AS & A Level · Practice questions · Motion on an inclined plane

Resolve along the slope.

Six original Cambridge-style questions on motion on smooth and rough inclined planes, 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

Perpendicular for R, along for F = ma.

Weight: \(mg\sin\alpha\) down the plane, \(mg\cos\alpha\) into it.
Perpendicular: \(R = mg\cos\alpha \pm\) components of other forces.
Friction: \(\mu R\), opposite to the motion (it switches direction when the particle turns round).
Along the plane: resultant \(= ma\), then suvat.
At rest at the top? It stays if \(\mu \ge \tan\alpha\).

Recall

[3 marks]

A particle is released from rest on a smooth plane inclined at \(30^\circ\). Find its acceleration, its speed after sliding 4 m, and the time taken.

\(a = 10\sin 30^\circ = 5\ \text{m s}^{-2}\). ✓
\(v = \sqrt{2 \times 5 \times 4} = \sqrt{40} = 6.32\ \text{m s}^{-1}\). ✓
\(t = 6.325 \div 5 = 1.26\) s. ✓

Calculation

[4 marks]

A particle of mass 2 kg is released from rest on a rough plane inclined at \(\alpha\), where \(\sin\alpha = 0.6\). The coefficient of friction is 0.25. Find its speed after 2 s and the distance it has travelled.

\(R = 16\) N, friction 4 N; \(12 - 4 = 2a\), so \(a = 4\ \text{m s}^{-2}\). ✓
\(v = 4 \times 2 = 8\ \text{m s}^{-1}\). ✓
\(s = \tfrac{1}{2}(4)(2^2) = 8\) m. ✓

Calculation

[4 marks]

A particle slides from rest down a rough plane inclined at \(30^\circ\) and travels 4 m in 2 s. Find the coefficient of friction, giving your answer in exact form.

\(4 = \tfrac{1}{2}a(2^2)\), so \(a = 2\ \text{m s}^{-2}\). ✓
Per unit mass: \(10\sin 30^\circ - 10\mu\cos 30^\circ = 2\), so \(5 - 5\sqrt{3}\mu = 2\). ✓
\(\mu = \dfrac{3}{5\sqrt{3}} = \dfrac{\sqrt{3}}{5}\) (0.346). ✓

Analysis

[6 marks]

A particle of mass 0.5 kg is projected up a line of greatest slope of a rough plane inclined at \(20^\circ\), with speed \(6\ \text{m s}^{-1}\). The coefficient of friction is 0.4.

(a) Find the distance it travels up the plane before coming to rest.

(b) Show that it then remains at rest.

(a) \(R = 5\cos 20^\circ = 4.698\) N; friction \(= 1.879\) N down the plane. ✓
\(-(5\sin 20^\circ + 1.879) = 0.5a\), so \(-(1.710 + 1.879) = 0.5a\) and \(a = -7.18\ \text{m s}^{-2}\). ✓
\(0 = 36 + 2(-7.179)s\), so \(s = 2.51\) m. ✓
(b) Weight component \(1.710\) N is less than the maximum friction \(1.879\) N (equivalently \(\mu = 0.4 > \tan 20^\circ = 0.364\)), so it stays at rest. ✓

Analysis

[5 marks]

A particle of mass 3 kg is pushed up a rough plane inclined at \(30^\circ\) by a horizontal force of 40 N. The coefficient of friction is 0.2. Find the acceleration of the particle.

\(R = 30\cos 30^\circ + 40\sin 30^\circ = 45.98\) N. ✓
Friction \(= 0.2 \times 45.98 = 9.196\) N down the plane. ✓
\(40\cos 30^\circ - 30\sin 30^\circ - 9.196 = 3a\). ✓
\(10.44 = 3a\), so \(a = 3.48\ \text{m s}^{-2}\). ✓

Analysis

[6 marks]

A skier of mass 70 kg starts from rest and skis straight down a slope inclined at \(15^\circ\) to the horizontal. The coefficient of friction between skis and snow is 0.1, and there is a constant air resistance of 40 N. Find the skier's speed after travelling 100 m.

\(R = 700\cos 15^\circ = 676.1\) N, so friction \(= 67.61\) N. ✓
Down the slope: \(700\sin 15^\circ - 67.61 - 40 = 70a\). ✓
\(181.17 - 67.61 - 40 = 73.56 = 70a\), so \(a = 1.051\ \text{m s}^{-2}\). ✓
\(v^2 = 2(1.051)(100) = 210.2\), so \(v = 14.5\ \text{m s}^{-1}\). ✓

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.