| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Impulse from velocity change |
| Difficulty | Moderate -0.3 This is a straightforward two-part impulse-momentum question requiring standard vector manipulation. Part (a) uses impulse-momentum theorem to find initial velocity then calculates magnitude. Part (b) uses dot product to find angle between velocity vectors. Both are routine M2 techniques with no conceptual challenges beyond applying memorized formulas. |
| Spec | 1.10c Magnitude and direction: of vectors6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(0.4\mathbf{u} + (-5\mathbf{i} + 3\mathbf{j}) = 0.4(12\mathbf{i} + 15\mathbf{j})\) | M1 | Impulse-momentum equation. Requires all 3 terms. Must be dimensionally correct. Condone sign error(s). |
| A1 | Correct unsimplified equation | |
| \(\mathbf{u} = \frac{9.8\mathbf{i} + 3\mathbf{j}}{0.4} = 24.5\mathbf{i} + 7.5\mathbf{j}\) | — | — |
| Speed \(= \sqrt{24.5^2 + 7.5^2} = 25.6\) m s\(^{-1}\) | M1 | Use of Pythagoras' theorem to find magnitude of their \(\mathbf{u}\) |
| A1 (4) | 25.6 or better |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\tan^{-1}\!\left(\frac{15}{12}\right)\), \(\tan^{-1}\!\left(\frac{7.5}{24.5}\right)\) | M1 | Use trig. to find useful angles. Follow their \(\mathbf{u}\) |
| A1 | Correct unsimplified. (51.3° & 17.0°, 38.7° & 73°) | |
| \(= 34.319…°\) or \(0.59899…\) rad | A1 (3) | Combine correctly to find the required angle. 34°, 0.60 rads or better |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\cos\theta = \dfrac{\begin{pmatrix}24.5\\7.5\end{pmatrix}\cdot\begin{pmatrix}12\\15\end{pmatrix}}{\sqrt{24.5^2+7.5^2}\sqrt{12^2+15^2}}\) | M1 | Use scalar product. Follow their \(\mathbf{u}\) |
| A1 | Correctly substituted | |
| \(\theta = 34.319…°\) or \(0.59899…\) rad | A1 | — |
# Question 2:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $0.4\mathbf{u} + (-5\mathbf{i} + 3\mathbf{j}) = 0.4(12\mathbf{i} + 15\mathbf{j})$ | M1 | Impulse-momentum equation. Requires all 3 terms. Must be dimensionally correct. Condone sign error(s). |
| | A1 | Correct unsimplified equation |
| $\mathbf{u} = \frac{9.8\mathbf{i} + 3\mathbf{j}}{0.4} = 24.5\mathbf{i} + 7.5\mathbf{j}$ | — | — |
| Speed $= \sqrt{24.5^2 + 7.5^2} = 25.6$ m s$^{-1}$ | M1 | Use of Pythagoras' theorem to find magnitude of their $\mathbf{u}$ |
| | A1 (4) | 25.6 or better |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\tan^{-1}\!\left(\frac{15}{12}\right)$, $\tan^{-1}\!\left(\frac{7.5}{24.5}\right)$ | M1 | Use trig. to find useful angles. Follow their $\mathbf{u}$ |
| | A1 | Correct unsimplified. (51.3° & 17.0°, 38.7° & 73°) |
| $= 34.319…°$ or $0.59899…$ rad | A1 (3) | Combine correctly to find the required angle. 34°, 0.60 rads or better |
**Total: [7]**
### Alt(b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\cos\theta = \dfrac{\begin{pmatrix}24.5\\7.5\end{pmatrix}\cdot\begin{pmatrix}12\\15\end{pmatrix}}{\sqrt{24.5^2+7.5^2}\sqrt{12^2+15^2}}$ | M1 | Use scalar product. Follow their $\mathbf{u}$ |
| | A1 | Correctly substituted |
| $\theta = 34.319…°$ or $0.59899…$ rad | A1 | — |
---\begin{enumerate}
\item A ball of mass 0.4 kg is moving in a horizontal plane when it is struck by a bat. The bat exerts an impulse $( - 5 \mathbf { i } + 3 \mathbf { j } ) \mathrm { N }$ s on the ball. Immediately after receiving the impulse the ball has velocity $( 12 \mathbf { i } + 15 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.
\end{enumerate}
Find\\
(a) the speed of the ball immediately before the impact,\\
(b) the size of the angle through which the direction of motion of the ball is deflected by the impact.\\
\hfill \mbox{\textit{Edexcel M2 2014 Q2 [7]}}