S5 Physics Practice Paper: Uniformly Accelerated Motion, Newton Laws, Momentum, Work And Energy, Circular Motion And Power

S5 Physics Practice Paper: Uniformly Accelerated Motion, Newton Laws, Momentum, Work and Energy, Circular Motion and Power

2026-11-12

S5 Physics – Mechanics and Energy

S5 Physics practice paper on accelerated motion, Newton second law with friction, momentum in collisions, energy on an incline, circular motion and power.

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Paper details　S5 Physics practice paper with model answers on kinematics, Newton laws with friction, momentum, work and energy, circular motion and power.

Part 1: Uniformly accelerated motion (16 marks)

Four questions. Write the formula used, substitute values and state units to 3 significant figures.

#	Question	Answer and marking notes
1	A car starts from rest and accelerates at 2.5 m/s2 along a straight line for 8.0 s. Find the velocity at the end of this stage and the displacement covered.	Use v = u + at with u = 0, so v = 0 + 2.5 x 8.0 = 20.0 m/s. The displacement is s = ut + 0.5 a t2 = 0 + 0.5 x 2.5 x 64.0 = 80.0 m. Check: mean velocity = 10.0 m/s times 8.0 s gives 80.0 m.
2	The car then travels at a constant 20.0 m/s for 5.0 s. Find the displacement.	With a = 0 the displacement is s = v t = 20.0 x 5.0 = 100.0 m. The velocity is constant, so the displacement is proportional to the time.
3	The car then decelerates uniformly to rest in 4.0 s. Find the acceleration, including its direction.	Use a = (v – u) / t with v = 0, u = 20.0 m/s and t = 4.0 s, so a = (0 – 20.0) / 4.0 = -5.0 m/s2. The minus sign means the acceleration points opposite to the motion, that is a deceleration of 5.0 m/s2.
4	Find the total displacement and the average velocity for the whole journey.	Total displacement = 80.0 + 100.0 + 40.0 = 220.0 m and total time = 17.0 s, so the average velocity is 220.0 / 17.0 = 12.94 m/s, that is 12.9 m/s to 3 significant figures. The braking displacement is 0.5 x 20.0 x 4.0 = 40.0 m in the same direction.

Part 2: Newton second law and friction (18 marks)

A block of mass 4.0 kg rests on a horizontal table. A pulling force of 40 N acts at 37 degrees above the horizontal. The coefficient of kinetic friction is 0.25. Take g = 10 m/s2, sin37 = 0.6 and cos37 = 0.8.

#	Question	Answer and marking notes
5	Find the normal force exerted by the table on the block.	There is no vertical acceleration, so N + F sin37 = m g and therefore N = 40.0 – 24.0 = 16.0 N. The upward component of the pull makes the normal force smaller than the weight of 40.0 N.
6	Find the friction force acting on the block.	The block slides, so the friction is kinetic: f = mu N = 0.25 x 16.0 = 4.0 N, directed horizontally opposite to the motion.
7	Find the acceleration of the block.	The net horizontal force is F cos37 – f = 32.0 – 4.0 = 28.0 N, so by Newton second law a = 28.0 / 4.0 = 7.0 m/s2.
8	If the pull continues, find the speed and displacement of the block after 2.0 s.	Using v = u + at with u = 0 gives v = 7.0 x 2.0 = 14.0 m/s and s = 0.5 a t2 = 0.5 x 7.0 x 4.0 = 14.0 m. The block is already sliding, so kinetic friction applies throughout.

Part 3: Momentum conservation and collision (18 marks)

Trolley A of mass 2.0 kg moves right at 6.0 m/s on a smooth horizontal track and collides with a stationary trolley B of mass 3.0 kg. The trolleys lock together after the collision.

#	Question	Answer and marking notes
9	Find the common velocity of the trolleys after the collision.	External forces are negligible, so momentum is conserved: 2.0 x 6.0 = 12.0 kg m/s = 5.0 v, giving v = 2.4 m/s to the right.
10	State the total momentum of the system before the collision.	Total momentum = 2.0 x 6.0 + 3.0 x 0.0 = 12.0 kg m/s to the right, and it is unchanged by the collision.
11	Calculate the kinetic energy lost in the collision.	Before: 0.5 x 2.0 x 6.0 squared = 36.0 J. After: 0.5 x 5.0 x 2.4 squared = 14.4 J. The loss is 21.6 J, that is 60 per cent of the initial kinetic energy, appearing as heat, sound and deformation.
12	What type of collision is this? Explain in terms of momentum and kinetic energy.	Momentum is conserved but kinetic energy decreases, so this is a perfectly inelastic collision. The trolleys move together with zero relative velocity, so the coefficient of restitution is zero and the kinetic energy loss is the largest possible.

Part 4: Work and energy on an incline (16 marks)

A block of mass 5.0 kg slides from rest down an incline of height 4.0 m and slope length 5.0 m. Friction does 40 J of work during the slide. Take g = 10 m/s2.

#	Question	Answer and marking notes
13	Find the decrease in gravitational potential energy.	The loss of potential energy is m g h = 5.0 x 10.0 x 4.0 = 200.0 J. Only the height change matters, not the path taken.
14	Find the magnitude of the friction force.	Friction does work f L, so f = 40.0 / 5.0 = 8.0 N acting up the incline.
15	Find the kinetic energy of the block at the bottom.	By energy conservation the loss of potential energy equals the gain in kinetic energy plus the energy dissipated by friction, so the kinetic energy is 200.0 – 40.0 = 160.0 J.
16	Find the speed at the bottom, and the speed if the incline were smooth.	From KE = 0.5 m v2, v = sqrt(2 x 160.0 / 5.0) = 8.0 m/s. Without friction, v = sqrt(2 x 10.0 x 4.0) = 8.9 m/s, which is larger.

Part 5: Horizontal circular motion (16 marks)

A ball of mass 0.5 kg moves at 2.0 m/s in a circle of radius 0.8 m on a smooth horizontal table, held by a string through a hole at the centre.

#	Question	Answer and marking notes
17	Find the centripetal acceleration of the ball.	a = v squared over r = 2.0 x 2.0 / 0.8 = 5.0 m/s2 directed towards the centre of the circle.
18	Find the tension in the string.	The tension alone provides the centripetal force, so T = m a = 0.5 x 5.0 = 2.5 N. The string length is fixed, so the ball moves in a circle in the horizontal plane.
19	Find the time taken for one complete revolution.	The period is 2 pi r / v = 2 pi x 0.8 / 2.0 = 2.51 s. The angular speed is v / r = 2.5 rad/s and 2 pi divided by it gives the same period.
20	If the speed increases to 4.0 m/s at the same radius, what is the required centripetal force?	F = 0.5 x 4.0 squared / 0.8 = 10.0 N. Doubling the speed makes the centripetal force four times larger, since it varies with the square of the speed.

Part 6: Power and efficiency (16 marks)

A pump lifts 60 kg of water through a height of 5.0 m in 20 s. The efficiency of the pump is 75 per cent. Take g = 10 m/s2.

#	Question	Answer and marking notes
21	Find the useful work done by the pump.	Useful work = m g h = 60.0 x 10.0 x 5.0 = 3000.0 J, all of it stored as gravitational potential energy of the water.
22	Find the output power of the pump.	P = 3000.0 / 20.0 = 150.0 W.
23	Find the input power of the pump.	Efficiency = output over input, so the input power is 150.0 / 0.75 = 200.0 W.
24	Find the energy dissipated as heat and friction in 20 s.	Input energy = 200.0 x 20.0 = 4000.0 J, so the dissipated energy is 4000.0 – 3000.0 = 1000.0 J, that is 25 per cent of the input.

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