| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Session | Specimen |
| Marks | 9 |
| 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 M2 mechanics question requiring standard application of impulse-momentum theorem (impulse = change in momentum) and kinetic energy formula. All three parts involve direct substitution into formulas with vector arithmetic—no problem-solving insight or novel approach needed, though the vector components and multi-part structure make it slightly more involved than the most basic recall questions. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03a Linear momentum: p = mv6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\mathbf{I} = m\mathbf{v} - m\mathbf{u}\) | M1 | |
| \(= 0.5 \times 20\mathbf{i} - 0.5(10\mathbf{i} + 24\mathbf{j})\) | A1 | |
| \(= 5\mathbf{i} - 12\mathbf{j}\) | ||
| \( | 5\mathbf{i} - 12\mathbf{j} | = 13\) Ns |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\tan\theta = \frac{12}{5}\) | M1 | |
| \(\theta = 67.4°\) | A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| K.E. lost \(= \frac{1}{2} \times 0.5(10^2 + 24^2) - \frac{1}{2} \times 0.5 \times 20^2\) | M1 A1 | |
| \(= 69\) J | A1 | (3 marks) |
# Question 5:
## Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| $\mathbf{I} = m\mathbf{v} - m\mathbf{u}$ | M1 | |
| $= 0.5 \times 20\mathbf{i} - 0.5(10\mathbf{i} + 24\mathbf{j})$ | A1 | |
| $= 5\mathbf{i} - 12\mathbf{j}$ | | |
| $|5\mathbf{i} - 12\mathbf{j}| = 13$ Ns | M1 A1 | (4 marks) |
## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $\tan\theta = \frac{12}{5}$ | M1 | |
| $\theta = 67.4°$ | A1 | (2 marks) |
## Part (c):
| Working | Marks | Notes |
|---------|-------|-------|
| K.E. lost $= \frac{1}{2} \times 0.5(10^2 + 24^2) - \frac{1}{2} \times 0.5 \times 20^2$ | M1 A1 | |
| $= 69$ J | A1 | (3 marks) |
---5. [In this question $\mathbf { i }$ and $\mathbf { j }$ are perpendi cular unit vectors in a horizontal plane.]
A ball of mass 0.5 kg is moving with velocity $( 10 \mathbf { i } + 24 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ when it is struck by a bat. Immediately after the impact the ball is moving with velocity $20 \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
Find
\begin{enumerate}[label=(\alph*)]
\item the magnitude of the impulse of the bat on the ball,
\item the size of the angle between the vector $\mathbf { i }$ and the impulse exerted by the bat on the ball,
\item the kinetic energy lost by the ball in the impact.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q5 [9]}}