| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Vector impulse: find deflection angle or impulse magnitude from angle |
| Difficulty | Standard +0.3 This is a standard 2D impulse-momentum question requiring vector decomposition and Pythagoras/trigonometry. While it involves multiple steps (resolving velocities, finding impulse components, calculating magnitude and angle), these are routine M2 techniques with no novel problem-solving required. The 8-mark allocation reflects computational work rather than conceptual difficulty, making it slightly easier than the average A-level question. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{I}\uparrow = 0.25 \times 40 \sin 60 = 5\sqrt{3}\) (8.66) | one component both | M1 A1 |
| \(\text{I}\leftarrow = 0.25(-20 + 30) = 2.5\) | ||
| \( | I | = \sqrt{75 + 6.25} = 9.01\) (Ns) |
| Answer | Marks |
|---|---|
| \(\frac{\sin\theta}{40} = \frac{\sin 60°}{\sqrt{1300}}\) | M1 A1 |
| \(\theta = 106°\) (3 s.f.) | |
| or \(\tan\theta = \pm\frac{5\sqrt{3}}{2.5}\) oce \(\theta = 106°\) | M1 A1 (4) |
| Answer | Marks |
|---|---|
| Use of \(\mathbf{I} = m(\mathbf{v} - \mathbf{u})\) | M1 |
| \(30^2 + 40^2 - 2 \times 30 \times 40\cos 60°\) (\(= 1300\)) | M1 A1 |
| \(I = 0.25\sqrt{1300} = 9.01\) N s (3 s.f.) | A1 |
| Answer | Marks |
|---|---|
| \(\mathbf{u} = 30\mathbf{i}\), \(\mathbf{v} = 40\cos 60\mathbf{i} + 40\sin 60\mathbf{j} = 20\mathbf{i} + 20\sqrt{3}\mathbf{j}\) | M1 |
| \(\mathbf{I} = \frac{1}{4}(-10\mathbf{i} + 20\sqrt{3}\mathbf{j}) = -2.5\mathbf{i} + 5\sqrt{3}\mathbf{j}\) | M1 A1 etc |
## (i)
$\text{I}\uparrow = 0.25 \times 40 \sin 60 = 5\sqrt{3}$ (8.66) | one component both | M1 A1 |
$\text{I}\leftarrow = 0.25(-20 + 30) = 2.5$ | | |
$|I| = \sqrt{75 + 6.25} = 9.01$ (Ns) | | M1 A1 (4) |
## (ii)
$\frac{\sin\theta}{40} = \frac{\sin 60°}{\sqrt{1300}}$ | M1 A1 |
$\theta = 106°$ (3 s.f.) | |
or $\tan\theta = \pm\frac{5\sqrt{3}}{2.5}$ oce $\theta = 106°$ | M1 A1 (4) |
**Total: [8]**
### Alternative to 4(i)
Use of $\mathbf{I} = m(\mathbf{v} - \mathbf{u})$ | M1 |
$30^2 + 40^2 - 2 \times 30 \times 40\cos 60°$ ($= 1300$) | M1 A1 |
$I = 0.25\sqrt{1300} = 9.01$ N s (3 s.f.) | A1 |
### 2nd Alternative to 4(i)
$\mathbf{u} = 30\mathbf{i}$, $\mathbf{v} = 40\cos 60\mathbf{i} + 40\sin 60\mathbf{j} = 20\mathbf{i} + 20\sqrt{3}\mathbf{j}$ | M1 |
$\mathbf{I} = \frac{1}{4}(-10\mathbf{i} + 20\sqrt{3}\mathbf{j}) = -2.5\mathbf{i} + 5\sqrt{3}\mathbf{j}$ | M1 A1 etc |
---\includegraphics{figure_1}
The points $A$, $B$ and $C$ lie in a horizontal plane. A batsman strikes a ball of mass $0.25$ kg. Immediately before being struck, the ball is moving along the horizontal line $AB$ with speed $30 \text{ ms}^{-1}$. Immediately after being struck, the ball moves along the horizontal line $BC$ with speed $40 \text{ ms}^{-1}$. The line $BC$ makes an angle of $60°$ with the original direction of motion $AB$, as shown in Figure 1.
Find, to 3 significant figures,
\begin{enumerate}[label=(\roman*)]
\item the magnitude of the impulse given to the ball,
\item the size of the angle that the direction of this impulse makes with the original direction of motion $AB$.
\end{enumerate}
[8]
\hfill \mbox{\textit{Edexcel M2 2010 Q4 [8]}}