S5 Physics Practice Paper: Newton’s Laws of Motion and Momentum

A 4.0 kg trolley at rest on a frictionless horizontal track is pushed by a constant horizontal force of 18 N.

(a) Find the acceleration of the trolley. (2 marks)
(b) Find its speed after 3.0 s. (2 marks)
(c) Find the distance travelled in 3.0 s. (1 mark)

(a) F = ma, so a = F / m = 18 / 4.0 = 4.5 m s⁻². (2 marks)
(b) From rest with constant acceleration, v = at = 4.5 × 3.0 = 13.5 m s⁻¹. (2 marks)
(c) s = ½at² = 0.5 × 4.5 × 3.0² = 20.2 m. (1 mark)

A 6.0 kg ball moves to the right with momentum 24 kg m s⁻¹.

(a) Find the speed of the ball. (2 marks)
(b) Find its kinetic energy. (2 marks)
(c) If the ball is brought to rest uniformly in 4.0 s, find the average net force on it and state its direction. (3 marks)

(a) p = mv, v = p / m = 24 / 6.0 = 4.0 m s⁻¹. (2 marks)
(b) E_k = ½mv² = 0.5 × 6.0 × 4² = 48 J. (2 marks)
(c) F = Δp / Δt = 24 / 4.0 = 6.0 N, acting opposite to the motion (to the left). (3 marks)

A student of mass 60 kg stands on a scale inside a lift (take g = 9.81 m s⁻²).

(a) The lift accelerates upwards at 2.0 m s⁻². Find the reading of the scale. (3 marks)
(b) Compare the reading with the student’s weight and explain the difference. (2 marks)
(c) The lift then moves upwards at a constant speed of 3.0 m s⁻¹. Find the reading. (2 marks)
(d) If the cable suddenly breaks and the lift falls freely, find the reading. (1 mark)

(a) Taking upwards as positive: R − mg = ma, R = m(g + a) = 60 × (9.81 + 2.0) = 708.6 N (about 709 N). (3 marks)
(b) The weight is W = mg = 588.6 N; the reading exceeds the weight by 120 N because the lift accelerates upwards and the scale must provide an extra upward net force. (2 marks)
(c) At constant speed a = 0, so R = mg = 588.6 N. (2 marks)
(d) In free fall a = g downwards, so R = m(g − g) = 0 N. (1 mark)

A 10 kg crate on a horizontal floor is pushed by a horizontal force of 45 N. The coefficient of kinetic friction between the crate and the floor is 0.25 (take g = 9.81 m s⁻²).

(a) Find the magnitude of the friction acting on the crate. (3 marks)
(b) Find the acceleration of the crate. (3 marks)
(c) If the pushing force is increased to 60 N, find the new acceleration. (2 marks)

(a) Normal reaction N = mg = 10 × 9.81 = 98.1 N, friction f = μN = 0.25 × 98.1 = 24.5 N. (3 marks)
(b) Net force = 45 − 24.5 = 20.5 N, a = net force / m = 2.05 m s⁻². (3 marks)
(c) New net force = 60 − 24.5 = 35.5 N, a = 3.55 m s⁻². (2 marks)

On a frictionless horizontal track, a 2.0 kg trolley moving at 6.0 m/s collides with a stationary 4.0 kg trolley and the two move together after the collision.

(a) Find the common velocity after the collision. (3 marks)
(b) Find the total kinetic energy before and after the collision, and the energy lost. (3 marks)
(c) Explain why momentum is conserved but kinetic energy is not. (2 marks)

(a) Conservation of momentum: 2.0 × 6.0 = (2.0 + 4.0) × v, v = 12 / 6.0 = 2.00 m s⁻¹. (3 marks)
(b) Initial E_k = ½ × 2.0 × 6.0² = 36 J; final E_k = ½ × 6.0 × 2.0² = 12 J; loss = 24 J (66.7%). (3 marks)
(c) No external horizontal force acts on the system, so momentum is conserved; however the collision is completely inelastic and some kinetic energy becomes heat and deformation energy, so kinetic energy is not conserved. (2 marks)

A small ball is projected horizontally at 5.0 m/s from a table of height 1.8 m (ignore air resistance, take g = 9.81 m s⁻²).

(a) Find the time of flight. (3 marks)
(b) Find the horizontal distance from the table edge to the landing point. (2 marks)
(c) Find the magnitude and direction of the velocity on landing. (3 marks)

(a) Vertically: h = ½gt², t = √(2h/g) = √(2 × 1.8 / 9.81) = 0.61 s. (3 marks)
(b) Horizontally at constant speed: x = 5.0 × 0.606 = 3.03 m. (2 marks)
(c) v_y = gt = 9.81 × 0.606 = 5.94 m s⁻¹; v = √(5.0² + 5.94²) = 7.77 m s⁻¹, at θ = tan⁻¹(5.94/5.0) = 50° below the horizontal. (3 marks)

A car travels around a horizontal bend of radius 50 m. The coefficient of friction between the tyres and the road is 0.60 (take g = 9.81 m s⁻²).

(a) Find the maximum safe speed, giving your answer in km/h. (3 marks)
(b) Explain what provides the centripetal force needed for the car to turn. (3 marks)
(c) After rain the coefficient drops to 0.35. Find the new maximum safe speed. (2 marks)

(a) Static friction provides the centripetal force: μmg = mv²/r, so v = √(μgr) = √(0.60 × 9.81 × 50) = 17.2 m s⁻¹ (= 61.8 km/h). (3 marks)
(b) The centripetal force is provided by static friction between the tyres and the road, directed towards the centre of the bend; without enough friction the car would slide along the tangent. (3 marks)
(c) v = √(0.35 × 9.81 × 50) = 13.1 m s⁻¹, clearly lower, so the car must slow down on a wet road. (2 marks)

A ball of mass 0.15 kg hits a wall perpendicularly at 20 m/s and rebounds at 15 m/s. The contact time is 0.020 s (take g = 9.81 m s⁻²).

(a) Find the magnitude of the change in momentum of the ball. (3 marks)
(b) Find the average force exerted by the wall on the ball. (3 marks)
(c) The weight of the ball is only about 1.5 N. Explain why the average force is much larger than its weight. (2 marks)

(a) Taking away from the wall as positive: Δp = m(v − u) = 0.15 × (15 − (−20)) = 5.25 kg m s⁻¹. (3 marks)
(b) F = Δp / Δt = 5.25 / 0.020 = 262 N (about 263 N). (3 marks)
(c) The weight is only 1.47 N, but the momentum changes greatly in a very short time, so by F = Δp/Δt the average force is over a hundred times the weight. (2 marks)

Blocks of mass 3.0 kg and 5.0 kg are connected by a light string over a smooth fixed pulley. The 5.0 kg block moves downwards (take g = 9.81 m s⁻²).

(a) Find the acceleration of the system. (4 marks)
(b) Find the tension in the string. (3 marks)
(c) Explain why the tension lies between the weights of the two blocks. (3 marks)

(a) For 5.0 kg: 5g − T = 5a; for 3.0 kg: T − 3g = 3a. Adding gives a = (5 − 3)g / (5 + 3) = 2.45 m s⁻². (4 marks)
(b) T = 3(g + a) = 3 × (9.81 + 2.45) = 36.8 N. (3 marks)
(c) Both blocks share the same acceleration: the 3.0 kg block accelerates upwards, so T > 3g (29.4 N); the 5.0 kg block accelerates downwards, so T is less than 5g (49.1 N). Hence the tension lies between the two weights. (3 marks)

A car of mass 1200 kg travelling at 25 m/s on a level road hits a wall and comes to rest.

(a) Find the momentum and kinetic energy of the car before the collision. (3 marks)
(b) If the car stops in 0.12 s, find the average force exerted by the wall. (3 marks)
(c) If the car has a crumple zone and stops in 0.80 s, find the average force. (2 marks)
(d) Explain how a crumple zone reduces the chance of injury to passengers. (2 marks)

(a) p = mu = 1200 × 25 = 30000 kg m s⁻¹ (3.0 × 10⁴); E_k = ½mu² = 0.5 × 1200 × 25² = 375000 J (3.75 × 10⁵ J). (3 marks)
(b) F = Δp / Δt = 30000 / 0.12 = 250000 N (2.5 × 10⁵ N). (3 marks)
(c) F = 30000 / 0.80 = 37500 N (3.75 × 10⁴ N). (2 marks)
(d) The crumple zone lengthens the collision time; for the same change in momentum the average force is much smaller (about one sixth), so the acceleration experienced by the passengers and the risk of injury are reduced. (2 marks)

A bullet of mass 0.020 kg travelling horizontally at 400 m/s embeds itself in a 3.0 kg block at rest on a smooth horizontal surface. The block then compresses a spring of spring constant k = 800 N/m.

(a) Find the common velocity of the block and bullet just after the impact. (3 marks)
(b) Find the maximum compression of the spring. (4 marks)
(c) Find the percentage of the original kinetic energy lost during the collision. (3 marks)

(a) Conservation of momentum: 0.020 × 400 = (0.020 + 3.0) × v, v = 8.0 / 3.02 = 2.65 m s⁻¹. (3 marks)
(b) All kinetic energy after the impact becomes elastic potential energy: ½(M+m)v² = ½kx², x = v√((M+m)/k) = 2.649 × √(3.02/800) = 0.163 m (about 16 cm). (4 marks)
(c) Initial E_k = ½ × 0.020 × 400² = 1600 J; E_k after impact = 10.6 J; loss = 1589 J, i.e. 99.3%, mostly converted into heat and deformation energy. (3 marks)

A ticker timer records the motion of a trolley on an inclined plane. The tape is marked every 0.10 s and four successive displacements are 4.9 cm, 6.9 cm, 8.9 cm and 10.9 cm.

(a) Find the increase in displacement per interval. (2 marks)
(b) Find the acceleration of the trolley. (3 marks)
(c) If the mass of the trolley is 0.50 kg, find the net force acting on it. (2 marks)
(d) State two sources of experimental error and one improvement for each. (3 marks)

(a) Increase per interval = 6.9 − 4.9 = 2.0 cm, and the next two intervals also increase by 2.0 cm, showing uniform acceleration. (2 marks)
(b) Δs = aT², a = 0.020 / 0.10² = 2.0 m s⁻². (3 marks)
(c) F = ma = 0.50 × 2.0 = 1.0 N. (2 marks)
(d) Sources of error: (i) friction between the tape and the timer, improved by keeping the tape straight and replacing the carbon paper regularly; (ii) random error in measuring displacements, improved by averaging over more intervals and using a finer ruler. (Other reasonable answers accepted; 1.5 marks each, 3 marks total)