| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Velocity after impulse (find unknown constant) |
| Difficulty | Standard +0.3 This is a straightforward impulse-momentum question requiring application of the impulse-momentum theorem in one dimension for part (i), then vector resolution in part (ii). The key insight that maximum/minimum speeds occur when impulse is parallel/antiparallel is standard M3 content, and the vector triangle in part (ii) involves basic trigonometry with given numerical values. Slightly above average due to the two-dimensional aspect and multi-step nature, but all techniques are routine for M3 students. |
| Spec | 6.03e Impulse: by a force6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Realising impulse must be in same direction as velocity, or opposite | M1 | |
| Max speed \(2.8\ \text{ms}^{-1}\), min speed \(1.2\ \text{ms}^{-1}\) | A1 [3] | \(0.8 +/- 0.6/0.3\); \(-1.2\) is wrong; various methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Impulse momentum diagram | M1 | Allow M1 if positions wrong; diagram must be correct |
| \(\cos\theta = \frac{0.6^2 + 0.24^2 - 0.75^2}{2 \times 0.6 \times 0.24}\) | A1, M1 | Triangle with sides labelled \(0.24\), \(0.6\) and \(0.75\) or \(0.8\), \(2\) and \(2.5\); \(v_x = 0.8 + 2\cos\theta\) M1 either; \(v_y = 2\sin\theta\) and correct diag A1 both; Square, add, giving \(1.61 = 3.2\cos\theta\) M1 |
| \(\theta = 120°\ (2.098\ \text{rad})\), angle shown correctly | A1 [4] | Accept \(59.8°\ (1.04\ \text{rad})\); consistent with their \(\theta\); dep M1A1M1; \(120.(21)\)...A1 |
## Question 1:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Realising impulse must be in same direction as velocity, or opposite | M1 | |
| Max speed $2.8\ \text{ms}^{-1}$, min speed $1.2\ \text{ms}^{-1}$ | A1 [3] | $0.8 +/- 0.6/0.3$; $-1.2$ is wrong; various methods |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Impulse momentum diagram | M1 | Allow M1 if positions wrong; diagram must be correct |
| $\cos\theta = \frac{0.6^2 + 0.24^2 - 0.75^2}{2 \times 0.6 \times 0.24}$ | A1, M1 | Triangle with sides labelled $0.24$, $0.6$ and $0.75$ or $0.8$, $2$ and $2.5$; $v_x = 0.8 + 2\cos\theta$ M1 either; $v_y = 2\sin\theta$ and correct diag A1 both; Square, add, giving $1.61 = 3.2\cos\theta$ M1 |
| $\theta = 120°\ (2.098\ \text{rad})$, angle shown correctly | A1 [4] | Accept $59.8°\ (1.04\ \text{rad})$; consistent with their $\theta$; dep M1A1M1; $120.(21)$...A1 |
---1 A particle $P$ of mass 0.3 kg is moving on a smooth horizontal surface with speed $0.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when it is struck by a horizontal impulse. The magnitude of the impulse is 0.6 Ns .
\begin{enumerate}[label=(\roman*)]
\item (a) Find the greatest possible speed of $P$ after the impulse acts.\\
(b) Find the least possible speed of $P$ after the impulse acts.
\item In fact the speed of $P$ after the impulse acts is $2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find the angle the impulse makes with the original direction of travel of $P$ and draw a sketch to make this direction clear.
\end{enumerate}
\hfill \mbox{\textit{OCR M3 2014 Q1 [7]}}