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school. “Paper 42” means Paper 4, variant 2.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Momentum, impulse & collisions · Core · Paper 2 style · 6 marks
On a wet road in Lahore, a car of mass 1200 kg travelling at 15 m s⁻¹ runs into the back of a stationary car of mass 900 kg. The two cars lock together and move off along the same line.
(a)Define linear momentum.[1]
(b)Calculate the speed of the two cars immediately after the collision.[2]
(c)Calculate the loss of kinetic energy in the collision.[3]
Show answer
(a) Momentum is the product of mass and velocity.
B1: mass × velocity (velocity, not speed).
(b) 8.57 m s⁻¹
M1: 1200 × 15 = (1200 + 900)v. A1: v = 18 000 / 2100 = 8.57 m s⁻¹.
(c) 57.9 kJ
C1: initial KE = ½ × 1200 × 15² = 135 kJ. C1: final KE = ½ × 2100 × 8.57² = 77.1 kJ. A1: loss = 57.9 kJ.
Q2Momentum, impulse & collisions · Core · Paper 1 style · 1 mark
(a)Which of the following is a correct unit for impulse?[1]
Show answer
(a) B kg m s⁻¹
Impulse = force × time, so N s = (kg m s⁻²)(s) = kg m s⁻¹, the same unit as momentum. A is the unit of force, not impulse.
Q3Momentum, impulse & collisions · Core · Paper 2 style · 5 marks
A cricket ball of mass 0.16 kg is bowled at 32 m s⁻¹ towards a batter. It is hit straight back along the same line at 41 m s⁻¹. The bat is in contact with the ball for 1.2 ms.
(a)State Newton's second law of motion in terms of momentum.[1]
(b)Calculate the magnitude of the change in momentum of the ball.[2]
(c)Calculate the average force exerted on the ball by the bat.[2]
Show answer
(a) The resultant force on a body is proportional to (equal to) its rate of change of momentum.
B1: resultant force ∝ / = rate of change of momentum (and in the same direction).
(b) 11.7 N s
C1: take towards the batter as positive: Δp = 0.16 × (−41) − 0.16 × 32. A1: |Δp| = 0.16 × 73 = 11.7 N s (kg m s⁻¹).
(c) 9730 N
C1: F = Δp/Δt = 11.7 / 1.2 × 10⁻³. A1: F = 9730 N (opposite to the ball's original direction).
Q4Momentum, impulse & collisions · Core · Paper 2 style · 5 marks
A football of mass 0.43 kg is at rest. During a kick, the force on the ball rises steadily from zero to 1800 N in 4.0 ms and then falls steadily back to zero in a further 6.0 ms. The force always acts in the same direction.
(a)State what is represented by the area under a force–time graph.[1]
(b)Calculate the impulse given to the ball.[2]
(c)Calculate the speed of the ball as it leaves the foot.[2]
Show answer
(a) The impulse of the force (the change in momentum).
B1: impulse / change of momentum.
(b) 9.00 N s
C1: area of triangle = ½ × 1800 × 0.010. A1: 9.00 N s.
(c) 20.9 m s⁻¹
C1: impulse = mv − 0, v = 9.00 / 0.43. A1: 20.9 m s⁻¹.
Q5Momentum, impulse & collisions · Core · Paper 1 style · 1 mark
(a)Two spheres collide head-on in a perfectly elastic collision. No external forces act. Which quantity is NOT necessarily the same before and after the collision?[1]
Show answer
(a) D the momentum of each individual sphere
Total momentum is always conserved; in an elastic collision total KE is conserved and the relative speed of approach equals the relative speed of separation. Momentum is transferred between the spheres, so each sphere's momentum changes. B is conserved here because the collision is elastic.
Q6Momentum, impulse & collisions · Standard · Paper 2 style · 7 marks
On a horizontal frictionless air track, trolley A of mass 0.50 kg moves at 0.60 m s⁻¹ towards trolley B of mass 0.30 kg, which is at rest. The trolleys collide head-on and the collision is perfectly elastic.
(a)State what is meant by a perfectly elastic collision.[1]
(b)Use conservation of momentum and the relative speeds of the trolleys to calculate the velocity of each trolley after the collision.[4]
(c)Show that your answers to (b) are consistent with the collision being perfectly elastic.[2]
Show answer
(a) A collision in which the total kinetic energy is conserved.
B1: total kinetic energy is the same before and after (momentum is always conserved).
(b) A: 0.150 m s⁻¹, B: 0.750 m s⁻¹, both in the original direction of A
C1: 0.50 × 0.60 = 0.50 vA + 0.30 vB. C1: relative speed of approach = relative speed of separation: 0.60 = vB − vA. M1: solve: 0.30 = 0.50 vA + 0.30(vA + 0.60) → 0.80 vA = 0.12. A1: vA = 0.150 m s⁻¹, vB = 0.750 m s⁻¹ (same direction as A's initial motion).
(c) KE before = 0.0900 J; KE after = 0.0900 J, so KE is conserved
M1: before: ½ × 0.50 × 0.60² = 0.0900 J. A1: after: ½ × 0.50 × 0.15² + ½ × 0.30 × 0.75² = 0.0900 J, equal.
Q7Momentum, impulse & collisions · Standard · Paper 2 style · 7 marks
A radon-222 nucleus, initially at rest, decays by emitting an α-particle of kinetic energy 5.49 MeV. The daughter nucleus is polonium-218. Take the mass of the α-particle as 4u and of the daughter nucleus as 218u, where u = 1.66 × 10⁻²⁷ kg.
(a)Explain why the daughter nucleus and the α-particle move off in exactly opposite directions.[2]
(b)Calculate the speed of the α-particle.[2]
(c)Calculate the speed of the daughter nucleus.[2]
(d)Calculate the kinetic energy of the daughter nucleus, in MeV.[1]
Show answer
(a) Total momentum before is zero, so the momenta after must be equal in magnitude and opposite in direction.
B1: initial momentum is zero (nucleus at rest) and momentum is conserved. B1: so the two momenta must cancel: equal and opposite.
(b) 1.63 × 10⁷ m s⁻¹
C1: KE = 5.49 × 10⁶ × 1.60 × 10⁻¹⁹ = 8.78 × 10⁻¹³ J, m = 4 × 1.66 × 10⁻²⁷ kg. A1: v = √(2KE/m) = 1.63 × 10⁷ m s⁻¹.
A1: KE ∝ p²/m with equal p, so KE_d = (4/218) × 5.49 = 0.101 MeV (or ½mv² with values from (c)).
Q8Momentum, impulse & collisions · Standard · Paper 2 style · 7 marks
On smooth ice, puck A of mass 0.17 kg moving at 6.0 m s⁻¹ strikes an identical puck B at rest. After the collision, A moves at 3.0 m s⁻¹ in a direction at 60° to its original direction. Friction is negligible.
(a)Calculate the components of the velocity of puck B after the collision, parallel and perpendicular to the original direction of A.[3]
(b)Hence find the speed of B and the angle its path makes with the original direction of A.[2]
(c)Determine whether the collision is elastic.[2]
Show answer
(a) 4.5 m s⁻¹ parallel; 2.6 m s⁻¹ perpendicular (on the opposite side of the line from A)
C1: parallel: 0.17 × 6.0 = 0.17 × 3.0 cos 60° + 0.17 vx → vx = 4.5 m s⁻¹. C1: perpendicular: 0 = 0.17 × 3.0 sin 60° + 0.17 vy. A1: vy = −2.6 m s⁻¹ (opposite side).
(b) 5.20 m s⁻¹ at 30° to the original direction
A1: v = √(4.5² + 2.60²) = 5.20 m s⁻¹. A1: tan θ = 2.60/4.5, θ = 30° (on the other side of the line from A).
(c) Yes: total KE before = 3.06 J = total KE after
M1: KE before = ½ × 0.17 × 6.0² = 3.06 J; after = ½ × 0.17 × (3.0² + 5.20²) = 0.765 + 2.30 = 3.06 J. A1: KE conserved so elastic.
Q9Momentum, impulse & collisions · Standard · Paper 1 style · 1 mark
(a)[1]
Show answer
(a) C 1.6 N s
Velocity reverses, so Δp = 0.20 × (3.0 − (−5.0)) = 1.6 N s. A (0.40 N s) comes from subtracting speeds and ignoring the change of direction.
Q10Momentum, impulse & collisions · Standard · Paper 2 style · 7 marks
A rifle of mass 4.2 kg fires a bullet of mass 12 g horizontally at 750 m s⁻¹. The rifle is initially at rest.
(a)Calculate the recoil speed of the rifle, assuming it is free to move.[2]
(b)The rifle is brought to rest over a distance of 0.15 m. Calculate the average force needed.[3]
(c)Explain why the bullet and rifle have equal magnitudes of momentum but very different kinetic energies.[2]
Show answer
(a) 2.14 m s⁻¹
C1: 0 = 0.012 × 750 − 4.2v. A1: v = 2.14 m s⁻¹.
(b) 64.3 N
C1: KE of rifle = ½ × 4.2 × 2.14² = 9.64 J. C1: work done = F × 0.15. A1: F = 64.3 N.
(c) Momenta are equal and opposite (total zero); KE = p²/2m, so the much smaller mass of the bullet gives it far more KE (3380 J compared with 9.64 J).
B1: momentum conserved from zero, so equal magnitudes. B1: KE = p²/2m, smaller mass → larger KE (bullet ≈ 3400 J).
Q11Momentum, impulse & collisions · Standard · Paper 2 style · 6 marks
A fire-fighter in Karachi directs a horizontal jet of water at a wall. Water leaves the hose at 18 m s⁻¹ and the mass of water leaving per second is 2.5 kg s⁻¹. The water hits the wall at right angles.
(a)Calculate the force exerted by the water on the wall when the water does not rebound.[2]
(b)The wall is replaced by a surface from which the water splashes straight back at 4.0 m s⁻¹. Calculate the new force.[2]
(c)Use Newton's third law to explain why the person holding the hose feels a force.[2]
Show answer
(a) 45.0 N
C1: F = rate of change of momentum = (mass per second) × Δv. A1: F = 2.5 × 18 = 45.0 N.
(b) 55.0 N
C1: Δv = 18 − (−4.0) = 22 m s⁻¹. A1: F = 2.5 × 22 = 55.0 N.
(c) The hose exerts a forward force on the water to accelerate it; the water exerts an equal and opposite (backward) force on the hose.
B1: hose pushes water forwards (increasing its momentum). B1: water pushes the hose backwards with an equal force (third-law pair, same type of force).
Q12Momentum, impulse & collisions · Challenge · Paper 2 style · 7 marks
A ballistic pendulum is used to measure the speed of an air-gun pellet. A pellet of mass 5.0 g is fired horizontally into a wooden block of mass 1.200 kg hanging from long strings. The pellet stays in the block, and the block rises through a vertical height of 8.5 cm. Air resistance is negligible.
(a)Calculate the speed of the block and pellet immediately after the impact.[2]
(b)Calculate the speed of the pellet just before it hits the block.[2]
(c)Show that the percentage of the pellet's initial kinetic energy that is converted to other forms in the impact is more than 99%, and state two of these other forms.[3]
Show answer
(a) 1.29 m s⁻¹
C1: ½(M + m)v² = (M + m)gh. A1: v = √(2 × 9.81 × 0.085) = 1.29 m s⁻¹.
(b) 311 m s⁻¹
C1: m u = (M + m)v, 0.005u = 1.205 × 1.29. A1: u = 311 m s⁻¹.
(c) 99.6% is lost, as thermal energy (heating) and sound / work deforming the block
M1: fraction of KE remaining = m/(M + m) (or calculate both KEs: 242 J and 1.01 J). A1: lost = 99.6%. B1: any two of: internal (thermal) energy, sound, work done in plastic deformation.
Q13Momentum, impulse & collisions · Challenge · Paper 2 style · 6 marks
Two gliders of masses 0.40 kg and 0.60 kg are held at rest on a level air track with a compressed spring between them. When released, the spring transfers 0.30 J of elastic potential energy to the kinetic energy of the gliders. The spring has negligible mass and falls away.
(a)Explain why the gliders move off in opposite directions with equal magnitudes of momentum.[2]
(b)Calculate the speed of each glider after they separate.[4]
Show answer
(a) Total momentum is zero before release and is conserved (no external horizontal force), so the momenta are equal and opposite.
B1: initial total momentum zero. B1: momentum conserved so p₁ = −p₂.
(b) 0.40 kg glider: 0.949 m s⁻¹; 0.60 kg glider: 0.632 m s⁻¹
C1: 0.40v₁ = 0.60v₂, so v₁ = 1.5v₂. C1: ½ × 0.40v₁² + ½ × 0.60v₂² = 0.30. M1: 0.45v₂² + 0.30v₂² = 0.30. A1: v₂ = 0.632 m s⁻¹, v₁ = 0.949 m s⁻¹.
Q14Momentum, impulse & collisions · Challenge · Paper 2 style · 6 marks
A firework shell is momentarily at rest at the top of its flight when it explodes into three fragments that all move horizontally. A fragment of mass 0.20 kg moves east at 30 m s⁻¹ and a fragment of mass 0.30 kg moves north at 20 m s⁻¹. The third fragment has mass 0.25 kg. Ignore gravity during the very short explosion.
(a)Calculate the speed of the third fragment and state its direction.[4]
(b)Calculate the total kinetic energy released by the explosion.[2]
Show answer
(a) 33.9 m s⁻¹ towards the south-west (45° south of west)
C1: momenta of first two: 6.0 N s east and 6.0 N s north. C1: third must cancel: magnitude √(6.0² + 6.0²) = 8.49 N s. A1: v = 8.49/0.25 = 33.9 m s⁻¹. B1: direction south-west, at 45° to both west and south.
Q15Momentum, impulse & collisions · Challenge · Paper 2 style · 7 marks
In a nuclear reactor, fast neutrons are slowed down by collisions with carbon-12 nuclei. Consider a neutron (mass 1u) moving at speed u that makes a head-on, perfectly elastic collision with a stationary carbon-12 nucleus (mass 12u).
(a)Using conservation of momentum and the fact that relative speed of approach equals relative speed of separation, show that the neutron rebounds with a speed of (11/13)u.[3]
(b)Calculate the fraction of its kinetic energy that the neutron keeps in one such collision.[1]
(c)A neutron starts with kinetic energy 2.0 MeV. Calculate the minimum number of such head-on collisions needed to reduce its kinetic energy below 1.0 keV.[3]
Show answer
(a) Neutron rebounds at 11u/13 (carbon moves forward at 2u/13).
M1: 1 × u = 1 × v_n + 12 × v_C. M1: u = v_C − v_n. A1: substitute: u = v_n + 12(u + v_n) → 13v_n = −11u, so speed 11u/13 (direction reversed).
Q16Drag, non-uniform motion & terminal velocity · Core · Paper 1 style · 1 mark
(a)[1]
Show answer
(a) C The resultant force is zero and the acceleration is zero.
At terminal velocity drag equals weight, so the resultant force and the acceleration are both zero. B describes the motion before terminal velocity is reached.
Q17Drag, non-uniform motion & terminal velocity · Core · Paper 2 style · 5 marks
A steel ball bearing is released from rest just below the surface of a tall cylinder of oil.
(a)State the three forces acting on the ball bearing as it falls through the oil.[1]
(b)Explain why the acceleration of the ball bearing decreases as it falls.[2]
(c)State and explain what happens to the ball bearing's speed once the acceleration becomes zero.[2]
Show answer
(a) Weight, upthrust and viscous drag.
B1: all three named (weight downwards, upthrust and drag upwards).
(b) As speed increases the drag increases; the resultant downward force (weight − upthrust − drag) decreases, so acceleration decreases.
B1: drag increases with speed. B1: resultant force decreases (weight and upthrust constant), so a = F/m decreases.
(c) It stays constant (terminal velocity) because drag + upthrust = weight, so the resultant force is zero.
Q18Drag, non-uniform motion & terminal velocity · Core · Paper 2 style · 5 marks
A skydiver of total mass 75 kg falls vertically before opening her parachute. At one instant her speed is 30 m s⁻¹ and the air resistance on her is 410 N.
(a)Calculate the acceleration of the skydiver at this instant.[2]
(b)State the air resistance on the skydiver when she reaches terminal velocity.[1]
(c)The air resistance is proportional to the square of the speed. Calculate the terminal speed.[2]
Show answer
(a) 4.34 m s⁻²
C1: resultant force = 75 × 9.81 − 410 = 326 N. A1: a = 326/75 = 4.34 m s⁻² (downwards).
(b) 736 N
B1: equal to weight, 75 × 9.81 = 736 N.
(c) 40.2 m s⁻¹
C1: 736/410 = (v_t/30)². A1: v_t = 40.2 m s⁻¹.
Q19Drag, non-uniform motion & terminal velocity · Standard · Paper 2 style · 6 marks
On the Lahore–Islamabad motorway, the air resistance F on a car is given by F = kv², where v is the speed of the car and k = 0.42 in SI units.
(a)Express the unit of k in SI base units.[1]
(b)Calculate the drag force at 25 m s⁻¹ and the power needed just to overcome it.[3]
(c)Show that increasing the speed from 25 m s⁻¹ to 30 m s⁻¹ multiplies this power by about 1.7.[2]
Show answer
(a) kg m⁻¹
B1: k = F/v² → kg m s⁻² / (m² s⁻²) = kg m⁻¹.
(b) 263 N; 6.56 kW
C1: F = 0.42 × 25² = 262.5 ≈ 263 N. C1: P = Fv. A1: P = 262.5 × 25 = 6.56 kW.
(c) factor 1.73
M1: P = kv³. A1: (30/25)³ = 1.73.
Q20Drag, non-uniform motion & terminal velocity · Standard · Paper 1 style · 1 mark
(a)[1]
Show answer
(a) B She decelerates, with a deceleration that decreases in magnitude, until she reaches a lower terminal velocity.
Opening the parachute makes drag greater than weight, so she decelerates; as she slows the drag falls, so the deceleration gets smaller until drag = weight again at a lower speed. She does not move upwards (A); she is still moving down.
Q21Drag, non-uniform motion & terminal velocity · Standard · Paper 3 style · 8 marks
A student drops steel spheres of different radius r into a tall tube of glycerol. She times each sphere between two marks 0.300 m apart, the upper mark being 15 cm below the surface.Results (r / mm, t / s): 1.0, 15.2; 1.5, 6.6; 2.0, 3.8; 2.5, 2.4; 3.0, 1.7.
(a)Calculate the terminal velocity v for each sphere. Give your values to an appropriate number of significant figures.[2]
(b)It is suggested that v = Kr², where K is a constant. Plot a suitable graph (describe the axes) and use it to estimate K in m s⁻¹ mm⁻².[3]
(c)Suggest one reason why the value of t for the largest sphere is the least reliable reading, and one improvement to deal with it.[2]
(d)Explain why the upper mark is placed well below the surface of the glycerol.[1]
Show answer
(a) 0.0197, 0.0455, 0.0789, 0.125, 0.176 m s⁻¹
M1: v = 0.300/t for each row. A1: all correct to 2 or 3 s.f. (0.0197, 0.0455, 0.0789, 0.125, 0.176).
(b) v against r² gives a straight line through the origin; K ≈ 0.020 m s⁻¹ mm⁻² (accept 0.019–0.021)
B1: plot v (y-axis) against r² (x-axis). M1: gradient from a large triangle on the line. A1: K = gradient ≈ 0.020 m s⁻¹ mm⁻².
(c) Time is short (≈1.7 s) so reaction time gives a large percentage uncertainty; use a longer fall distance / video with a timer / light gates.
B1: short time → large percentage uncertainty from reaction time. B1: specific improvement: longer distance between marks, or video the fall with a stopwatch in view, or light gates.
(d) So that the sphere has reached terminal velocity before timing starts.
B1: sphere must already be at constant (terminal) speed when it passes the first mark.
Q22Drag, non-uniform motion & terminal velocity · Challenge · Paper 2 style · 8 marks
A spherical raindrop of radius 0.50 mm falls through still air. The density of water is 1000 kg m⁻³. The air resistance is kv², where v is the speed and k is a constant for this drop. The terminal speed of the drop is 4.0 m s⁻¹. Ignore upthrust.
(a)Calculate the constant k.[3]
(b)Calculate the acceleration of the drop when its speed is 2.0 m s⁻¹.[2]
(c)For drops of different radius r, k is proportional to r². Show that the terminal speed is proportional to √r, and calculate the terminal speed of a drop of radius 1.0 mm.[3]
Show answer
(a) 3.21 × 10⁻⁷ kg m⁻¹
C1: m = (4/3)π(0.50 × 10⁻³)³ × 1000 = 5.24 × 10⁻⁷ kg. C1: at terminal velocity kv_t² = mg. A1: k = mg/4.0² = 3.21 × 10⁻⁷ kg m⁻¹.
(b) 7.36 m s⁻²
C1: a = (mg − kv²)/m = g(1 − v²/v_t²). A1: a = 9.81 × (1 − 0.25) = 7.36 m s⁻².
(c) v_t ∝ √r; 5.66 m s⁻¹
M1: mg ∝ r³, so v_t² = mg/k ∝ r³/r² = r. M1: v_t ∝ √r. A1: v_t = 4.0 × √2 = 5.66 m s⁻¹.
Q23Drag, non-uniform motion & terminal velocity · Challenge · Paper 2 style · 7 marks
A ball of mass 0.15 kg is thrown vertically upwards on a playing field in Peshawar. Air resistance acts on it at all times while it is moving. When it is moving upwards at 12 m s⁻¹ the air resistance is 0.60 N. When it is moving downwards at 8.0 m s⁻¹ the air resistance is 0.27 N.
(a)Calculate the acceleration of the ball at the instant it is moving upwards at 12 m s⁻¹.[2]
(b)Calculate the acceleration of the ball at the instant it is moving downwards at 8.0 m s⁻¹.[2]
(c)State the acceleration of the ball at the top of its path.[1]
(d)Explain why the ball takes longer to fall back to the thrower than it took to rise.[2]
Show answer
(a) 13.8 m s⁻² downwards
C1: weight and drag both act downwards: F = 1.47 + 0.60 = 2.07 N. A1: a = 13.8 m s⁻² downwards.
(b) 8.01 m s⁻² downwards
C1: drag now acts upwards: F = 1.47 − 0.27 = 1.20 N. A1: a = 8.01 m s⁻² downwards.
(c) 9.81 m s⁻² downwards
B1: speed is zero so drag is zero; a = g = 9.81 m s⁻².
(d) Air resistance removes energy, so at each height the ball is slower on the way down; also the downward acceleration is less than g on the way down but greater than g on the way up.
B1: work done against drag, so speed at any height lower on the way down. B1: same distance covered at lower average speed (acceleration < g down, > g up) → longer time.
Q24Newton's laws, mass & weight · Core · Paper 1 style · 1 mark
(a)[1]
Show answer
(a) C the gravitational force of the book on the Earth
Third-law pairs act on different bodies and are the same type of force: the Earth pulls the book, so the book pulls the Earth (gravitational). A is equal and opposite to the weight but acts on the same body and is a different type of force.
Q25Newton's laws, mass & weight · Core · Paper 2 style · 6 marks
A passenger of mass 62 kg stands in a lift in an office tower in Islamabad. The lift starts from rest and accelerates upwards at 1.5 m s⁻².
(a)Distinguish between mass and weight.[2]
(b)Calculate the force exerted on the passenger by the floor of the lift.[2]
(c)State the force exerted by the floor when the lift moves upwards at constant speed. Explain your answer.[2]
Show answer
(a) Mass is the property of a body that resists change in motion (quantity of matter), measured in kg; weight is the gravitational force on it, measured in N.
B1: mass: property resisting changes in motion / amount of matter, scalar, kg. B1: weight: force due to a gravitational field on the mass (W = mg), vector, N.
(b) 701 N
C1: R − mg = ma. A1: R = 62 × (9.81 + 1.5) = 701 N.
(c) 608 N (= weight)
B1: 62 × 9.81 = 608 N. B1: constant velocity so resultant force zero (Newton's first law).
Q26Newton's laws, mass & weight · Core · Paper 2 style · 5 marks
On a farm near Multan, a tractor of mass 3200 kg pulls a loaded trolley of mass 1800 kg along a straight level track using a light tow bar. The driving force on the tractor is 7500 N. Resistive forces of 900 N act on the tractor and 600 N act on the trolley.
(a)Calculate the acceleration of the tractor and trolley.[2]
(b)Calculate the tension in the tow bar.[2]
(c)State the force exerted by the trolley on the tractor through the tow bar.[1]
Show answer
(a) 1.20 m s⁻²
C1: resultant force on whole system = 7500 − 900 − 600 = 6000 N. A1: a = 6000/5000 = 1.20 m s⁻².
(b) 2760 N
C1: for trolley: T − 600 = 1800 × 1.20. A1: T = 2760 N.
(c) 2760 N backwards
B1: equal and opposite (Newton's third law): 2760 N towards the back.
Q27Newton's laws, mass & weight · Standard · Paper 2 style · 6 marks
A block of mass 2.0 kg rests on a smooth horizontal bench. It is attached by a light inextensible string, passing over a frictionless pulley at the edge of the bench, to a block of mass 0.50 kg hanging freely. The system is released from rest.
(a)Calculate the acceleration of the blocks.[2]
(b)Calculate the tension in the string.[2]
(c)Calculate the time taken for the block on the table to move 0.80 m from rest.[2]
Show answer
(a) 1.96 m s⁻²
C1: 0.50 × 9.81 = (2.0 + 0.50)a. A1: a = 1.96 m s⁻².
(b) 3.92 N
C1: T = 2.0 × a (only force on the block on the table). A1: T = 3.92 N (less than 4.9 N, the weight of the hanging block).
(c) 0.903 s
C1: s = ½at². A1: t = √(2 × 0.80 / 1.96) = 0.903 s.
Q28Newton's laws, mass & weight · Standard · Paper 2 style · 5 marks
A box of mass 30 kg is released on a ramp inclined at 25° to the horizontal at a warehouse loading bay. A constant frictional force of 60 N acts up the ramp while the box slides down.
(a)Calculate the component of the weight of the box acting down the ramp.[1]
(b)Calculate the acceleration of the box.[2]
(c)The box starts from rest. Calculate its speed after sliding 4.0 m down the ramp.[2]
Show answer
(a) 124 N
A1: 30 × 9.81 × sin 25° = 124 N.
(b) 2.15 m s⁻²
C1: resultant = 124.4 − 60 = 64.4 N. A1: a = 64.4/30 = 2.15 m s⁻².
(c) 4.14 m s⁻¹
C1: v² = 2as = 2 × 2.15 × 4.0. A1: v = 4.14 m s⁻¹.
Q29Newton's laws, mass & weight · Challenge · Paper 2 style · 7 marks
A student stands on bathroom scales (which measure the normal contact force in newtons) in a lift. The lift starts from rest and makes one journey. The readings are:stage 1 (at rest): 588 Nstage 2 (speeding up): 650 Nstage 3 (constant speed): 588 Nstage 4 (slowing down): 540 Nstage 5 (at rest): 588 N
(a)Calculate the mass of the student.[1]
(b)Calculate the magnitude of the acceleration during stages 2 and 4.[3]
(c)Explain how the readings show that the lift travelled upwards rather than downwards.[2]
(d)State the reading on the scales if the lift were in free fall.[1]
Show answer
(a) 59.9 kg
A1: m = 588/9.81 = 59.9 kg.
(b) stage 2: 1.03 m s⁻² upwards; stage 4: 0.801 m s⁻² downwards
C1: R − mg = ma. A1: stage 2: a = (650 − 588)/59.9 = 1.03 m s⁻² (upward). A1: stage 4: a = (588 − 540)/59.9 = 0.801 m s⁻² (downward).
(c) Starting from rest, the reading rose above the weight, so the acceleration while speeding up was upwards; speeding up from rest means the velocity is in the direction of the acceleration, i.e. upwards. In stage 4 the downward acceleration then slows this upward motion.
B1: stage 2 reading > weight → upward resultant force/acceleration. B1: lift speeds up from rest, so velocity is in the same direction as this acceleration: upwards (a downward journey would have given a reading below 588 N at the start).
(d) 0 N
B1: zero (R = m(g − g) = 0).
Q30Newton's laws, mass & weight · Challenge · Paper 2 style · 6 marks
At a cement plant, sand falls vertically onto a horizontal conveyor belt at a steady rate of 15 kg s⁻¹. The belt moves at a constant speed of 2.0 m s⁻¹. The sand lands with no horizontal velocity and is quickly carried along at the speed of the belt.
(a)Use Newton's second law in terms of momentum to calculate the horizontal force the motor must apply to the belt to keep its speed constant.[2]
(b)Calculate the power supplied by this force.[1]
(c)Calculate the rate at which the sand gains kinetic energy and explain why it differs from your answer to (b).[3]
Show answer
(a) 30.0 N
C1: F = (Δm/Δt) × v. A1: F = 15 × 2.0 = 30.0 N.
(b) 60.0 W
A1: P = Fv = 30.0 × 2.0 = 60.0 W.
(c) 30.0 W; the other half is transferred to thermal energy (and sound) by friction as the sand slips on the belt before reaching belt speed
C1: ½ × 15 × 2.0² per second. A1: 30.0 W. B1: the remaining 30.0 W is dissipated as heat/internal energy by friction while sand slides relative to the belt.
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