S4 Physics Paper: Forces And Motion (100 Marks)

S4 Physics Paper: Forces and Motion (100 marks)

2026-11-11

S4 Physics – forces and motion

S4 Physics paper: 6 questions, 22 parts, 100 marks, 75 minutes on acceleration, Newton second law with friction, momentum, projectiles and Hooke law, with marking notes.

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Paper details　S4 Physics practice paper on forces and motion: acceleration, Newton second law with friction, momentum and impulse, projectiles, Hooke law, circular motion, work and power. 6 questions, 22 parts, 100 marks, 75 minutes, with full marking notes.

Part A Question 1: Uniform Acceleration and Average Speed

A car starts from rest and accelerates uniformly along a straight line. Take the direction of motion as positive. Give units in every answer.

#	Question	Answer and marking notes
1	The car speeds up from rest to 15 m/s in 30 s. Find its acceleration.	Answer: a = (v – u) / t = (15 – 0) / 30 = 0.5 m/s squared.
2	Find the distance travelled in the first 30 s.	Answer: s = 0.5 (u + v) t = 0.5 x 15 x 30 = 225 m, which matches s = 0.5 a t squared = 225 m.
3	The car then travels at a constant 15 m/s for 20 s. Find the distance covered.	Answer: s = v t = 15 x 20 = 300 m.
4	Find the average speed over the whole 50 s.	Answer: total distance = 225 + 300 = 525 m, so average speed = 525 / 50 = 10.5 m/s.

Part B Question 2: Newton Second Law and Friction

Take g as 10 m/s squared. All forces act horizontally and the body moves on a rough horizontal surface.

#	Question	Answer and marking notes
5	A body of mass 4 kg experiences a net force of 12 N. Find its acceleration.	Answer: a = F / m = 12 / 4 = 3 m/s squared.
6	Starting from rest under this net force for 5 s, find the speed at 5 s.	Answer: v = u + a t = 0 + 3 x 5 = 15 m/s.
7	If friction of 4 N opposes the motion while a forward force of 16 N is applied, find the acceleration.	Answer: net force = 16 – 4 = 12 N, so a = 12 / 4 = 3 m/s squared.
8	In that case, find the distance travelled from rest in 5 s.	Answer: s = 0.5 a t squared = 0.5 x 3 x 5 squared = 37.5 m.

Part C Question 3: Conservation of Momentum and Impulse

The collision occurs on a smooth horizontal surface. Take the original direction of motion as positive and use kg m s to the minus one for momentum.

#	Question	Answer and marking notes
9	A ball of mass 2 kg moving at 6 m/s hits a stationary ball of mass 4 kg and they stick together. Find their common velocity after the collision.	Answer: momentum is conserved, 2 x 6 = (2 + 4) v, so v = 2 m/s in the original direction.
10	Find the change in momentum of the 2 kg ball during the collision.	Answer: change in momentum = m v – m u = 2 x 2 – 2 x 6 = -8 kg m per second, the minus sign showing the opposite direction.
11	If the collision lasts 0.2 s, find the average force on that ball.	Answer: F = change in momentum / time = -8 / 0.2 = -40 N, that is 40 N opposite to the original motion.

Part D Question 4: Free Fall and Projectile Motion

Air resistance is negligible. Take g as 10 m/s squared and downward as positive.

#	Question	Answer and marking notes
12	A stone is released from rest at a height of 45 m. Find the time to reach the ground.	Answer: s = 0.5 g t squared, 45 = 0.5 x 10 x t squared, so t = 3 s.
13	Find the speed of the stone on landing.	Answer: v = g t = 10 x 3 = 30 m/s.
14	If the stone is thrown horizontally at 20 m/s from the same height, find the horizontal displacement on landing.	Answer: horizontal motion is uniform, R = u t = 20 x 3 = 60 m.
15	Find the magnitude of the landing velocity in the horizontal throw (3 significant figures).	Answer: v = square root of (u squared plus (g t) squared) = square root of (20 squared plus 30 squared), about 36.1 m/s.

Part E Question 5: Hooke Law and Elastic Potential Energy

The spring obeys Hooke law, g is 10 m/s squared and the mass of the spring is negligible.

#	Question	Answer and marking notes
16	A spring of force constant 200 N/m extends by 0.15 m. Find the tension in the spring.	Answer: F = k x = 200 x 0.15 = 30 N.
17	A 3 kg mass is hung on the spring and held at rest. Find the extension.	Answer: F = m g = 3 x 10 = 30 N, so x = F / k = 30 / 200 = 0.15 m.
18	Find the elastic potential energy stored in the spring at that extension.	Answer: E = 0.5 k x squared = 0.5 x 200 x 0.15 squared = 2.25 J.

Part F Question 6: Circular Motion, Work and Power

Take g as 10 m/s squared. The body moves uniformly along a horizontal circle. Give units in every answer.

#	Question	Answer and marking notes
19	A body of mass 0.5 kg moves at 4 m/s along a circle of radius 0.8 m. Find the centripetal acceleration.	Answer: a = v squared / r = 4 squared / 0.8 = 20 m/s squared.
20	Find the centripetal force acting on the body.	Answer: F = m a = 0.5 x 20 = 10 N.
21	A 10 kg load is lifted vertically through 3 m. Find the work done.	Answer: W = m g h = 10 x 10 x 3 = 300 J.
22	If the lift takes 5 s, find the average power.	Answer: P = W / t = 300 / 5 = 60 W.

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