| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Velocity after impulse (direct calculation) |
| Difficulty | Moderate -0.5 This is a straightforward application of the impulse-momentum theorem in 2D. Students need to apply impulse = change in momentum, find the new velocity vector, then calculate its magnitude using Pythagoras. It's more routine than average A-level questions since it requires only direct formula application with no problem-solving insight, but the vector arithmetic and multi-step calculation make it slightly more involved than pure recall. |
| Spec | 1.10c Magnitude and direction: of vectors6.03a Linear momentum: p = mv6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{I} = m\mathbf{v} - m\mathbf{u}\) | M1A1 | |
| \(5\mathbf{i} - 3\mathbf{j} = \frac{1}{4}\mathbf{v} - \frac{1}{4}(3\mathbf{i} + 7\mathbf{j})\) | ||
| \(\mathbf{v} = 23\mathbf{i} - 5\mathbf{j}\) | A1 | |
| \( | \mathbf{v} | = \sqrt{23^2 + 5^2} = 23.5\) |
## Question 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{I} = m\mathbf{v} - m\mathbf{u}$ | M1A1 | |
| $5\mathbf{i} - 3\mathbf{j} = \frac{1}{4}\mathbf{v} - \frac{1}{4}(3\mathbf{i} + 7\mathbf{j})$ | | |
| $\mathbf{v} = 23\mathbf{i} - 5\mathbf{j}$ | A1 | |
| $|\mathbf{v}| = \sqrt{23^2 + 5^2} = 23.5$ | M1A1 | |
**Total: [5]**
---\begin{enumerate}
\item A particle of mass 0.25 kg is moving with velocity $( 3 \mathbf { i } + 7 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ when it receives the impulse $( 5 \mathbf { i } - 3 \mathbf { j } )$ N s.
\end{enumerate}
Find the speed of the particle immediately after the impulse.\\
\hfill \mbox{\textit{Edexcel M2 2009 Q1 [5]}}