| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| 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 | Standard +0.3 This is a straightforward impulse-momentum vector problem requiring resolution in two perpendicular directions and basic trigonometry. Part (i) involves standard application of impulse = change in momentum with vector components, while part (ii) uses symmetry to deduce the answer quickly. The calculations 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: relation |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.3 \times 0.4 + l, l = 0.3 \times 0.6\cos30°\) | M1, M1, A1, A1 4 | Similar with velocities. 0.096890. 68.26. Direction must be clear in words or on diagram. Similar with velocities. Or slip in figure |
| Answer | Marks | Guidance |
|---|---|---|
| 68.26° from original direction | (M1), (A1), (A1), (A1) | Allow for wrong angle CAO. Accept 111.74° if clear on diagram. |
| Answer | Marks | Guidance |
|---|---|---|
| Direction opposite to (i) clearly shown and labelled or stated | B1 ft, B1 ft 2 | Accept direction at angle 141.7° with direction of 0.6 m s⁻¹ motion |
## (i)
$0.3 \times 0.4 + l, l = 0.3 \times 0.6\cos30°$ | M1, M1, A1, A1 4 | Similar with velocities. 0.096890. 68.26. Direction must be clear in words or on diagram. Similar with velocities. Or slip in figure
$l_c = (\pm) 0.035884572$
$l_m = (\pm) 0.09$
Magnitude of $I = 0.0969$ (Ns)
68.3° from original direction
OR correct triangle of momentum
$I^2 = 0.18^2 + 0.12^2 - 2 \times 0.18 \times 0.12 \times \cos 30°$
$I = 0.096890$ (Ns)
68.26° from original direction | (M1), (A1), (A1), (A1) | Allow for wrong angle CAO. Accept 111.74° if clear on diagram.
## (ii)
Magnitude = 0.0969 N
Direction opposite to (i) clearly shown and labelled or stated | B1 ft, B1 ft 2 | Accept direction at angle 141.7° with direction of 0.6 m s⁻¹ motion | If total 0 scored, allow SC1 for 38.3° or 68.3° seen
---\includegraphics{figure_1}
A particle $P$ of mass $0.3$ kg is moving with speed $0.4$ m s$^{-1}$ in a straight line on a smooth horizontal surface when it is struck by a horizontal impulse. After the impulse acts $P$ has speed $0.6$ m s$^{-1}$ and is moving in a direction making an angle $30°$ with its original direction of motion (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Find the magnitude of the impulse and the angle its line of action makes with the original direction of motion of $P$. [4]
\end{enumerate}
Subsequently a second impulse acts on $P$. After this second impulse acts, $P$ again moves from left to right with speed $0.4$ m s$^{-1}$ in a direction parallel to its original direction of motion.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item State the magnitude of the second impulse, and show the direction of the second impulse on a diagram. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR M3 2016 Q1 [6]}}