| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2023 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Impulse from velocity change |
| Difficulty | Standard +0.3 This is a standard M2 impulse-momentum question requiring vector resolution in two perpendicular directions and solving simultaneous equations. While it involves multiple steps (resolving impulse, applying momentum conservation in two directions, using Pythagoras), these are routine techniques for M2 students with no novel insight required. Slightly easier than average due to the straightforward setup and clear numerical values. |
| Spec | 1.10c Magnitude and direction: of vectors6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Form impulse-momentum equation | M1 | Dimensionally correct. Accept vector form or separate components. Condone sine/cosine confusion. |
| One correct equation | A1 | e.g. one correct component of \(\begin{pmatrix}I\cos 60°\\ I\sin 60°\end{pmatrix} = \frac{1}{4}\left[\begin{pmatrix}12\cos\alpha\\ 12\sin\alpha\end{pmatrix} - \begin{pmatrix}8\\ 0\end{pmatrix}\right] = \begin{pmatrix}3\cos\alpha - 2\\ 3\sin\alpha\end{pmatrix}\) or \(8\sin 60° = 12\cos(30°+\alpha)\) or \(I = 0.25(12\sin(30°+\alpha) - 8\cos 60°)\) |
| Form a second impulse-momentum equation | M1 | |
| Correct second equation | A1 | |
| Complete method to solve for \(I\) | DM1 | Dependent on the two preceding M marks. e.g. from \(36 = (I+4)^2 + 3I^2\ \ (4I^2 + 8I - 20 = 0)\) |
| \(I = \sqrt{6} - 1\) (or 1.45 or 1.4) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(I = mv - mu\) to draw a vector triangle. Dimensionally consistent. | M1 | |
| Correct diagram with sides 2, 3, \(I\) and angle \(120°\) | A1 | Correct diagram |
| Form equation in \(I\), e.g. by using cosine rule | M1 | |
| \(4 + I^2 - 4I\cos 120° = 9\) | A1 | Correct unsimplified equation. A correct cosine rule equation can imply the first M1A1 if no diagram seen |
| Solve for \(I\) | DM1 | Dependent on 2 preceding M marks. \(I^2 + 2I - 5 = 0\) |
| \(I = \sqrt{6} - 1\) (or 1.45 or 1.4) | A1 |
# Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Form impulse-momentum equation | M1 | Dimensionally correct. Accept vector form or separate components. Condone sine/cosine confusion. |
| One correct equation | A1 | e.g. one correct component of $\begin{pmatrix}I\cos 60°\\ I\sin 60°\end{pmatrix} = \frac{1}{4}\left[\begin{pmatrix}12\cos\alpha\\ 12\sin\alpha\end{pmatrix} - \begin{pmatrix}8\\ 0\end{pmatrix}\right] = \begin{pmatrix}3\cos\alpha - 2\\ 3\sin\alpha\end{pmatrix}$ or $8\sin 60° = 12\cos(30°+\alpha)$ or $I = 0.25(12\sin(30°+\alpha) - 8\cos 60°)$ |
| Form a second impulse-momentum equation | M1 | |
| Correct second equation | A1 | |
| Complete method to solve for $I$ | DM1 | Dependent on the two preceding M marks. e.g. from $36 = (I+4)^2 + 3I^2\ \ (4I^2 + 8I - 20 = 0)$ |
| $I = \sqrt{6} - 1$ (or 1.45 or 1.4) | A1 | |
**Alternative method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $I = mv - mu$ to draw a vector triangle. Dimensionally consistent. | M1 | |
| Correct diagram with sides 2, 3, $I$ and angle $120°$ | A1 | Correct diagram |
| Form equation in $I$, e.g. by using cosine rule | M1 | |
| $4 + I^2 - 4I\cos 120° = 9$ | A1 | Correct unsimplified equation. A correct cosine rule equation can imply the first M1A1 if no diagram seen |
| Solve for $I$ | DM1 | Dependent on 2 preceding M marks. $I^2 + 2I - 5 = 0$ |
| $I = \sqrt{6} - 1$ (or 1.45 or 1.4) | A1 | |
**Total: [6]**
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f180f5f0-43c5-4365-b0d8-7284220b481e-08_424_752_246_667}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A particle $Q$ of mass 0.25 kg is moving in a straight line on a smooth horizontal surface with speed $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when it receives an impulse of magnitude $I \mathrm { Ns }$.
The impulse acts parallel to the horizontal surface and at $60 ^ { \circ }$ to the original direction of motion of $Q$.
Immediately after receiving the impulse, the speed of $Q$ is $12 \mathrm {~ms} ^ { - 1 }$\\
As a result of receiving the impulse, the direction of motion of $Q$ is turned through $\alpha ^ { \circ }$, as shown in Figure 2.
Find the value of $I$
\hfill \mbox{\textit{Edexcel M2 2023 Q3 [6]}}