| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Angle change from impulse |
| Difficulty | Standard +0.3 This is a standard M2 impulse-momentum question requiring vector resolution in two perpendicular directions, followed by Pythagoras and inverse tangent to find final speed and direction. The angle (90°+α) with tan α = 3/4 is a routine setup that simplifies to sin/cos = 3/5 and 4/5. Straightforward application of impulse-momentum theorem with no conceptual challenges beyond standard M2 technique. |
| Spec | 6.03e Impulse: by a force6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| \(12.5\sin\alpha = \frac{1}{4}(v_1 - 30)\) | M1 | NB In a Q with parts labelled (i) & (ii) marks are awarded when seen – they do not belong to a particular part of the Q. Impulse = change in momentum parallel to the initial direction |
| \(\text{or } \frac{-12.5\sin\alpha = \frac{1}{4}(v_1 - 30)}{(v_1 = 0)}\) | A1 | Correct equation |
| \(12.5\cos\alpha = \frac{1}{4}(v_2 - 0)\) | M1 | Impulse = change in momentum perpendicular to the initial direction. Condone sin/cos confusion |
| \((v_2 = 40)\) | A1 | Correct equation |
| NB could be in the form: \(\begin{pmatrix} -12.5\sin\alpha \\ 12.5\cos\alpha \end{pmatrix} = 0.25v - 0.25 \begin{pmatrix} 30 \\ 0 \end{pmatrix}\) | ||
| speed is \(40\) m s\(^{-1}\); perpendicular to original direction | A1 | cwo. Correct magnitude of speed after impulse. NB Must be speed, not velocity |
| A1 | cwo. Correct direction (relative to the line given on the diagram – e.g. accept "vertically", "North", \(\mathbf{j}\) direction, "up") | |
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Using a vector triangle: \((\frac{1}{4}v)^2 = 7.5^2 + 12.5^2 - 2 \times 7.5 \times 12.5\cos(90° - \alpha)\) | M1 | Use cosine rule to find \(\frac{1}{4}v\). Terms must be of correct form, but accept unsimplified or slips e.g. their \(\frac{1}{4} \times 30\) |
| A1 | Correct equation cao (penultimate mark on open) | |
| \(v = 40\) m s\(^{-1}\) | A1 | cao (final mark on open) |
| \(\frac{12.5}{sin\theta} = \frac{7.5}{\sin\alpha}\) | M1 | Use sine rule to find angle between initial and final directions |
| A1 | Correct equation in \(\alpha\) and \(\theta\) | |
| \(\theta = 90°\) | A1 | cao (final mark on open) |
| 6 |
| $12.5\sin\alpha = \frac{1}{4}(v_1 - 30)$ | M1 | NB In a Q with parts labelled (i) & (ii) marks are awarded when seen – they do not belong to a particular part of the Q. Impulse = change in momentum parallel to the initial direction |
| $\text{or } \frac{-12.5\sin\alpha = \frac{1}{4}(v_1 - 30)}{(v_1 = 0)}$ | A1 | Correct equation |
| $12.5\cos\alpha = \frac{1}{4}(v_2 - 0)$ | M1 | Impulse = change in momentum perpendicular to the initial direction. Condone sin/cos confusion |
| $(v_2 = 40)$ | A1 | Correct equation |
| | | NB could be in the form: $\begin{pmatrix} -12.5\sin\alpha \\ 12.5\cos\alpha \end{pmatrix} = 0.25v - 0.25 \begin{pmatrix} 30 \\ 0 \end{pmatrix}$ |
| speed is $40$ m s$^{-1}$; perpendicular to original direction | A1 | cwo. Correct magnitude of speed after impulse. NB Must be speed, not velocity |
| | A1 | cwo. Correct direction (relative to the line given on the diagram – e.g. accept "vertically", "North", $\mathbf{j}$ direction, "up") |
| | 6 | |
**OR**
| Using a vector triangle: $(\frac{1}{4}v)^2 = 7.5^2 + 12.5^2 - 2 \times 7.5 \times 12.5\cos(90° - \alpha)$ | M1 | Use cosine rule to find $\frac{1}{4}v$. Terms must be of correct form, but accept unsimplified or slips e.g. their $\frac{1}{4} \times 30$ |
| | A1 | Correct equation cao (penultimate mark on open) |
| $v = 40$ m s$^{-1}$ | A1 | cao (final mark on open) |
| $\frac{12.5}{sin\theta} = \frac{7.5}{\sin\alpha}$ | M1 | Use sine rule to find angle between initial and final directions |
| | A1 | Correct equation in $\alpha$ and $\theta$ |
| $\theta = 90°$ | A1 | cao (final mark on open) |
| | 6 | |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{12cd7355-f632-4a84-825f-a269851c6ec4-08_330_570_242_657}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A small ball $B$ of mass 0.25 kg is moving in a straight line with speed $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ on a smooth horizontal plane when it is given an impulse. The impulse has magnitude 12.5 N s and is applied in a horizontal direction making an angle of $\left( 90 ^ { \circ } + \alpha \right)$, where $\tan \alpha = \frac { 3 } { 4 }$, with the initial direction of motion of the ball, as shown in Figure 3.\\
(i) Find the speed of $B$ immediately after the impulse is applied.\\
(ii) Find the direction of motion of $B$ immediately after the impulse is applied.\\
\hfill \mbox{\textit{Edexcel M2 2012 Q5 [6]}}