| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Impulse from velocity change |
| Difficulty | Moderate -0.8 This is a straightforward application of the impulse-momentum theorem (impulse = change in momentum) with vector arithmetic. Students need to rearrange mv_final - mv_initial = impulse and substitute given values. It requires only direct recall of a standard formula with basic vector subtraction, making it easier than average with no problem-solving insight needed. |
| Spec | 6.03a Linear momentum: p = mv6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{I} = \text{Anom.} \quad 12\mathbf{i} - 9\mathbf{j} = 0.6[5(\mathbf{i} + 3\mathbf{j}) - \mathbf{u}]\) | M1 A1 | |
| \(20\mathbf{i} - 15\mathbf{j} = 5\mathbf{i} + 3\mathbf{j} - \mathbf{u}\) | M1 | |
| \(\mathbf{u} = -15\mathbf{i} + 18\mathbf{j}\) | A1 | (4) |
$\mathbf{I} = \text{Anom.} \quad 12\mathbf{i} - 9\mathbf{j} = 0.6[5(\mathbf{i} + 3\mathbf{j}) - \mathbf{u}]$ | M1 A1 |
$20\mathbf{i} - 15\mathbf{j} = 5\mathbf{i} + 3\mathbf{j} - \mathbf{u}$ | M1 |
$\mathbf{u} = -15\mathbf{i} + 18\mathbf{j}$ | A1 | (4)\begin{enumerate}
\item A ball of mass 0.6 kg bounces against a wall and is given an impulse of $( 12 \mathbf { i } - 9 \mathbf { j } ) \mathrm { Ns }$ where $\mathbf { i }$ and $\mathbf { j }$ are perpendicular horizontal unit vectors. The velocity of the particle after the impact is $( 5 \mathbf { i } + 3 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.
\end{enumerate}
Find the velocity of the particle before the impact.\\
(4 marks)\\
\hfill \mbox{\textit{Edexcel M2 Q1 [4]}}