| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2022 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Vector impulse: find velocity or speed after impulse |
| Difficulty | Standard +0.3 This is a straightforward impulse-momentum vector problem requiring resolution of the impulse into components, application of momentum conservation in two perpendicular directions, and use of Pythagoras to find final speed. The given tan α = 4/3 makes the trigonometry simple (3-4-5 triangle). Standard FM1 technique with no novel insight required, slightly easier than average A-level due to clean numbers and direct method. |
| Spec | 6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Impulse-momentum equation(s) | M1 | Use of \(\mathbf{I}=m\mathbf{v}-m\mathbf{u}\) in two dimensions; allow combined vector equation or one component; condone sin/cos confusion; allow if \(m\) seen but not substituted |
| \(\begin{pmatrix}3\cos\alpha\\3\sin\alpha\end{pmatrix}=\frac{1}{2}\begin{pmatrix}v_x-2.8\\v_y\end{pmatrix}\) giving \(v_x=\frac{32}{5}\), \(v_y=\frac{24}{5}\) | A1 A1 | Equation for one component correct unsimplified; equations for both components correct unsimplified; allow A1A1 for correct unsimplified vector equation; allow if in terms of \(m\) and \(\alpha\) |
| \(v=\frac{1}{5}\sqrt{32^2+24^2}\) | M1 | Correct use of Pythagoras for components to obtain numerical speed; may be seen or implied (alert: 3,4,5 triangle) |
| \(v=8\ (\text{ms}^{-1})\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using cosine rule | M1 | |
| \(v^2=2.8^2+6^2-2\times2.8\times6\cos(\pi-\alpha)\) | A1 A1 | |
| Solve for \(v\) | M1 | |
| \(v=8\ (\text{ms}^{-1})\) | A1 |
# Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Impulse-momentum equation(s) | M1 | Use of $\mathbf{I}=m\mathbf{v}-m\mathbf{u}$ in two dimensions; allow combined vector equation or one component; condone sin/cos confusion; allow if $m$ seen but not substituted |
| $\begin{pmatrix}3\cos\alpha\\3\sin\alpha\end{pmatrix}=\frac{1}{2}\begin{pmatrix}v_x-2.8\\v_y\end{pmatrix}$ giving $v_x=\frac{32}{5}$, $v_y=\frac{24}{5}$ | A1 A1 | Equation for one component correct unsimplified; equations for both components correct unsimplified; allow A1A1 for correct unsimplified vector equation; allow if in terms of $m$ and $\alpha$ |
| $v=\frac{1}{5}\sqrt{32^2+24^2}$ | M1 | Correct use of Pythagoras for components to obtain numerical speed; may be seen or implied (alert: 3,4,5 triangle) |
| $v=8\ (\text{ms}^{-1})$ | A1 | |
**Alternative (cosine rule):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using cosine rule | M1 | |
| $v^2=2.8^2+6^2-2\times2.8\times6\cos(\pi-\alpha)$ | A1 A1 | |
| Solve for $v$ | M1 | |
| $v=8\ (\text{ms}^{-1})$ | A1 | |3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{86a37170-046f-46e5-9c8c-06d5f98ca4fe-10_302_442_244_813}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A particle $P$ of mass 0.5 kg is moving in a straight line with speed $2.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when it receives an impulse of magnitude 3 Ns .\\
The angle between the direction of motion of $P$ immediately before receiving the impulse and the line of action of the impulse is $\alpha$, where $\tan \alpha = \frac { 4 } { 3 }$, as shown in Figure 2.
Find the speed of $P$ immediately after receiving the impulse.
\hfill \mbox{\textit{Edexcel FM1 2022 Q3 [5]}}