| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Vector impulse: find deflection angle or impulse magnitude from angle |
| Difficulty | Moderate -0.3 This is a straightforward impulse-momentum question requiring application of standard mechanics principles. Part (i) involves resolving momentum components and using the given angle condition to find I (routine calculation with tan α given). Part (ii) requires combining perpendicular velocity components using Pythagoras. The question is slightly easier than average as it's a standard textbook-style problem with clear setup and no conceptual surprises, though it does require careful vector resolution. |
| Spec | 6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks |
|---|---|
| M1, A1 | correct orientation, \(\alpha\) and one side labelled correctly, right angle implied; first two marks may be implied by correct working |
| M1, A1, M1 | resolve parallel or perp to dir of motion both; attempt to manipulate |
| A1 | (AG - answer given) |
| Answer | Marks |
|---|---|
| M1, A1 [2] | or using Pythagoras, with 3 and 1.25 oe |
### Part (i)
**Answer:** Impulse momentum diagram: right-angled triangle with angle $\alpha$ and sides labelled 3, $v$ and $I/0.2$ or 0.6, 0.2$v$ and $I$
$\tan \alpha = I/(0.2 \times 3)$, $I = 0.25$ shown
OR $0.2 \times 3 = 0.2v \cos\alpha$ and $I = 0.2v \sin \alpha$
$\frac{I}{0.2 \times 3} = \tan \alpha$, $I = 0.25$
| M1, A1 | correct orientation, $\alpha$ and one side labelled correctly, right angle implied; first two marks may be implied by correct working |
| M1, A1, M1 | resolve parallel or perp to dir of motion both; attempt to manipulate |
| A1 | (AG - answer given) |
### Part (ii)
**Answer:** $\cos\alpha = 3/v$ (speed) = 3.25 m s$^{-1}$
| M1, A1 [2] | or using Pythagoras, with 3 and 1.25 oe |A particle $P$ of mass $0.2$ kg is moving on a smooth horizontal surface with speed $3\text{ ms}^{-1}$, when it is struck by an impulse of magnitude $I$ Ns. The impulse acts horizontally in a direction perpendicular to the original direction of motion of $P$, and causes the direction of motion of $P$ to change by an angle $\alpha$, where $\tan \alpha = \frac{5}{12}$.
\begin{enumerate}[label=(\roman*)]
\item Show that $I = 0.25$. [4]
\item Find the speed of $P$ after the impulse acts. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR M3 2015 Q1 [6]}}