| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Angle change from impulse |
| Difficulty | Challenging +1.2 This is a standard Further Mechanics impulse-momentum problem requiring resolution of velocities parallel and perpendicular to the barrier, application of Newton's experimental law (coefficient of restitution), and energy considerations. Part (a) involves straightforward component resolution and algebraic manipulation (3 marks). Part (b) requires combining the result from (a) with the energy loss condition to solve simultaneous equations (5 marks). While it requires careful bookkeeping of components and multiple techniques, the approach is methodical and follows standard procedures taught in Further Mechanics courses. The 90° deflection condition provides a clear constraint that guides the solution. |
| Spec | 6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
A particle $P$ of mass $m$ is moving with speed $u$ on a fixed smooth horizontal surface. The particle strikes a fixed vertical barrier. At the instant of impact the direction of motion of $P$ makes an angle $\alpha$ with the barrier. The coefficient of restitution between $P$ and the barrier is $e$. As a result of the impact, the direction of motion of $P$ is turned through $90°$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\tan^2 \alpha = \frac{1}{e}$. [3]
\end{enumerate}
The particle $P$ loses two-thirds of its kinetic energy in the impact.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the value of $\alpha$ and the value of $e$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q6 [8]}}