CAIE A-Level Further Math AS 3.6 Momentum Questions

Practise resolving direct or oblique impacts along the line of centres, combining momentum with restitution and checking kinetic-energy change or parameter bounds.

Exam points

conserve momentum along the line of centres with a consistent signed velocity convention
apply restitution to relative normal speeds while retaining unchanged tangential components
recombine components for speed or direction and compare total kinetic energy before and after

Covered syllabus points

3.6.1—Coefficient of restitution 3.6.2—Momentum and impacts

[Maximum number: 5]

Figure for Question 1 — CAIE A-Level Further Math AS

A particle P of mass m is moving with speed u on a fixed smooth horizontal surface. It collides at an angle $\alpha$ with a fixed smooth vertical barrier. After the collision, P moves at an angle $\theta$ with the barrier, where $\tan \theta=\frac{1}{2}$ (see diagram). The coefficient of restitution between P and the barrier is e. The particle P loses 20\% of its kinetic energy as a result of the collision.

Find the value of e.

Parallel to wall $v \cos \theta=u \cos \alpha$

Perpendicular to wall $v \sin \theta=e u \sin \alpha$
Both

Dividing, $e=\frac{1}{2 \tan \alpha}$
AEF

KE reduced by 20%, so $\frac{1}{2} m u^{2}\left(\cos ^{2} \alpha+e^{2} \sin ^{2} \alpha\right)=\frac{4}{5} \times \frac{1}{2} m u^{2}$
Dimensionally correct equation in u or v, but not
both. Must have either $\alpha$ or $\theta$ , but not both.
Must see $\frac{4}{5}$ on the correct side of the equation.

Eliminate $e: \cos \alpha=\frac{4}{5}$ $e=\frac{2}{3}$
PUBLISHED

2

Alternative method for question 2

Parallel to wall $v \cos \theta=u \cos \alpha$

Perpendicular to wall $v \sin \theta=e u \sin \alpha$
Both

$\left[\sin (\theta)=\frac{\sqrt{5}}{5}, \cos (\theta)=\frac{2 \sqrt{5}}{5}\right] \quad u \sin (\alpha)=\frac{\sqrt{5} v}{5 e}, u \cos (\alpha)=\frac{2 \sqrt{5} v}{5}$ $u^{2}=\left[u^{2} \cos ^{2}(\alpha)+u^{2} \sin ^{2}(\alpha)\right]=\frac{4 v^{2}}{5}+\frac{v^{2}}{5 e^{2}}$
AEF, e.g. $\frac{v^{2}}{5}\left(4+\frac{1}{e^{2}}\right)$ $\frac{1}{2} m v^{2}=\left[\frac{4}{5} \times \frac{1}{2} m u^{2}=\right] \frac{2}{5} m \frac{v^{2}}{5}\left(4+\frac{1}{e^{2}}\right)$
Dimensionally correct equation in v.

Must have either $\alpha$ or $\theta$ , but not both.

Must see $\frac{4}{5}$ or $\frac{2}{5}$ on the correct side of the equation.

$e=\frac{2}{3}$
5

[Maximum number: 5]

Figure for Question 2 — CAIE A-Level Further Math AS

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Covered syllabus points

Question 1

Question 2