 4.8) to (0.49, 0) M1 straight line continues with same slope to (0.98, –4.8) (labelled Y) A1 1(b)(ii) a horizontal line M1 from (0, 3.6) to (0.98, 3.6) (labelled X) A1 1(c) s = ut + ½at2 = (4.8 × 0.49) + (½ × –9.81 × 0.492) or s = ½(u + v)t or area under graph = ½ × (4.8 + 0) × 0.49 or s = vt – ½at2 = ½ × 9.81 × 0.492 or v2 = u2 + 2as s = 4.82 / (2 × 9.81) C1 s = 1.2 m A1 Question Answer Marks 1(d) (∆)E = mg(∆)h C1 E = ½mv2 C1 ratio = (½ × m × 3.62) / (m × 9.81 × 1.2) or ratio = [(½ × m × 6.02) – (m × 9.81 × 1.2)] / (m × 9.81 × 1.2) or ratio = (½ × m × 3.62) / (½ × m × 4.82) C1 ratio = 0.56 A1 1(e) (force due to) air resistance acts in opposite direction to the velocity or (with air resistance, average) resultant force is larger (than weight) B1

This question in 9702/22 Oct/Nov 2018

2 (a) State Newton’s second law of motion. … … [1] (b) A car of mass 850 kg tows a trailer in a straight line along a horizontal road, as shown in Fig. 2.1. car trailer tow-bar mass 850 kg horizontal road Fig. 2.1 The car and the trailer are connected by a horizontal tow-bar. The variation with time t of the velocity v of the car for a part of its journey is shown in Fig. 2.2. 15 14 v / m s–1 13 12 11 10 9 8 0 5 10 15 20 25 t / s Fig. 2.2 (i) Calculate the distance travelled by the car from time t = 0 to t = 10 s. distance = … m [2] (ii) At time t = 10 s, the resistive force acting on the car due to air resistance and friction is 510 N. The tension in the tow-bar is 440 N. For the car at time t = 10 s: 1. use Fig. 2.2 to calculate the acceleration acceleration = … m s−2 [2] 2. use your answer to calculate the resultant force acting on the car resultant force = … N [1] 3. show that a horizontal force of 1300 N is exerted on the car by its engine [1] 4. determine the useful output power of the engine. output power = … W [2] (c) A short time later, the car in (b) is travelling at a constant speed and the tension in the tow-bar is 480 N. The tow-bar is a solid metal rod that obeys Hooke’s law. Some data for the tow-bar are listed below. Young modulus of metal = 2.2 × 1011 Pa original length of tow-bar = 0.48 m cross-sectional area of tow-bar = 3.0 × 10−4 m2 Determine the extension of the tow-bar. extension = … m [3] (d) The driver of the car in (b) sees a pedestrian standing directly ahead in the distance. The driver operates the horn of the car from time t = 15 s to t = 17 s. The frequency of the sound heard by the pedestrian is 480 Hz. The speed of the sound in the air is 340 m s−1. Use Fig. 2.2 to calculate the frequency of the sound emitted by the horn. frequency = … Hz [2] [Total: 14]

14 marks

Mark scheme: 2(a) (resultant) force proportional/equal to/is rate of change of momentum B1 2(b)(i) distance = area under graph or s = ½ (u + v) t = ½ × (9 + 13) × 10 or s = ut + ½at 2 = (9 × 10) + (½ × 0.40 × 102) or s = vt – ½at 2 = (13 × 10) – (½ × 0.40 × 102) or v 2 = u 2 + 2as 132 = 92 + (2 × 0.40 × s) C1 distance = 110 m A1 Question Answer Marks 2(b)(ii) 1. a = gradient or a = (v – u) / t or a = ∆v / (∆)t e.g. a = (14 – 9) / 12.5 or (13 – 9) / 10 C1 a = 0.40 m s–2 A1 2. resultant force = 850 × 0.40 = 340 N A1 3. (F =) 510 + 440 + 340 = 1300 (N) A1 4. P = Fv C1 = 1300 × 13 = 1.7 × 104 W A1 2(c) E = σ / ε C1 E = (F / A) / (∆L / L) or E = FL / A∆L C1 ∆L = (480 × 0.48) / (3.0 × 10–4 × 2.2 × 1011) = 3.5 × 10–6 m A1 2(d) fo = fs v / (v – vs) 480 = fs × 340 / (340 – 14) C1 fs = 460 Hz A1

This question in 9702/22 May/June 2019

2 (a) A resultant force F moves an object of mass m through distance s in a straight line. The force gives the object an acceleration a so that its speed changes from initial speed u to final speed v. (i) State an expression for: 1. the work W done by the force, in terms of a, m and s W = … [1] 2. the distance s, in terms of a, u and v. s = … [1] (ii) Use your answers in (i) to show that the kinetic energy of the object is given by 1 kinetic energy = × mass × (speed)2. 2 Explain your working. [2] (b) A ball of mass 0.040 kg is projected into the air from horizontal ground, as illustrated in Fig. 2.1. Y path of ball h ball, mass 0.040 kg X ground Fig. 2.1 The ball is launched from a point X with a kinetic energy of 4.5 J. At point Y, the ball has a speed of 9.5 m s−1. Air resistance is negligible. (i) For the movement of the ball from X to Y, draw a solid line on Fig. 2.1 to show: 1. the distance moved (label this line D) 2. the displacement (label this line S). [2] (ii) By consideration of energy transfer, determine the height h of point Y above the ground. h = … m [3] (iii) On Fig. 2.2, sketch the variation of the kinetic energy of the ball with its vertical height above the ground for the movement of the ball from X to Y. Numerical values are not required. kinetic energy 0 0 h height Fig. 2.2 [2] [Total: 11]

11 marks

Mark scheme: 2(a)(i) 1. W = mas B1 2. s = (v 2 – u 2) / 2a B1 2(a)(ii) W/work equals energy transferred/gain or change in kinetic energy B1 W (= mas) = ma(v 2 – u 2) / 2a leading to W = m(v 2 – u 2) / 2 (so KE = ½mv 2) B1 2(b)(i) 1. solid curved line drawn from X to Y along path of ball and labelled D B1 2. solid straight line drawn from X to Y and labelled S B1 2(b)(ii) (∆)E = mg(∆)h C1 4.5 = (0.040 × 9.81 × h) + (½ × 0.040 × 9.52) C1 h = 6.9 m A1 2(b)(iii) line with a negative gradient starting from a non-zero value of kinetic energy when the vertical height is zero M1 straight line ends at a non-zero value of kinetic energy when the vertical height is h A1

This question in 9702/23 May/June 2019

4 Two vertical metal plates in a vacuum are separated by a distance of 0.12 m. Fig. 4.1 shows a side view of this arrangement. 0.080 m sand X particle 2.0 m 0 V + 900 V path of particle metal plate metal plate Y 0.12 m Fig. 4.1 (not to scale) Each plate has a length of 2.0 m. The potential difference between the plates is 900 V. The electric field between the plates is uniform. A negatively charged sand particle is released from rest at point X, which is a horizontal distance of 0.080 m from the top of the positively charged plate. The particle then travels in a straight line and collides with the positively charged plate at its lowest point Y, as illustrated in Fig. 4.1. (a) Describe the pattern of the field lines (lines of force) between the plates. … … … [2] (b) State the names of the two forces acting on the particle as it moves from X to Y. … [1] (c) By considering the vertical motion of the sand particle, show that the time taken for the particle to move from X to Y is 0.64 s. [2] (d) Calculate the horizontal component of the acceleration of the particle. horizontal component of acceleration = … m s−2 [2] (e) (i) Calculate the magnitude of the electric field strength. electric field strength = … N C−1 [2] (ii) The sand particle has mass m and charge q. Use your answers in (d) and (e)(i) to q determine the ratio m. ratio = … C kg−1 [2] q(f) Another particle has a smaller magnitude of the ratio than the sand particle. This particle is m also released from point X. For the movement of this particle, state the effect, if any, of the decreased magnitude of the ratio on: (i) the vertical component of the acceleration … [1] (ii) the horizontal component of the acceleration. … [1] [Total: 13]

13 marks

Mark scheme: 4(a) straight (horizontal) lines and from the +0.90 kV plate/to the 0 V plate B1 (lines are) equally spaced B1 4(b) weight/gravitational force and electric force B1 4(c) s = ½ at 2 or s = ut + ½at 2 and u = 0 C1 2.0 = ½ × 9.81 × t 2 so t = 0.64 s A1 4(d) 0.080 = ½ × a × 0.642 C1 a = 0.39 m s–2 A1 4(e)(i) E = (∆)V / (∆)d C1 E = 0.90 × 103 / 0.12 = 7.5 × 103 N C–1 A1 4(e)(ii) ma = Eq or F = ma and F = Eq C1 q / m = 0.39 / 7.5 × 103 = 5.2 × 10–5 C kg–1 A1 4(f)(i) no effect B1 4(f)(ii) decreases/smaller B1

This question in 9702/23 May/June 2019

2 A small charged glass bead of weight 5.4 × 10–5 N is initially at rest at point A in a vacuum. The bead then falls through a uniform horizontal electric field as it moves in a straight line to point B, as illustrated in Fig. 2.1. vertical glass bead weight 5.4 × 10–5 N A horizontal charge –3.7 × 10–9 C uniform horizontal path of the electric field, × 104 V m–1 falling bead field strength 1.3 B side view Fig. 2.1 (not to scale) The electric field strength is 1.3 × 104 V m–1. The charge on the bead is –3.7 × 10–9 C. (a) Describe how two metal plates could be used to produce the electric field. Numerical values are not required. … … … [2] (b) Determine the magnitude of the electric force acting on the bead. electric force = … N [2] (c) Use your answer in (b) and the weight of the bead to show that the resultant force acting on it is 7.2 × 10–5 N. [1] (d) Explain why the resultant force on the bead of 7.2 × 10–5 N is constant as the bead moves along path AB. … … … … [2] (e) (i) Calculate the magnitude of the acceleration of the bead along the path AB. acceleration = … m s–2 [2] (ii) The path AB has length 0.58 m. Use your answer in (i) to determine the speed of the bead at point B. speed = … m s–1 [2] [Total: 11]

11 marks

Mark scheme: 2(a) the (two) plates are vertical (and separated) B1 left plate positively charged and right plate negatively charged/earthed or right plate negatively charged and left plate positively charged/earthed B1 2(b) F = Eq C1 = 1.3 × 104 × 3.7 × 10–9 = 4.8 × 10–5 N A1 2(c) F2 = (4.8 × 10–5)2 + (5.4 × 10–5)2 so F = 7.2 × 10–5 N or F = [(4.8 × 10–5)2 + (5.4 × 10–5)2]0.5 so F = 7.2 × 10–5 N A1 2(d) electric force is constant (because field strength/E is constant) B1 weight is constant (and so resultant force constant) B1 2(e)(i) m = 5.4 × 10–5 / 9.81 (= 5.5 × 10–6) C1 a = 7.2 × 10–5 / (5.5 × 10–6) =13 m s–2 A1 2(e)(ii) v2 = u2 + 2as v2 = 2 × 13 × 0.58 C1 v = 3.9 m s–1 A1

This question in 9702/21 Oct/Nov 2019

2 (a) Define acceleration. … [1] (b) A steel ball of diameter 0.080 m is released from rest and falls vertically in air, as illustrated in Fig. 2.1. position of ball steel ball of when released diameter 0.080 m 0.280 m horizontal beam of light of position P negligible width of ball Fig. 2.1 (not to scale) A horizontal beam of light of negligible width is a vertical distance of 0.280 m below the bottom of the ball when it is released. The ball falls through and breaks the beam of light. (i) Explain why the force due to air resistance acting on the ball may be neglected when calculating the time taken for the ball to reach the beam of light. … … [1] (ii) Calculate the time taken for the ball to fall from rest to position P where the bottom of the ball touches the beam of light. time taken = … s [2] (iii) Determine the time interval during which the beam of light is broken by the ball. time interval = … s [2] (c) A different ball is released from the same position as the steel ball in (b). This ball has the same diameter but a much lower density. For this ball, the force due to air resistance cannot be neglected as the ball falls. State and explain the change, if any, to the time interval during which the beam of light is broken by the ball. … … … … [2] [Total: 8]

8 marks

Mark scheme: 2(a) change in velocity / time (taken) A1 2(b)(i) weight ≫ (force due to) air resistance or (force due to) air resistance is negligible compared to weight B1 2(b)(ii) s = ut + ½at 2 0.280 = ½ × 9.81 × t 2 C1 t = 0.24 s A1 Question Answer Marks 2(b)(iii) total distance fallen = 0.280 + 0.080 = 0.360 0.360 = ½ × 9.81 × t 2 t = 0.27 s C1 time taken = 0.27 – 0.24 = 0.03 s A1 or v = 9.81 × 0.239 or (2 × 9.81 × 0.280)0.5 or (2 × 0.280) / 0.239 v = 2.34 (m s–1) (C1) 0.080 = 2.34t + ½ × 9.81 × t 2 solving quadratic equation gives t = 0.03 s allow any correct method using equations of uniform accelerated motion (A1) 2(c) (average) resultant force/acceleration/speed/velocity (of low-density ball) is less B1 (so) time interval is longer B1

This question in 9702/22 Oct/Nov 2019

2 A dolphin is swimming under water at a constant speed of 4.50 m s–1. (a) The dolphin emits a sound as it swims directly towards a stationary submerged diver. The frequency of the sound heard by the diver is 9560 Hz. The speed of sound in the water is 1510 m s–1. Determine the frequency, to three significant figures, of the sound emitted by the dolphin. frequency = … Hz [2] (b) The dolphin strikes the bottom of a floating ball so that the ball rises vertically upwards from the surface of the water, as illustrated in Fig. 2.1. path of ball height of ball above ball surface surface of water speed 5.6 m s–1 Fig. 2.1 The ball leaves the water surface with speed 5.6 m s–1. Assume that air resistance is negligible. (i) Calculate the maximum height reached by the ball above the surface of the water. height = … m [2] (ii) The ball leaves the water at time t = 0 and reaches its maximum height at time t = T. On Fig. 2.2, sketch a graph to show the variation of the speed of the ball with time t from t = 0 to t = T. Numerical values are not required. speed 0 0 T time t Fig. 2.2 [1] (iii) The mass of the ball is 0.45 kg. Use your answer in (b)(i) to calculate the change in gravitational potential energy of the ball as it rises from the surface of the water to its maximum height. change in gravitational potential energy = … J [2] (iv) State and explain the variation in the magnitude of the acceleration of the ball as it falls back towards the surface of the water if air resistance is not negligible. … … … … … [2] [Total: 9]

9 marks

Mark scheme: 2(a) f0 = fS v / (v – vS) 9560 = f × 1510 / (1510 – 4.50) C1 f = 9530 Hz A1 2(b)(i) v 2 = u 2 + 2as height = 5.62 / (2 × 9.81) C1 = 1.6 m A1 2(b)(ii) downward sloping straight line starting from a point on the speed axis and ending at point (T, 0) B1 Question Answer Marks 2(b)(iii) (Δ)E = mg(Δ)h = 0.45 × 9.81 × 1.6 C1 = 7.1 J A1 2(b)(iv) air resistance increases (and weight constant) B1 (resultant force decreases so) acceleration decreases B1

This question in 9702/22 Feb/March 2020

2 (a) State Newton’s second law of motion. … … [1] (b) A delivery company suggests using a remote-controlled aircraft to drop a parcel into the garden of a customer. When the aircraft is vertically above point P on the ground, it releases the parcel with a velocity that is horizontal and of magnitude 5.4 m s–1. The path of the parcel is shown in Fig. 2.1. 5.4 m s–1 X parcel path of parcel h P Q horizontal ground d Fig. 2.1 (not to scale) The parcel takes a time of 0.81 s after its release to reach point Q on the horizontal ground. Assume air resistance is negligible. (i) On Fig. 2.1, draw an arrow from point X to show the direction of the acceleration of the parcel when it is at that point. [1] (ii) Determine the height h of the parcel above the ground when it is released. h = … m [2] (iii) Calculate the horizontal distance d between points P and Q. d = … m [1] (c) Another parcel is accidentally released from rest by a different aircraft when it is hovering at a great height above the ground. Air resistance is now significant. (i) On Fig. 2.2, draw arrows to show the directions of the forces acting on the parcel as it falls vertically downwards. Label each arrow with the name of the force. parcel velocity Fig. 2.2 [2] (ii) By considering the forces acting on the parcel, state and explain the variation, if any, of the acceleration of the parcel as it moves downwards before it reaches constant (terminal) speed. … … … … … … [3] (iii) Describe the energy conversion that occurs when the parcel is falling through the air at constant (terminal) speed. … … [1] [Total: 11]

11 marks

Mark scheme: 2(a) (resultant) force proportional to rate of change of momentum B1 2(b)(i) arrow drawn vertically downwards from point X B1 2(b)(ii) s = ut + ½at2 h = ½ × 9.81 × 0.812 C1 = 3.2 m A1 2(b)(iii) d = 5.4 × 0.81 = 4.4 m A1 2(c)(i) downward pointing arrow labelled weight B1 upward pointing arrow labelled air resistance B1 2(c)(ii) air resistance increases B1 weight constant or resultant force decreases B1 (so) acceleration decreases B1 2(c)(iii) gravitational potential energy to thermal/internal energy B1

This question in 9702/21 May/June 2020

2 A small block is lifted vertically upwards by a toy aircraft, as illustrated in Fig. 2.1. aircraft string velocity block Fig. 2.1 As the block is moving upwards, the string breaks at time t = 0. The block initially continues moving upwards and then falls and hits the ground at time t = 0.90 s. The variation with time t of the velocity v of the block is shown in Fig. 2.2. 1.96 v / m s–1 0 0 0.20 0.900.90 t / s –6.86 Fig. 2.2 Air resistance is negligible. (a) State the feature of the graph in Fig. 2.2 that shows the block has a constant acceleration. … [1] (b) Use Fig. 2.2 to determine the height of the block above the ground when the string breaks at time t = 0. height = … m [3] (c) The block has a weight of 0.86 N. Calculate the difference in gravitational potential energy of the block between time t = 0 and time t = 0.90 s. difference in gravitational potential energy = … J [2] (d) On Fig. 2.3, sketch a line to show the variation of the distance moved by the block with time t from t = 0 to t = 0.20 s. Numerical values of distance are not required. distance moved 0 0 0.20 t / s Fig. 2.3 [2] (e) A block of greater mass is now released from the same height with the same upward velocity. Air resistance is still negligible. State and explain the effect, if any, of the increased mass on the speed with which the block hits the ground. … … [1] [Total: 9]

9 marks

Mark scheme: 2(a) constant gradient B1 2(b) (displacement until 0.20 s =) ½ × 1.96 × 0.20 (= 0.196 m) or (displacement after 0.20 s =) ½ × 6.86 × 0.70 (= 2.401 m) C1 height = 2.401 – 0.196 C1 = 2.2 m (alternative methods are possible using equations of uniformly accelerated motion) A1 2(c) (Δ)E = mg(Δ)h or W(Δ)h C1 (Δ)E = 0.86 × 2.2 = 1.9 J A1 2(d) curved line from the origin M1 gradient of curved line decreases and is zero at t = 0.20 s only A1 2(e) acceleration (of free fall) is unchanged/is not dependent on mass and (so) no effect B1

This question in 9702/21 Oct/Nov 2020

3 A ball is fired horizontally with a speed of 41.0 m s–1 from a stationary cannon at the top of a hill. The ball lands on horizontal ground that is a vertical distance of 57 m below the cannon, as shown in Fig. 3.1. ball, initial speed cannon 41.0 m s–1 path of ball 57 m horizontal ground Fig. 3.1 (not to scale) Assume air resistance is negligible. (a) Show that the time taken for the ball to reach the ground, after being fired, is 3.4 s. [2] (b) Calculate the horizontal distance of the ball from the cannon at the point where the ball lands on the ground. horizontal distance = … m [1] (c) Determine the magnitude of the displacement of the ball from the cannon at the point where the ball lands on the ground. displacement = … m [2] (d) The ball leaves the cannon at time t = 0. On Fig. 3.2, sketch a graph to show the variation of the magnitude v of the vertical component of the velocity of the ball with time t from t = 0 to t = 3.4 s. Numerical values are not required. v 0 0 3.4 t / s Fig. 3.2 [1] (e) The cannon recoils horizontally with a speed of 0.340 m s–1 when it fires the ball. The total mass of the ball and the cannon is 1480 kg. Assume that no external horizontal forces act on the ball-cannon system. Determine, to three significant figures, the mass of the ball. mass = … kg [2] (f) The cannon now fires a ball of smaller mass. Assume that air resistance is still negligible. State and explain the change, if any, to the graph in Fig. 3.2 due to the decreased mass of the ball. … … … [2] [Total: 10]

10 marks

Mark scheme: 3(a) s = ½at 2 C1 57 = ½ × 9.81 × t 2 and t = 3.4 (s) A1 3(b) horizontal distance = 41 × 3.4 = 140 m A1 3(c) (displacement)2 = 572 + 1402 C1 displacement = ( 572 + 1402)0.5 = 150 m A1 3(d) straight line from the origin with positive gradient B1 3(e) (1480 – m) × 0.340 = m × 41.0 C1 m = 12.2 kg A1 or mc 0.34 = mb 41 and mc + mb = 1480 (C1) mc = (41 / 0.34)mb (41 / 0.34)mb + mb = 1480 mb = 12.2 kg (A1) 3(f) acceleration (of free fall) is unchanged/is not dependent on mass M1 (so) no change (to the graph) A1

This question in 9702/23 Oct/Nov 2020

2 A ball is thrown vertically downwards to the ground, as illustrated in Fig. 2.1. ball speed u path of ball 1.5 m speed 8.7 m s–1 ground Fig. 2.1 The ball is thrown with speed u from a height of 1.5 m. The ball then hits the ground with speed 8.7 m s–1. Assume that air resistance is negligible. (a) Calculate speed u. u = … m s–1 [2] (b) State how Newton’s third law applies to the collision between the ball and the ground. … … … … [2] (c) The ball is in contact with the ground for a time of 0.091 s. The ball rebounds vertically and leaves the ground with speed 5.4 m s–1. The mass of the ball is 0.059 kg. (i) Calculate the magnitude of the change in momentum of the ball during the collision. change in momentum = … N s [2] (ii) Determine the magnitude of the average resultant force that acts on the ball during the collision. average resultant force = … N [1] (iii) Use your answer in (c)(ii) to calculate the magnitude of the average force exerted by the ground on the ball during the collision. average force = … N [2] (d) The ball was thrown downwards at time t = 0 and hits the ground at time t = T. On Fig. 2.2, sketch a graph to show the variation of the speed of the ball with time t from t = 0 to t = T. Numerical values are not required. speed 0 0 T t Fig. 2.2 [1] (e) In practice, air resistance is not negligible. State and explain the variation, if any, with time t of the gradient of the graph in (d) when air resistance is not negligible. … … … … [2] [Total: 12]

12 marks

Mark scheme: 2(a) v2 = u2 + 2as u2 = 8.72 – (2 × 9.81 × 1.5) C1 u = 6.8 m s–1 A1 2(b) (magnitude of) force on ball (by ground) equal to force on ground (by ball) B1 (direction of) force on ball (by ground) opposite to force on ground (by ball) B1 2(c)(i) (p = ) 0.059 × 8.7 or 0.059 × 5.4 C1 change in momentum = 0.059 (8.7 + 5.4) = 0.83 N s A1 2(c)(ii) resultant force = 0.83 / 0.091 or 0.059 [(8.7 + 5.4) / 0.091] = 9.1 N A1 2(c)(iii) (W =) 0.059 × 9.81 C1 (W =) 0.58 (N) force = 9.1 + 0.58 = 9.7 N A1 2(d) straight line with a positive gradient and starting from a non-zero value of speed at t = 0 and ending when t = T B1 2(e) air resistance increases B1 resultant force/acceleration decreases so gradient (of curve) decreases B1

This question in 9702/22 May/June 2021

3 (a) Define power. … … [1] (b) A car of mass 1700 kg moves in a straight line along a slope that is at an angle θ to the horizontal, as shown in Fig. 3.1. B 25 m car, slope A θ mass 1700 kg horizontal Fig. 3.1 (not to scale) The car moves at constant velocity for a distance of 25 m from point A to point B. Air resistance and friction provide a total resistive force of 440 N that opposes the motion of the car. For the movement of the car from A to B: (i) state the change in the kinetic energy change in kinetic energy = … J [1] (ii) calculate the work done against the total resistive force. work done = … J [1] (c) The movement of the car in (b) from A to B causes its gravitational potential energy to increase by 4.8 × 104 J. Calculate: (i) the increase in vertical height h of the car for its movement from A to B h = … m [2] (ii) angle θ. θ = … ° [1] (d) The engine of the car in (b) produces an output power of 1.7 × 104 W to move the car along the slope. Calculate the time taken for the car to move from A to B. time = … s [2] [Total: 8]

8 marks

Mark scheme: 3(a) work (done) / time (taken) B1 3(b)(i) zero / 0 J A1 3(b)(ii) work done = 440 × 25 = 1.1 × 104 J A1 3(c)(i) (Δ)E(P) = mg(Δ)h C1 h = 4.8 × 104 / (1700 × 9.81) = 2.9 m A1 3(c)(ii) θ = sin–1 (2.9 / 25) = 6.7° A1 3(d) work done = 4.8 × 104 + 1.1 × 104 (= 5.9 × 104 J) C1 time = 5.9 × 104 / 1.7 × 104 = 3.5 s A1

This question in 9702/22 Oct/Nov 2021

2 A steel ball is projected horizontally from the top of a table, as shown in Fig. 2.1. ball table 4.9 m s–1 path of ball edge of table ground 180 cm Fig. 2.1 (not to scale) The ball is projected horizontally at a speed of 4.9 m s–1. The ball lands on the ground a horizontal distance of 180 cm from the edge of the table. Assume that air resistance is negligible. (a) (i) Calculate the time taken for the ball to reach the ground. time = … s [1] (ii) Calculate the vertical component of the velocity of the ball as it hits the ground. velocity = … m s–1 [2] (iii) Determine the magnitude and the angle to the horizontal of the velocity of the ball as it hits the ground. magnitude of velocity = … m s–1 angle to the horizontal = … ° [3] (b) The ball is projected by means of a compressed spring which is attached to a fixed block as shown in Fig. 2.2. ball x0 frictionless fixed track block spring Fig. 2.2 The ball is placed on a frictionless track in front of the spring. The ball is then pulled back so that the spring has compression x0. When the spring is released, the ball is projected horizontally as shown in Fig. 2.3. ball spring Fig. 2.3 The variation with compression x of the applied force F for the spring is shown in Fig. 2.4. 8 F / N 6 4 2 0 0 2 4 6 8 10 x / cm Fig. 2.4 The ball is a uniform sphere of steel of diameter 0.016 m and mass 0.017 kg. (i) Calculate the density of the steel. density = … kg m–3 [3] (ii) All of the elastic potential energy in the spring is converted into kinetic energy of the ball. The speed of the ball as it leaves the spring is 4.9 m s–1. Show that the maximum elastic potential energy of the spring is 0.20 J. [2] (iii) Use Fig. 2.4 to determine the spring constant k of the spring. k = … N m–1 [2] (iv) Use your answer in (b)(iii) and the value of energy given in (b)(ii) to determine the compression x0 of the spring. x0 = … m [2] (c) The steel ball is replaced by a polystyrene ball of the same diameter but of much lower mass. The spring is given compression x0 and is then released. Air resistance on this ball is not negligible after it leaves the spring. Explain: (i) why this ball leaves the spring with a greater speed than that of the steel ball … … … [1] (ii) why this ball takes a longer time to reach the ground than the steel ball. … … … [1] [Total: 17]

17 marks

Mark scheme: 2(a)(i) t = 1.8 / 4.9 A1 = 0.37 s 2(a)(ii) v = u + at C1 = 9.81  0.37 = 3.6 m s–1 A1 2(a)(iii) v 2 = 3.62 + 4.92 C1 v = 6.1 m s–1 A1 = tan–1 (3.6 / 4.9) A1 = 36° 2(b)(i) = m / V C1 4 C1 V = r3 3 4 A1 = 0.017 / [   (0.016 / 2)3 ] 3 = 7900 kg m–3 2(b)(ii) (E =) ½mv2 C1 (E =) ½  0.017  4.92 = 0.20 (J) A1 2(b)(iii) k = F / x or k = gradient C1 e.g. k = 6.4 / 10  10–2 A1 = 64 N m–1 (allow 63–65 N m–1) 2(b)(iv) E = ½kx2 C1 or E = ½Fx and F = kx x0 = [(2  0.20) / 64]0.5 A1 = 0.079 m or 0.080 m 2(c)(i) same elastic potential energy / same (initial) kinetic energy and (polystyrene ball has) smaller mass (so greater speed) B1 or same (average) force and (polystyrene ball has) smaller mass, (so greater average acceleration so greater speed) 2(c)(ii) (for the polystyrene ball there is) B1 less (average vertical) acceleration / smaller (average vertical component of) resultant force (so takes longer time to reach ground)

This question in 9702/21 Oct/Nov 2022

1 (a) Underline all the SI base units in the following list. ampere coulomb current kelvin newton [1] (b) A toy car moves in a horizontal straight line. The displacement s of the car is given by the equation v 2 s = 2a where a is the acceleration of the car and v is its final velocity. State two conditions that apply to the motion of the car in order for the above equation to be valid. 1 … 2 … [2] (c) An experiment is performed to determine the acceleration of the car in (b). The following measurements are obtained: s = 3.89 m ± 0.5% v = 2.75 m s–1 ± 0.8%. (i) Calculate the acceleration a of the car. a = … m s–2 [1] (ii) Determine the percentage uncertainty, to two significant figures, in a. percentage uncertainty = … % [2] (iii) Use your answers in (c)(i) and (c)(ii) to determine the absolute uncertainty in the calculated value of a. absolute uncertainty = … m s–2 [1] [Total: 7]

7 marks

Mark scheme: Question Answer Marks 1(a) only ampere and kelvin underlined B1 1(b) initial speed / velocity is zero B1 (non-zero magnitude of) acceleration is constant / uniform (and in a straight line) B1 1(c)(i) a = 2.752 / (2  3.89) A1 = 0.97 m s–2 1(c)(ii) percentage uncertainty = (2  0.8) + 0.5 C1 = 2.1% A1 1(c)(iii) absolute uncertainty = (2.1 / 100)  0.97 A1 = 0.02 m s–2

This question in 9702/22 Feb/March 2023

1 A well has a depth of 36 m from ground level to the surface of the water in the well, as shown in Fig. 1.1. ground 36 m well surface of water Fig. 1.1 (not to scale) A student wishes to find the depth of the well. The student plans to drop a stone down the well and record the time taken from releasing the stone to hearing the splash made by the stone as it enters the water. (a) Assume that air resistance is negligible and that the stone is released from rest. Calculate the time taken for the stone to fall from ground level to the surface of the water. time = … s [2] (b) The time recorded by the student using a stop-watch is not equal to the time in (a). Suggest three possible reasons, other than the effect of air resistance, for this difference. 1 … … 2 … … 3 … … [3] (c) The student repeats the experiment three times and uses the results to calculate the depth of the well. The values are shown in Table 1.1. Table 1.1 1st experiment 2nd experiment 3rd experiment depth / m 54.4 53.9 54.1 The true depth of the well is 36.0 m. Explain why these results may be described as precise but not accurate. … … … … [2] [Total: 7]

7 marks

Mark scheme: 1(a) t = √(2s / g) = √[(2  36) / 9.81] C1 = 2.7 s A1 1(b)  reaction time between hearing the splash and stopping the stop-watch  the sound (of the splash) takes time to reach the student or the stone hits the water at a different time to the sound being heard or the sound (of the splash) has to travel to the student  the student might not let go of the stone from ground level  the student might not let go of the stone and start the stop-watch at the same time  stop-watch may not be properly calibrated / has a zero error  (local value of) g is not (exactly) 9.81 (m s2)  stone given initial velocity / initial velocity not zero  stone does not fall (exactly) vertically / in a straight line Any three points, 1 mark each B3 1(c) precise: results are close together / have little scatter B1 not accurate: the values are not close to / 50% different / (very) different from the true value B1

This question in 9702/23 May/June 2023

3 A trolley A moves along a horizontal surface at a constant velocity towards another trolley B which is moving at a lower constant speed in the same direction. Fig. 3.1 shows the trolleys at time t = 0. A B horizontal surface Fig. 3.1 Table 3.1 shows data for the trolleys. Table 3.1 trolley mass / kg initial speed / m s–1 A 0.25 0.48 B 0.75 0.12 The two trolleys collide elastically and then separate. Resistive forces are negligible. Fig. 3.2 shows the variation with time t of the velocity v for trolley B. 0.5 v / m s–1 0.4 0.3 B 0.2 0.1 0 / s 0 0.1 0.2 0.3 0.4 0.5t –0.1 –0.2 –0.3 –0.4 –0.5 Fig. 3.2 (a) State what is represented by the area under a velocity–time graph. … [1] (b) Use Table 3.1 and Fig. 3.2 to determine: (i) the acceleration of trolley B during the collision acceleration of B = … m s–2 [2] (ii) the magnitude and direction of the final velocity of trolley A. magnitude = … m s–1 direction … [3] (c) On Fig. 3.2, sketch the variation of the velocity of trolley A with time t from t = 0 to t = 0.50 s. [3] [Total: 9]

9 marks

Mark scheme: 3(a) displacement A1 3(b)(i) a = gradient or a = v / ()t or a = (v – u) / t C1 e.g. a = (0.30 – 0.12) / (0.35 – 0.15) A1 a = 0.90 m s–2 3(b)(ii) (0.25  0.48) + (0.75  0.12) = (0.25 v) + (0.75  0.30) C1 or (0.48 – 0.12) = (0.30 – v) or (½  0.25  0.482) + (½  0.75  0.122) = (½  0.25 × v 2) + (½ × 0.75  0.302) v = (–)0.060 m s–1 A1 direction: to the left / from the right / opposite to (its) initial velocity / opposite to (initial / final) velocity of B B1 3(c) sketch: horizontal line from (0, 0.48) to (0.15, 0.48) B1 horizontal line from (0.35, –0.06) to (0.5, –0.06) B1 straight line between (0.15, 0.48) and (0.35, –0.06) B1

This question in 9702/21 Oct/Nov 2023

2 (a) Define acceleration. … … [1] (b) An Olympic diver stands on a platform above a pool of water, as shown in Fig. 2.1. 5.9 m s–1 diver 60° horizontal platform 9.0 m surface of water 1.2 m Fig. 2.1 (not to scale) When the diver is on the platform his centre of gravity is a vertical height of 9.0 m above the surface of the water. The diver jumps from the platform with a velocity of 5.9 m s–1 at an angle of 60° to the horizontal. Air resistance is negligible. When the diver hits the surface of the water, his centre of gravity is a vertical height of 1.2 m above the surface of the water. Calculate the speed of the diver at the instant he hits the surface of the water. speed = … m s–1 [3] (c) The diver in (b) enters the water and decelerates. (i) Describe and explain the variation of the viscous drag force acting on the diver in the water as he moves downwards. … … … … [2] (ii) The diver has a volume of 7.5 × 10–2 m3 . The density of the water is 1.0 × 103 kg m–3 . Show that the upthrust acting on the diver when he is entirely underwater is 740 N. [1] (iii) At a particular instant when the diver is entirely underwater his horizontal velocity is zero. The viscous drag force acting on him at this instant is 950 N vertically upwards. The diver has mass 78 kg. Determine the magnitude and direction of the acceleration of the diver. acceleration = … m s–2 direction … [4] [Total: 11]

11 marks

Mark scheme: 2(a) rate of change of velocity B1 2(b) ½ m()v2= mg()h C1 v2 = 5.92 + 2  9.81  7.8 C1 v2 = 188 v = 14 m s–1 A1 or by resolving components (C1) Vertically: v2 = u2 + 2as v2 = (5.9sin60)2 +2  –9.81  (1.2–9.0) vv = 13.4 horizontally: (C1) vh = 5.9cos60 vh = 2.95 resultant velocity = √(13.42 + 2.952) (A1) = 14 m s–1 2(c)(i) (As the diver moves down their) speed decreases B1 (So) viscous force / drag (force) decreases B1 2(c)(ii) (F =) gV A1 = 1000  9.81  7.5  10–2 = 740 (N)

This question in 9702/22 Feb/March 2024

2 (a) Define velocity. … … [1] (b) A student throws a ball over a vertical wall of height h, as shown in Fig. 2.1. path of ball wall 22 m s–1 ball 40° horizontal h ground 1.2 m 36 m Fig. 2.1 (not to scale) The ball leaves the hand of the student at a height of 1.2 m above the horizontal ground. The ball has an initial velocity of 22 m s–1 at an angle of 40° to the horizontal. The wall is a horizontal distance of 36 m from where the student releases the ball. Air resistance is negligible. (i) Determine the time taken for the ball to reach the wall. time taken = … s [2] (ii) Calculate the vertical component u of the initial velocity of the ball. u = … m s–1 [1] (iii) The ball just goes over the wall. Calculate the height h of the wall. h = … m [3] [Total: 7]

7 marks

Mark scheme: 2(a) change in displacement / time (taken) B1 2(b)(i) horizontal velocity = 22  cos 40° C1 time taken = 36 / (22  cos 40°) = 2.1 s A1 2(b)(ii) u = 22  sin 40° = 14 m s−1 A1 2(b)(iii) s = ut + ½ at2 = (14  2.1) + (½  −9.81  2.12) C1 = 7.8 (m) C1 (therefore) height of wall = 7.8 + 1.2 = 9.0 m A1 Question Answer Marks 2(b)(iii) or other methods possible e.g. time to maximum height = (0 – 14) / –9.81 (= 1.43 s) time from maximum height to wall = 2.1 – 1.43 (= 0.67 s) maximum height above release = (14  1.43) + (½  −9.81  1.432) = 9.99 (m) (C1) height from maximum to wall = 0.5  9.81  0.672 (= 2.2 m) height above release = 9.99 – 2.2 = 7.8 (m) (C1) height of wall = 1.2 + 7.8 = 9.0 m (A1)

This question in 9702/21 May/June 2024

2 (a) Define displacement from a point. … … [1] (b) An object is projected horizontally at a speed of 6.0 m s–1 from a slope, as shown in Fig. 2.1. 6.0 m s–1 object slope θ horizontal Fig. 2.1 (not to scale) The slope is at an angle θ to the horizontal. Air resistance is negligible. The object lands on the slope a time of 0.71 s later and stops without rolling or bouncing. (i) Determine the horizontal distance travelled by the object. distance = … m [1] (ii) Determine the vertical distance travelled by the object. distance = … m [2] (iii) Use your answers in (b)(i) and (b)(ii) to calculate θ. θ = … ° [2] (iv) Determine the magnitude of the displacement of the object from its original position. displacement = … m [2] (v) By considering energy, calculate the speed of the object just before it lands. speed = … m s–1 [3] [Total: 11]

11 marks

Mark scheme: 2(a) distance (from the point) in a straight line in a given direction B1 2(b)(i) distance = speed  time = 6.0  0.71 = 4.3 m A1 2(b)(ii) s = ut + ½ at2 = ½  9.81  0.712 C1 = 2.5 m A1 2(b)(iii) tan  = 2.5 / 4.3 or hypotenuse = √(4.32 + 2.52) ( = 4.97 m) cos  = 4.3 / 4.97 or sin  = 2.5 / 4.97 C1  = 30° A1 Question Answer Marks 2(b)(iv) displacement = √(4.32 + 2.52) C1 = 4.9 m or 5.0 m A1 or displacement = 2.5 / sin 30° or displacement = 4.3 / cos 30° (C1) = 5.0 m (A1) 2(b)(v) KE = ½mv2 or GPE = mgh C1 initial KE + loss in GPE = final KE (½  m  6.02) + (m  9.81  2.5) = (½  m  v2) C1 v = 9.2 m s–1 A1

This question in 9702/23 May/June 2024

1 (a) Define acceleration. … … [1] (b) A small aircraft is flying horizontally at a speed of 42 m s–1 at a height of 63 m above horizontal ground, as shown in Fig. 1.1. speed 42 m s–1 63 m ground Fig. 1.1 The aircraft drops a small parcel. The parcel is released from the aircraft at the instant shown in Fig. 1.1. Air resistance is negligible. (i) On Fig. 1.1, draw a line to show the path of the parcel as it falls from the aircraft to the ground. [1] (ii) Calculate the time taken from the instant of release to the instant the parcel reaches the ground. time = … s [2] (iii) Calculate the vertical component of the velocity of the parcel immediately before it reaches the ground. vertical component of velocity = … m s–1 [1] (iv) Determine the speed at which the parcel reaches the ground. speed = … m s–1 [2] [Total: 7]

7 marks

Mark scheme: Question Answer Marks 1(a) rate of change of velocity B1 1(b)(i) curved path from aircraft to ground, starting horizontal at aircraft and then with increasing negative gradient as it moves B1 towards the ground 1(b)(ii) s = ut + ½at2 C1 63 = ½  9.81  t2 time = 3.6 s A1 1(b)(iii) v2 = 2  9.81  63 A1 or v = 0 + (9.81  3.6) or 63 = (v  3.6) – (½  9.81  3.62) or 63 = ½  (0 + v)  3.6 v = 35 m s–1 1(b)(iv) speed2 = 352 + 422 C1 speed = 55 m s–1 A1

This question in 9702/23 Oct/Nov 2024

3 (a) A truck R of mass 9400 kg moves with constant acceleration in a straight line down a slope, as illustrated in Fig. 3.1. R A 180 m B Fig. 3.1 At point A the speed of the truck is 13 m s–1 and at point B the speed of the truck is 22 m s–1. A and B are a distance of 180 m apart. (i) Calculate the acceleration of the truck between A and B. acceleration = … m s–2 [2] (ii) Determine the gain in kinetic energy of the truck between A and B. gain in kinetic energy = … J [3] (b) A short time after passing point B truck R moves in a straight line on horizontal ground. The driver of the truck applies the brakes. Fig. 3.2 shows the variation with time of the momentum of the truck. 25 20 momentum / 104 kg m s–1 15 10 5 0 0 5 10 15 20 25 time / s Fig. 3.2 (i) Define force. … … [1] (ii) Show that the average resultant force F acting on truck R between time t = 0 and t = 15 s is –1.2 × 104 N. [1] (iii) An identical truck S has the same initial momentum as truck R. Truck S experiences a constant force equal to the force F in (b)(ii). State and explain whether truck S will take more, less or the same amount of time to come to rest as truck R. … … … … … … … [3] [Total: 10]

10 marks

Mark scheme: 3(a)(i) a = (v2 – u2) / 2s C1 = (222 – 132) / (2  180) = 0.88 m s–2 A1 OR (C1) [t = (180  2) / (22 + 13) = 10.3] 180 = 13  10.3 + ½ a  10.32 or 180 = 22  10.3 – ½ a  10.3 2 or 22 = 13 + a  10.3 a = 0.88 m s–2 (A1) 3(a)(ii) ()E = ½m()v2 C1 gain in KE = ½m(v2 – u2) C1 = ½  9400  (222 – 132) = 1.5  106 J A1 OR (C1) W = Fs = ma  d gain in KE = 9400  0.88  180 (C1) = 1.5  106 J (A1) 3(b)(i) rate of change of momentum B1 3(b)(ii) (Force =) (2.5  104 – 21  104) / 15 = – 1.2  104 (N) A1 3(b)(iii) The change of momentum (for S and R to come to rest) is the same B1 (Average) force on R (to come to rest) is B1 (–21  104 / 23 =) 0.91  104 N or (Average) force on R (to come to rest) is less than the force on S / less than F (S will come to rest in) less time (than R). B1 OR (B1) The change of momentum (for S and R to come to rest) is the same Time for S to come to rest is (–21  104 / 1.2  104 =) 17.5 s (and time for R to come to rest is 23 s) (B1) (S comes to rest in) less time (than R). (B1)

This question in 9702/22 Feb/March 2025

1 (a) Define velocity. … … [1] (b) In an experiment, two objects A and B are released from the side of a building, as shown in Fig. 1.1. building A 10.0 m 3.0 m s–1 B h ground Fig. 1.1 (not to scale) Object A is released from rest at a height of 10.0 m above horizontal ground. Object B is released with an initial upward velocity of 3.0 m s−1 at a height h above the ground. Both objects take the same time to reach the ground and they do not collide with each other. Air resistance is negligible. Calculate h. h = … m [3] (c) In a second experiment, object B is released from the same height as in (b) but with a speed of 6.0 m s−1 at an angle of 60° to the vertical, as shown in Fig. 1.2. 60° building 6.0 m s–1 B Fig. 1.2 (i) State and explain whether the time taken for object B to reach the ground is less than, the same as, or greater than the time taken in the first experiment. … … … [2] (ii) By considering energy, state and explain whether the speed at which object B reaches the ground is less than, the same as, or greater than in the first experiment. … … … [2] [Total: 8]

8 marks

Mark scheme: Question Answer Marks 1(a) Rate of change of displacement B1 1(b) For object A: C1 1 2 s = ut + at 2 2  10 so t = 9.81 =1.4 Then for object B: C1 1 2 s = ut + at 2 h = −(3 1.4) + (0.5  9.81  1.42) = 5.7 m A1 OR (C1) h = −(3 1.4) + 10 = 5.7 m (A1) 1(c)(i) time taken (to reach the ground is) same B1 The initial vertical (component of the) velocity is the same (as in part (1b)) B1 1(c)(ii) The (total) initial energy is greater (than in part (1b)) B1 change in gravitational potential energy is same, so speed is greater B1

This question in 9702/23 May/June 2025

1 (a) Define acceleration. … … [1] (b) A rocket is launched vertically from the surface of the Earth. Fig. 1.1 shows the variation of the velocity of the rocket with time for the first 20 s after its launch. 400 velocity / m s–1 200 0 0 5 10 15 20 time / s Fig. 1.1 (i) Determine the acceleration of the rocket. acceleration = … m s–2 [1] (ii) Show that the height of the rocket above the surface of the Earth at a time of 20 s after launch is 3.2 km. [2] (c) The mass of the rocket in (b) is 2.9 × 106 kg. Assume that this mass remains constant. For this rocket, from launch to its height at a time of 20 s after launch: (i) calculate the gain in gravitational potential energy ΔEP ΔEP = … J [2] (ii) calculate the gain in kinetic energy ΔEK ΔEK = … J [2] (iii) determine the average power output of the rocket engines. Assume that resistive forces are negligible. power = … W [2] [Total: 10]

10 marks

Mark scheme: Question Answer Marks 1(a) rate of change of velocity B1 1(b)(i) acceleration = 320 / 20 A1 = 16 m s–2 1(b)(ii) 1 C1 s = ((u) + v)t or distance = area under graph 2 1 A1 (height) =  20  320 = 3200 m or 3.2 km 2 1(c)(i) (EP) = mgh C1 = 2.9  106  9.81  3200 A1 = 9.1  1010 J 1(c)(ii) 1 C1 (EK) = m(v2) 2 1 A1 =  2.9  106  3202 2 = 1.5  1011 J 1(c)(iii) (power =) work done / time C1 power = ((1.5 + 0.91)  1011) / 20 A1 = 1.2  1010 W

This question in 9702/23 Oct/Nov 2025