Successive oblique impactsEdexcel A-Level Further Maths: Revision notes

Section 1

One wall at a time

A sphere or particle can hit smooth fixed surfaces in turn. Each impact is an oblique impact with a smooth plane: the component of velocity parallel to the wall is unchanged, and the component perpendicular to it is reversed and multiplied by $e$ . The velocity after one impact is the velocity before the next. Keep the working in a table of components, one line per impact.

Key termssmooth planenormal component

Exam tip

For each wall, say which component is perpendicular to it before writing any numbers.

Section 2

Two walls at right angles

Take $\mathbf{i}$ and $\mathbf{j}$ along the walls. A wall parallel to $\mathbf{i}$ changes only the $\mathbf{j}$ -component, and a wall parallel to $\mathbf{j}$ changes only the $\mathbf{i}$ -component. So the order of the impacts does not matter for the final velocity, which is $(-e_2u_x)\mathbf{i}+(-e_1u_y)\mathbf{j}$ . Example: $(10\mathbf{i}+4\mathbf{j})$ with $e=\frac12$ at both walls becomes $(10\mathbf{i}-2\mathbf{j})$ then $(-5\mathbf{i}-2\mathbf{j})$ . When the two coefficients are equal the final velocity is $-e$ times the initial velocity: parallel to the original path, reversed, with speed multiplied by $e$ . With unequal coefficients it is not parallel.

Key termscorner

Common mistake

Applying $e$ to both components at each wall.

Section 3

Two parallel walls

For a sphere bouncing between parallel walls, the component parallel to the walls never changes, and the component perpendicular to them is multiplied by $e$ at every impact, so it is $e^nu_\perp$ after $n$ impacts. Example: components $8$ (parallel) and $6$ (perpendicular) with $e=\frac12$ . After the first impact $(8,3)$ , speed $\sqrt{73}=8.54$ m s $^{-1}$ . After the second $(8,1.5)$ , speed $\sqrt{66.25}=8.14$ m s $^{-1}$ .

Key termsparallel walls

Section 4

Angles after successive impacts

If $\theta$ is the angle between the path and a wall then $\tan\theta=\frac{u_\perp}{u_\parallel}$ . After an impact, $\tan\theta_{\text{after}}=e\tan\theta_{\text{before}}$ . After a second impact with a parallel wall, $\tan\theta=e^2\tan\theta_0$ . Example: $\theta=60^\circ$ , $e=\frac23$ gives $\tan\theta=\frac23\sqrt3$ after one impact, so $\theta=49.1^\circ$ . The angle falls at each impact: the path gets closer to the wall direction.

Key termsangle with the wall

Section 5

Speed and kinetic energy

After the final impact, the speed is $\sqrt{(\text{parallel})^2+(\text{perpendicular})^2}$ using the latest components. The kinetic energy lost over several impacts is $\frac12m(u^2-v^2)$ with $u$ the initial and $v$ the final speed, so only the first and last states are needed. Example: $(8\mathbf{i}+6\mathbf{j})$ becoming $(-4\mathbf{i}-3\mathbf{j})$ loses $\frac12m(100-25)$ .

Key termskinetic energy lost

Common mistake

Adding up energy losses impact by impact, then forgetting one. Compare only the initial and final states.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Successive oblique impacts

A smooth particle of mass $0.5$ kg moves on a smooth horizontal table. Two fixed smooth vertical walls $W_1$ and $W_2$ meet at right angles: $W_1$ is parallel to $\mathbf{i}$ and $W_2$ is parallel to $\mathbf{j}$ . The particle has velocity $(8\mathbf{i}+6\mathbf{j})$ m s $^{-1}$ , hits $W_1$ and then hits $W_2$ . The coefficient of restitution between the particle and each wall is $0.5$ .
Find the total kinetic energy lost by the particle in the two impacts.2 marks
A smooth sphere of mass $0.2$ kg moves on a smooth horizontal surface between two fixed parallel smooth vertical walls $W_1$ and $W_2$ . It hits $W_1$ with speed $12$ m s $^{-1}$ , its direction of motion making an angle of $60^\circ$ with the wall, and then crosses to hit $W_2$ . The coefficient of restitution between the sphere and each wall is $\frac23$ .
Find the total kinetic energy lost by the sphere in the two impacts.2 marks
A smooth particle $P$ of mass $0.4$ kg moves on a smooth horizontal surface with velocity $(6\mathbf{i}+8\mathbf{j})$ m s $^{-1}$ . It hits a fixed smooth vertical wall $W_1$ that is parallel to $\mathbf{j}$ , and then hits a fixed smooth vertical wall $W_2$ that is parallel to $\mathbf{i}$ . The coefficient of restitution between $P$ and $W_1$ is $\frac13$ , and between $P$ and $W_2$ is $\frac12$ .
Find the speed of $P$ immediately after it hits $W_1$ , and the angle between its path and $W_1$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).