S5 Physics Paper: Newton Laws, Momentum And Circular Motion (100 Marks)

S5 Physics Paper: Newton Laws, Momentum and Circular Motion (100 marks)

2026-11-11

S5 Physics – mechanics and circular motion

S5 Physics paper: Newton second law, friction, momentum, projectile and circular motion, work, power and apparent weight, with full answers and marking points.

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Paper details　S5 Physics practice paper with six structured questions on Newton second law and friction, conservation of momentum and collisions, projectile motion, circular motion and safe speed, work, energy and power on an incline, and apparent weight in a lift, with full answers.

Question 1 Newton second law and friction (16 marks)

A 4.0 kg box rests on horizontal ground. A person pushes it with a horizontal force of 20 N. Take g = 9.8 m s^-2.

#	Question	Answer and marking notes
1	(a) If the coefficient of friction is 0.20, find the friction.	Answer: f = mumg = 0.20 x 4.0 x 9.8 = 7.84 N.
2	(b) Find the acceleration of the box.	Answer: net force = 20 – 7.84 = 12.16 N, a = 12.16 / 4.0 = 3.04 m s^-2.
3	(c) If the push lasts 3.0 s, find the speed at that moment.	Answer: v = a*t = 3.04 x 3.0 = 9.12 m s^-1.
4	(d) Find the distance travelled in these 3.0 s.	Answer: s = 0.5at^2 = 0.5 x 3.04 x 9.0 = 13.68 m.
5	(e) If the coefficient rises to 0.30 with the same push, find the new acceleration and state whether the box can start moving.	Answer: f = 0.30 x 4.0 x 9.8 = 11.76 N, a = (20 – 11.76) / 4.0 = 2.06 m s^-2; the net force is positive so the box starts moving.

Question 2 Conservation of momentum and collisions (16 marks)

A 2.0 kg body A moves at 6.0 m s^-1 on a smooth horizontal surface and collides with a stationary 4.0 kg body B.

#	Question	Answer and marking notes
6	(a) Find the total momentum before the collision.	Answer: p = m_A*u_A = 2.0 x 6.0 = 12.0 kg m s^-1 along the motion of A.
7	(b) If they stick together after the collision, find the common velocity.	Answer: v = p / (m_A + m_B) = 12.0 / 6.0 = 2.0 m s^-1.
8	(c) Find the kinetic energy lost in this perfectly inelastic collision.	Answer: initial KE 36.0 J, final KE 12.0 J, loss 24.0 J.
9	(d) If the collision is perfectly elastic, find the velocities of A and B afterwards.	Answer: v_A = (m_A – m_B)/(m_A + m_B) * u_A = -2.0 m s^-1 (reversed), v_B = 2m_A/(m_A + m_B) u_A = 4.0 m s^-1.
10	(e) Verify by kinetic energy that (d) conserves energy.	Answer: final KE = 0.5 x 2.0 x 4.0 + 0.5 x 4.0 x 16.0 = 36.0 J, equal to the initial 36.0 J, so KE is conserved.

Question 3 Projectile motion (17 marks)

A ball is projected at 20 m s^-1 at 30 degrees above the horizontal. Ignore air resistance, g = 9.8 m s^-2.

#	Question	Answer and marking notes
11	(a) Find the horizontal and vertical components of the initial velocity.	Answer: u_x = 20 cos30 = 17.32 m s^-1; u_y = 20 sin30 = 10.00 m s^-1.
12	(b) Find the time of flight.	Answer: t = 2*u_y / g = 20.0 / 9.8 = 2.04 s.
13	(c) Find the horizontal range.	Answer: R = u_x*t = 17.32 x 2.04 = 35.35 m.
14	(d) Find the maximum height reached.	Answer: H = u_y^2 / 2g = 100 / 19.6 = 5.10 m.

Question 4 Circular motion and safe speed (17 marks)

A 1200 kg car goes round a flat bend of radius 50 m at 15 m s^-1. Take g = 9.8 m s^-2.

#	Question	Answer and marking notes
15	(a) Find the centripetal acceleration of the car.	Answer: a = v^2 / r = 225 / 50 = 4.50 m s^-2 towards the centre.
16	(b) Find the required centripetal force.	Answer: F = m*a = 1200 x 4.50 = 5400 N, supplied by friction.
17	(c) Find the minimum coefficient of friction needed at this speed.	Answer: mu = v^2 / (r*g) = 225 / (50 x 9.8) = 0.46.
18	(d) If the maximum coefficient is 0.60, find the greatest safe speed.	Answer: v = sqrt(murg) = sqrt(0.60 x 9.8 x 50) = sqrt(294) = 17.15 m s^-1.

Question 5 Work, energy and power on an incline (17 marks)

A 20 kg crate is pulled 5.0 m up a 30 degree incline by a force of 150 N along the incline. The coefficient of friction is 0.25, g = 9.8 m s^-2.

#	Question	Answer and marking notes
19	(a) Find the component of weight along the incline, the normal force and the friction.	Answer: component = mgsin30 = 98.0 N; N = mgcos30 = 169.74 N; f = mu*N = 42.44 N.
20	(b) Find the work done by the pulling force.	Answer: W = F*s = 150 x 5.0 = 750 J.
21	(c) Find the energy dissipated by friction.	Answer: W_f = f*s = 42.44 x 5.0 = 212.2 J, converted to internal energy.
22	(d) Find the gain in kinetic energy and compare it with the work done by the net force.	Answer: rise h = 2.5 m, PE gain 490 J, KE gain = 750 – 212.2 – 490 = 47.8 J; net force = 9.56 N, work by net force = 9.56 x 5.0 = 47.8 J, the same.

Question 6 Apparent weight in a lift (17 marks)

A 60 kg student stands on a scale inside a lift. Take g = 9.8 m s^-2.

#	Question	Answer and marking notes
23	(a) Find the true weight of the student.	Answer: W = m*g = 60 x 9.8 = 588 N.
24	(b) If the lift accelerates upwards at 2.0 m s^-2, find the reading.	Answer: N = m*(g + a) = 60 x (9.8 + 2.0) = 708 N, greater than the true weight.
25	(c) If the lift accelerates downwards at 2.0 m s^-2, find the reading.	Answer: N = m*(g – a) = 60 x (9.8 – 2.0) = 468 N, less than the true weight.
26	(d) If the cable snaps and the lift falls freely, find the reading.	Answer: a = g, so N = m*(g – g) = 0 N; the reading is zero.
27	(e) Use Newton second law to explain the different readings in (b) and (c).	Answer: net force N – mg = ma, so N = m*(g + a); a is positive upwards and negative downwards, so the reading increases and decreases respectively.

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