S4 Physics Paper: Newton Laws, Momentum And Collisions (100 Marks)

S4 Physics Paper: Newton Laws, Momentum and Collisions (100 marks)

2026-11-11

S4 Physics – Newton laws and momentum

S4 Physics paper: 6 questions, 22 parts, 100 marks, 60 min. Newton laws, inertia, velocity-time graphs, friction, momentum and collisions, work, power and circular motion.

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Section A – Newton laws of motion and inertia

Answer in full sentences and name the forces acting on each body.

#	Question	Answer and marking notes
1	State Newton first law of motion and explain how inertia relates to mass.	If the net force on a body is zero, a body at rest stays at rest and a moving body keeps moving with constant velocity in a straight line. Inertia is the resistance of a body to a change in its state of motion; the larger the mass, the larger the inertia and the larger the net force needed to change its motion. (5 marks: statement 3; inertia and mass 2)
2	How do a seat belt and an air bag reduce injury to a passenger during emergency braking?	When the car slows sharply the passenger keeps moving forward because of inertia. The belt exerts a backward force so the passenger decelerates with the car, spreading the force over the shoulder and pelvis. The air bag inflates on impact and lengthens the stopping time; since F = change in momentum / time, the same momentum change over a longer time gives a smaller average force. (5 marks: inertia 2; belt spreads force 1; bag lengthens time 1; formula 1)
3	Using a person pushing a wall, state Newton third law and explain why an action-reaction pair does not cancel out.	The person pushes the wall and the wall pushes back on the person with an equal and opposite force. Newton third law: the forces two bodies exert on each other are equal in magnitude, opposite in direction, act along the same line and appear at the same time. The pair acts on two different bodies, the wall and the person, so they never cancel; only forces acting on the same body combine into a net force. (4 marks: statement 2; same time and line 1; different bodies 1)
4	Distinguish between mass and weight, and state how the mass and the weight of a body differ on the Moon.	Mass is the amount of matter in a body, measured in kg; it is a scalar and does not change with position. Weight is the gravitational force on the body, equal to mg, measured in N; it is a vector pointing towards the centre of the Earth and depends on the gravitational field strength. Since g on the Moon is about one sixth of that on the Earth, the mass is unchanged but the weight is about one sixth. (4 marks: mass 1; weight 1; scalar and vector 1; Moon values 1)

Section B – Uniform acceleration and velocity-time graphs

A car starts from rest, accelerates uniformly at 2.5 m/s2 for 8.0 s, then decelerates uniformly at 4.0 m/s2 until it stops.

#	Question	Answer and marking notes
5	Find the speed at the end of 8.0 s and the distance travelled during acceleration.	v = u + at = 0 + 2.5 x 8.0 = 20.0 m/s; s = ut + 0.5 at2 = 0 + 0.5 x 2.5 x 8.0^2 = 80.0 m. (5 marks: speed 2; distance formula 2; value and unit 1)
6	Find the braking distance and the time taken to stop.	From v2 = u2 + 2as, 0 = 20.0^2 – 2 x 4.0 x s, so s = 20.0^2 / 8.0 = 50.0 m. From v = u + at, t = 20.0 / 4.0 = 5.0 s. (5 marks: formula 2; distance 1; time 2)
7	Find the total distance and the average speed for the whole journey.	Total distance = 80.0 + 50.0 = 130.0 m; total time = 8.0 + 5.0 = 13.0 s; average speed = 130.0 / 13.0 = 10.0 m/s. (4 marks: distance 1; time 1; value 1; unit 1)
8	Describe the two stages on one velocity-time graph and state what the area under the line represents.	The first stage rises from the origin to (8.0 s, 20.0 m/s) with gradient 2.5 m/s2; the second falls from that point to (13.0 s, 0) with gradient -4.0 m/s2. The area between the line and the time axis is the displacement, 80.0 + 50.0 = 130.0 m, matching the calculation. (4 marks: first stage 1; second stage 1; gradient 1; area 1)

Section C – Newton second law and friction

Take g = 10 m/s2. A 5.0 kg block on a horizontal floor, kinetic friction coefficient 0.30, is pushed by a 25 N horizontal force.

#	Question	Answer and marking notes
9	Find the friction force on the block and its acceleration.	Normal reaction = mg = 5.0 x 10 = 50 N, so friction f = 0.30 x 50 = 15.0 N. Net force = 25 – 15.0 = 10.0 N and a = 10.0 / 5.0 = 2.0 m/s2. (5 marks: normal reaction 1; friction 2; net force 1; acceleration 1)
10	If the push is kept constant, find the speed and the displacement after 4.0 s.	v = at = 2.0 x 4.0 = 8.0 m/s; s = 0.5 at2 = 0.5 x 2.0 x 4.0^2 = 16.0 m. (4 marks: speed 2; displacement 2)
11	If the horizontal force is raised to 35 N, find the new acceleration and explain why friction is unchanged.	The normal reaction is still 50 N and the coefficient is the same, so friction stays at 15.0 N. Net force = 35 – 15.0 = 20.0 N, so a = 20.0 / 5.0 = 4.0 m/s2. (5 marks: reason 1; net force 2; value 1; unit 1)
12	On a smooth floor, what is the displacement in 4.0 s under the same 25 N push? Compare with the rough case.	With no friction a = 25 / 5.0 = 5.0 m/s2 and s = 0.5 x 5.0 x 4.0^2 = 40.0 m, which is 24.0 m more than the 16.0 m found with friction, because the net force is larger. (4 marks: acceleration 1; displacement 1; comparison 1; reason 1)

Section D – Conservation of momentum and collisions

The following collisions happen on a smooth horizontal track; friction is negligible.

#	Question	Answer and marking notes
13	Define momentum with its unit and explain why it is a vector.	Momentum p = mv, the product of mass and velocity, measured in kg m/s (or N s). Velocity is a vector, so momentum has the direction of the velocity; a positive direction must be chosen and momentum in the opposite direction is taken as negative. (5 marks: definition 1; unit 1; vector reason 2; sign convention 1)
14	A 2.0 kg trolley moving at 3.0 m/s hits a stationary 4.0 kg trolley and they stick together. Find the common velocity.	Total momentum before = 2.0 x 3.0 + 4.0 x 0 = 6.0 kg m/s. After the collision the combined mass is 6.0 kg, so by conservation 6.0 = 6.0 v and v = 1.0 m/s in the original direction. (5 marks: initial momentum 2; equation 2; value and direction 1)
15	For the collision above, find the kinetic energy before and after, and the percentage of kinetic energy lost.	Kinetic energy before = 0.5 x 2.0 x 3.0^2 = 9.0 J; after = 0.5 x 6.0 x 1.0^2 = 3.0 J. Loss = 6.0 J, that is 6.0 / 9.0 x 100% = 66.7%. (4 marks: before 1; after 1; loss 1; percentage 1)
16	If the 2.0 kg trolley hits a stationary 1.0 kg trolley elastically, find both velocities and verify that momentum and kinetic energy are conserved.	Conservation of momentum gives 2.0 x 3.0 = 2.0 v1 + 1.0 v2; for an elastic collision kinetic energy is conserved and the approach speed equals the separation speed, 3.0 = v2 – v1. Solving gives v1 = 1.0 m/s and v2 = 4.0 m/s. Check: total momentum = 2.0 x 1.0 + 1.0 x 4.0 = 6.0 kg m/s and total kinetic energy = 0.5 x 2.0 x 1.0^2 + 0.5 x 1.0 x 4.0^2 = 9.0 J, the same as before. (4 marks: equations 2; values 1; check 1)

Section E – Work, energy and power

Take g = 10 m/s2 and ignore air resistance.

#	Question	Answer and marking notes
17	A 60 kg runner climbs stairs of vertical height 12 m in 8.0 s. Find the work done against gravity and the average power.	W = mgh = 60 x 10 x 12 = 7200 J; P = W / t = 7200 / 8.0 = 900 W. (5 marks: formula 1; work 2; power formula 1; power value 1)
18	A 0.50 kg ball falls freely from a height of 1.8 m. Find its speed and kinetic energy on reaching the ground.	From mgh = 0.5 mv2, v = sqrt(2gh) = sqrt(2 x 10 x 1.8) = 6.0 m/s; kinetic energy = 0.5 x 0.50 x 6.0^2 = 9.0 J, equal to mgh = 0.50 x 10 x 1.8 = 9.0 J. (5 marks: equation 1; speed 2; formula 1; value 1)
19	The ball rebounds to a height of 0.80 m. Find the kinetic energy after rebound, the energy lost and the coefficient of restitution.	Rebound speed = sqrt(2 x 10 x 0.80) = 4.0 m/s, so kinetic energy = 0.5 x 0.50 x 4.0^2 = 4.0 J. Loss = 9.0 – 4.0 = 5.0 J, appearing as sound and internal energy. Coefficient of restitution = 4.0 / 6.0 = 0.667. (4 marks: kinetic energy 1; loss 1; energy form 1; coefficient 1)

Section F – Circular motion

A 1.5 kg object moves in a horizontal circle of radius 0.80 m on a smooth table at a constant speed of 4.0 m/s, held by a string.

#	Question	Answer and marking notes
20	Find the centripetal acceleration of the object and the tension in the string.	a = v2 / r = 4.0^2 / 0.80 = 20.0 m/s2, directed towards the centre. The tension supplies the centripetal force, F = ma = 1.5 x 20.0 = 30.0 N. (5 marks: formula 1; value 1; direction 1; tension 2)
21	Find the time for one revolution and explain why the speed is constant but the velocity keeps changing.	T = 2 pi r / v = 2 pi x 0.80 / 4.0 = 1.26 s. In uniform circular motion the magnitude of the velocity is constant but its direction changes continuously along the tangent; velocity is a vector, so a change of direction is a change of velocity, hence there is an acceleration towards the centre. (5 marks: formula 1; value 1; direction change 2; centripetal acceleration 1)
22	If the string can withstand at most 45 N, find the greatest speed the object may have without breaking it.	From F = mv2 / r, 45 = 1.5 x v2 / 0.80, so v2 = 45 x 0.80 / 1.5 = 24 and v = 4.90 m/s. (4 marks: formula 1; rearrangement 1; value 1; unit 1)

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