S4 Physics Paper: Mechanics, Newton Laws, Momentum And Circular Motion (100 Marks)

S4 Physics Paper: Mechanics, Newton Laws, Momentum and Circular Motion (100 marks)

2026-11-12

Physics – Mechanics and Momentum

S4 Physics: 6 questions, 24 parts, 100 marks, 75 min. Covers braking, slope friction, momentum, impulse, circular motion and power.

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Paper details　

Section A: Uniform Acceleration and Braking Distance (16 marks)

A car starts from rest and accelerates at 2.0 m/s2 along a straight road for 5.0 s. The driver then brakes and the car decelerates uniformly to rest in 4.0 s. Take the direction of motion as positive in all parts.

#	Question	Answer and marking notes
1	(a) Find the velocity at the end of the acceleration stage and the distance travelled in that stage. (4 marks)	Answer: v = u + at = 0 + 2.0 x 5.0 = 10.0 m/s; s = ut + 0.5at^2 = 0 + 0.5 x 2.0 x 25 = 25.0 m. Marking: 2 marks for the velocity, 2 for the distance.
2	(b) Find the acceleration during braking, giving magnitude and direction. (4 marks)	Answer: a = (v – u)/t = (0 – 10.0)/4.0 = -2.5 m/s2, so the deceleration is 2.5 m/s2 opposite to the motion. Marking: 2 marks for the substitution, 1 for the value, 1 for the direction.
3	(c) Find the braking distance and the total distance travelled. (4 marks)	Answer: Using v^2 = u^2 + 2as, 0 = 100 – 2 x 2.5 x s, so the braking distance is 100/5.0 = 20.0 m. Total distance = 25.0 + 20.0 = 45.0 m. Marking: 2 marks for the equation, 1 for the braking distance, 1 for the total.
4	(d) On a wet road the braking distance of the same car is clearly longer. Explain with Newton second law, and distinguish reaction distance from braking distance. (4 marks)	Answer: A wet road has a smaller coefficient of friction, so the maximum friction force is smaller; by Newton second law a = F/m the deceleration is smaller and the braking distance for the same initial speed is longer. The reaction distance is the distance covered at the original speed while the driver reacts and applies the brake, so it depends on reaction time, whereas the braking distance runs from the moment the brakes act until the car stops and depends on the initial speed and the deceleration. Marking: 2 marks for friction and deceleration, 1 for reaction distance, 1 for braking distance.

Section B: Friction and Motion on an Inclined Plane (17 marks)

A 5.0 kg block slides from rest down an inclined plane. The plane makes 30 degrees with the horizontal and the coefficient of kinetic friction between the block and the plane is 0.25. Take g = 9.81 m/s2.

#	Question	Answer and marking notes
5	(a) Find the component of the weight along the plane and the normal force from the plane. (4 marks)	Answer: Component along the plane = mg sin(theta) = 5.0 x 9.81 x 0.5 = 24.52 N; normal force = mg cos(theta) = 5.0 x 9.81 x cos 30 = 42.48 N. Marking: 2 marks for the parallel component, 2 for the normal force.
6	(b) Find the magnitude of the kinetic friction acting on the block. (4 marks)	Answer: f = muN = 0.25 x 42.48 = 10.62 N acting up the plane. Marking: 2 marks for the formula, 1 for the value, 1 for the direction.
7	(c) Find the acceleration of the block down the plane. (4 marks)	Answer: Net force = 24.52 – 10.62 = 13.91 N, so a = F/m = 13.91/5.0 = 2.78 m/s2 down the plane. Marking: 2 marks for the net force, 1 for Newton second law, 1 for the value and direction.
8	(d) How long does the block take to slide 4.0 m? Explain why a larger coefficient of friction increases this time. (5 marks)	Answer: From s = 0.5at^2, 4.0 = 0.5 x 2.78 x t^2, so t^2 = 8.0/2.78 = 2.876 and t = 1.70 s. A larger coefficient of friction gives a larger friction force, hence a smaller net force and a smaller acceleration, so more time is needed for the same distance. Marking: 2 marks for the equation, 2 for the time, 1 for the explanation.

Section C: Conservation of Momentum and Collisions (17 marks)

A 2.0 kg laboratory trolley moves at 3.0 m/s on a horizontal track and collides with a stationary 4.0 kg trolley. After the collision the two trolleys move together with a common velocity. Friction on the track is negligible.

#	Question	Answer and marking notes
9	(a) Find the common velocity after the collision. (4 marks)	Answer: Momentum is conserved: 2.0 x 3.0 + 0 = (2.0 + 4.0) x v, so v = 6.0/6.0 = 1.00 m/s in the original direction. Marking: 2 marks for the equation, 1 for the value, 1 for the direction.
10	(b) Find the total kinetic energy before and after the collision, and the kinetic energy lost. (4 marks)	Answer: Before = 0.5 x 2.0 x 9.0 = 9.0 J; after = 0.5 x 6.0 x 1.0 = 3.0 J; energy lost = 6.0 J. Marking: 1 mark each for before, after and the loss, 1 for units.
11	(c) Find the impulse received by the 4.0 kg trolley during the collision. (4 marks)	Answer: Impulse = change in momentum = 4.0 x (1.0 – 0) = 4.00 N s in the original direction. Equivalently the 2.0 kg trolley loses 4.0 N s, so the two impulses are equal and opposite. Marking: 2 marks for the definition, 1 for the value, 1 for the direction or the third law.
12	(d) Kinetic energy falls by 66.7 per cent yet momentum is conserved. Explain why both statements can hold, and state where the lost energy goes. (5 marks)	Answer: Momentum is a vector and the net external force on the system during the collision is zero, so total momentum is conserved; kinetic energy is a scalar and can be converted into other forms. This is a perfectly inelastic collision, the trolleys stay joined, and part of the kinetic energy becomes deformation, sound and internal energy. Momentum and kinetic energy are separate conditions, and conservation of one does not require conservation of the other. Marking: 2 marks for the momentum argument, 2 for the energy argument, 1 for where the energy goes.

Section D: Impulse, Change in Momentum and Impact Force (17 marks)

A 0.50 kg ball hits a wall at 8.0 m/s and rebounds along the same line at 6.0 m/s. The contact time between ball and wall is 0.020 s. Take the direction towards the wall as positive.

#	Question	Answer and marking notes
13	(a) Find the change in momentum of the ball and state its direction. (4 marks)	Answer: Initial momentum = 0.50 x 8.0 = 4.0 N s; final momentum = 0.50 x (-6.0) = -3.0 N s; change = -3.0 – 4.0 = -7.00 N s, that is 7.0 N s directed away from the wall. Marking: 1 mark each for the two momenta, 1 for the subtraction, 1 for the direction.
14	(b) Find the average force exerted by the wall on the ball. (5 marks)	Answer: F = change in momentum / time = 7.00/0.020 = 350.0 N directed away from the wall. Marking: 2 marks for impulse equals change in momentum, 1 for substitution, 1 for the value, 1 for the direction.
15	(c) If the contact time is extended to 0.040 s for the same change in momentum, find the average force. (4 marks)	Answer: For the same change in momentum, F is inversely proportional to the contact time, so F = 7.00/0.040 = 175.0 N, about half the previous value. Marking: 2 marks for the inverse relation, 1 for the value, 1 for the comparison.
16	(d) Use this to explain how airbags and crumple zones protect passengers, and why following through increases the force in a tennis stroke. (4 marks)	Answer: Seat belts, airbags and crumple zones lengthen the contact time so that the same change in momentum is spread over a longer interval, reducing the average force on the body. Following through lengthens the contact time between racket and ball and also allows a larger change in the momentum of the racket, so the impulse on the ball is larger and its speed is higher. Marking: 2 marks for lengthening the time, 1 for an example, 1 for the follow-through.

Section E: Circular Motion and Centripetal Force (16 marks)

A 1200 kg car travels at 15 m/s around a flat bend of radius 50 m. The coefficient of friction between the tyres and the road is 0.60. Take g = 9.81 m/s2.

#	Question	Answer and marking notes
17	(a) Find the centripetal acceleration of the car. (4 marks)	Answer: a = v^2/r = 225/50 = 4.50 m/s2 directed towards the centre of the bend. Marking: 2 marks for the formula, 1 for the value, 1 for the direction.
18	(b) Find the required centripetal force and state which force provides it. (4 marks)	Answer: F = ma = 1200 x 4.50 = 5400.0 N, provided by the sideways static friction between road and tyres. Marking: 1 mark for Newton second law, 2 for the value, 1 for naming friction.
19	(c) Find the maximum speed at which the car can take this bend. (4 marks)	Answer: Maximum static friction = mu mg = 0.60 x 1200 x 9.81 = 7063 N. Setting mv^2/r equal to this gives v^2 = mu g r = 0.60 x 9.81 x 50 = 294.3, so v = 17.16 m/s. Marking: 2 marks for the maximum friction, 1 for equating it to the centripetal force, 1 for the answer.
20	(d) If a wet road lowers the coefficient to 0.40, what is the new maximum speed? Explain why a passenger feels pushed towards the outside of the bend. (4 marks)	Answer: v = square root of (0.40 x 9.81 x 50) = square root of 196.2 = 14.0 m/s. By inertia the passenger tends to continue in a straight line along the tangent, so as the car turns the door or seat pushes the passenger inwards; what the passenger feels is the centripetal force from the car, not a real outward force on the body. Marking: 2 marks for the new speed, 2 for the inertia explanation.

Section F: Work, Energy and Power (17 marks)

A 1500 kg car accelerates uniformly from rest to 20 m/s along a straight road in 8.0 s while a constant resistive force of 400 N acts on it.

#	Question	Answer and marking notes
21	(a) Find the increase in the kinetic energy of the car. (4 marks)	Answer: Increase in kinetic energy = 0.5 x 1500 x 400 = 300000.0 J. Marking: 2 marks for the formula, 1 for the value, 1 for the unit.
22	(b) Find the distance travelled in 8.0 s and the work done against the resistive force. (5 marks)	Answer: Average speed = (0 + 20)/2 = 10 m/s, so the distance is 10 x 8.0 = 80.0 m. Work against resistance = 400 x 80.0 = 32000.0 J. Marking: 2 marks for the average speed, 1 for the distance, 2 for the work.
23	(c) Find the total work done by the engine and its average power. (4 marks)	Answer: Total work = increase in kinetic energy + work against resistance = 300000.0 + 32000.0 = 332000.0 J. Average power = 332000.0/8.0 = 41500.0 W, that is 41.5 kW. Marking: 2 marks for energy conservation, 1 for the total work, 1 for the power.
24	(d) Find the instantaneous power of the engine at t = 8.0 s and explain why it exceeds the average power. (4 marks)	Answer: Acceleration = 20/8.0 = 2.5 m/s2, net force = 1500 x 2.5 = 3750 N, so the driving force needed is 3750 + 400 = 4150 N. Instantaneous power = Fv = 4150 x 20 = 83000.0 W, that is 83 kW. Power equals force times speed and the speed rises from zero to 20 m/s, so the power at the end is far above the average for the whole interval. Marking: 2 marks for the driving force, 1 for the power, 1 for the explanation.

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