| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| 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 subtraction, followed by a standard angle calculation using dot product. Both parts are direct recall of formulas with minimal problem-solving required, making it easier than average for M2. |
| Spec | 6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form |
| Answer | Marks |
|---|---|
| Use of \(I = mv - mu\) | M1 |
| \(= 0.7((3\mathbf{i} + 4\mathbf{j}) - (\mathbf{i} - 2\mathbf{j}))\) | A1 |
| \(= 0.7(2\mathbf{i} + 6\mathbf{j}) = 1.4\mathbf{i} + 4.2\mathbf{j}\) (N s) | A1 |
| (3) |
| Answer | Marks |
|---|---|
| Required angle \(= \tan^{-1} 3 + \tan^{-1} 2\) | M1 |
| Follow their impulse | A1 ft |
| \(= 135°\) (Allow \(225°\)) | A1 |
| (3) |
| Answer | Marks |
|---|---|
| Correct use of scalar product with relevant vectors | M1 |
| \(\cos\theta = \frac{(\mathbf{i} - 2\mathbf{j}) \cdot (\mathbf{i} + 3\mathbf{j})}{\sqrt{1 + 4}\sqrt{1 + 9}} = \frac{-5}{5\sqrt{2}}\) | A1 ft |
| \(\theta = 135°\) | A1 |
| Answer | Marks |
|---|---|
| Correct use of cosine rule with relevant sides | M1 |
| \(\cos\theta = \frac{(1.4^2 + 4.2^2) + (\mathbf{i}^2 + (-2)^2) - (0.4^2 + 6.2^2)}{2\sqrt{(1.4^2 + 4.2^2)}\sqrt{(\mathbf{i}^2 + (-2)^2)}}\) | A1 ft |
| \(\theta = 135°\) | A1 |
## 1a.
Use of $I = mv - mu$ | M1 |
$= 0.7((3\mathbf{i} + 4\mathbf{j}) - (\mathbf{i} - 2\mathbf{j}))$ | A1 |
$= 0.7(2\mathbf{i} + 6\mathbf{j}) = 1.4\mathbf{i} + 4.2\mathbf{j}$ (N s) | A1 |
(3) | |
## 1b.
Required angle $= \tan^{-1} 3 + \tan^{-1} 2$ | M1 |
Follow their impulse | A1 ft |
$= 135°$ (Allow $225°$) | A1 |
(3) | |
**Alternative for last 3 marks using scalar product:**
Correct use of scalar product with relevant vectors | M1 |
$\cos\theta = \frac{(\mathbf{i} - 2\mathbf{j}) \cdot (\mathbf{i} + 3\mathbf{j})}{\sqrt{1 + 4}\sqrt{1 + 9}} = \frac{-5}{5\sqrt{2}}$ | A1 ft |
$\theta = 135°$ | A1 |
**Alternative for last 3 marks using cosine rule:**
Correct use of cosine rule with relevant sides | M1 |
$\cos\theta = \frac{(1.4^2 + 4.2^2) + (\mathbf{i}^2 + (-2)^2) - (0.4^2 + 6.2^2)}{2\sqrt{(1.4^2 + 4.2^2)}\sqrt{(\mathbf{i}^2 + (-2)^2)}}$ | A1 ft |
$\theta = 135°$ | A1 |
**Notes for Q1:**
- M1 for Use of $I = mv - mu$ (M0 if $m$ omitted or $g$ included) with $v$ and $u$ substituted but allow $m$ and allow terms reversed
- First A1 for a correct equation
- Second A1 for a correct answer, $\mathbf{i}$'s and $\mathbf{j}$'s need to be collected
- Isw if they go on and give a magnitude.
- 1(b): M1 for a complete correct method to find required angle for their $\mathbf{I}$ (M0 if they are finding the wrong angle)
- First A1 ft for a correct expression from their $\mathbf{I}$ and $\mathbf{u}$
- Second A1 for $135°$ (Accept $134.9...°$ or $225.1...$)
---\begin{enumerate}
\item A particle $P$ of mass 0.7 kg is moving with velocity ( $\mathbf { i } - 2 \mathbf { j }$ ) $\mathrm { m } \mathrm { s } ^ { - 1 }$ when it receives an impulse. Immediately after receiving the impulse, $P$ is moving with velocity $( 3 \mathbf { i } + 4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.\\
(a) Find the impulse.\\
(b) Find, in degrees, the size of the angle between the direction of the impulse and the direction of motion of $P$ immediately before receiving the impulse.\\
(3)\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2018 Q1 [6]}}