AS & A Level · Practice questions · Newton's laws of motion

Force, mass, acceleration.

Six original Cambridge-style questions on Newton's laws of motion with constant forces, using \(g = 10\), with full worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Diagram, resolve, F = ma, suvat.

Diagram: all forces, plus an arrow for the acceleration.
Perpendicular to motion: forces balance; this usually gives \(R\).
Friction: moving means \(F = \mu R\), opposing the motion.
Along the motion: resultant force \(= ma\).
Motion: use suvat with the acceleration you found (or the reverse).

Recall

[2 marks]

State Newton's first law. A particle of mass 3 kg on a smooth horizontal surface is acted on by a horizontal force of 12 N. Find its acceleration.

A particle stays at rest or moves with constant velocity unless a resultant force acts on it. ✓
\(12 = 3a\), so \(a = 4\ \text{m s}^{-2}\). ✓

Calculation

[3 marks]

A car of mass 1200 kg brakes uniformly from \(20\ \text{m s}^{-1}\) to rest in a distance of 40 m. Find the braking force, assuming no other resistance.

\(0 = 20^2 + 2a(40)\), so \(a = -5\ \text{m s}^{-2}\). ✓
Braking force \(= 1200 \times 5 = 6000\) N. ✓

Calculation

[4 marks]

A boy of mass 60 kg stands in a lift. Find the normal reaction from the floor when the lift (a) accelerates upwards at \(1.5\ \text{m s}^{-2}\), (b) moves at constant speed, (c) moves downwards while slowing down at \(1.5\ \text{m s}^{-2}\).

(a) \(R - 600 = 60(1.5)\), so \(R = 690\) N. ✓
(b) \(a = 0\), so \(R = 600\) N. ✓
(c) Moving down and slowing means the acceleration is UPWARDS, \(1.5\ \text{m s}^{-2}\): \(R = 690\) N, the same as (a). ✓

Analysis

[5 marks]

A block of mass 8 kg starts from rest on rough horizontal ground, \(\mu = 0.25\). It is pulled by a force of 50 N at \(30^\circ\) above the horizontal. Find the distance it travels in the first 4 s.

\(R = 80 - 50\sin 30^\circ = 55\) N; friction \(= 0.25 \times 55 = 13.75\) N. ✓
\(50\cos 30^\circ - 13.75 = 8a\), so \(a = 3.694\ \text{m s}^{-2}\). ✓
\(s = \tfrac{1}{2}(3.694)(4^2) = 29.6\) m. ✓

Analysis

[4 marks]

A puck is projected at \(8\ \text{m s}^{-1}\) across rough horizontal ice, \(\mu = 0.32\). Find how far it slides and how long it takes to stop.

Friction \(= 0.32 \times 10m = 3.2m\), so the deceleration is \(3.2\ \text{m s}^{-2}\). ✓
\(0 = 64 - 2(3.2)s\), so \(s = 10\) m. ✓
\(0 = 8 - 3.2t\), so \(t = 2.5\) s. ✓

Analysis

[6 marks]

A crate of mass 50 kg is raised from rest by a vertical rope. For the first 3 s the tension is 560 N. The tension then changes to a constant value \(T\) N and the crate comes to rest after a further 2 s.

(a) Find the speed of the crate after 3 s.

(b) Find \(T\).

(a) \(560 - 500 = 50a\), \(a = 1.2\); \(v = 1.2 \times 3 = 3.6\ \text{m s}^{-1}\). ✓
(b) Deceleration \(= 3.6 \div 2 = 1.8\ \text{m s}^{-2}\): \(T - 500 = 50(-1.8)\), so \(T = 410\). ✓
(c) Heights: \(\tfrac{1}{2}(3.6)(3) = 5.4\) m and \(\tfrac{1}{2}(3.6)(2) = 3.6\) m. ✓
Total \(= 9\) m. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.