| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Particle-barrier collision with angle |
| Difficulty | Moderate -0.3 This is a straightforward M3 impact question requiring standard application of coefficient of restitution and impulse formulas. Part (i) uses Pythagoras to find the vertical component (given the answer to show), then e = separation/approach speed. Part (ii) requires recognizing impulse acts vertically (smooth surface means no horizontal friction) and calculating change in momentum. All steps are routine for M3 students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.03e Impulse: by a force6.03f Impulse-momentum: relation6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
A smooth sphere of mass 0.3 kg bounces on a fixed horizontal surface. Immediately before the sphere bounces the components of its velocity horizontally and vertically downwards are $4 \text{ m s}^{-1}$ and $6 \text{ m s}^{-1}$ respectively. The speed of the sphere immediately after it bounces is $5 \text{ m s}^{-1}$.
\begin{enumerate}[label=(\roman*)]
\item Show that the vertical component of the velocity of the sphere immediately after impact is $3 \text{ m s}^{-1}$, and hence find the coefficient of restitution between the surface and the sphere. [3]
\item State the direction of the impulse on the sphere and find its magnitude. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR M3 2009 Q1 [6]}}