| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2020 |
| Session | January |
| Marks | 7 |
| 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 resolution of impulse into components, application of impulse-momentum principle in two perpendicular directions, then using Pythagoras and trigonometry to find final speed and angle. It's slightly easier than average because it follows a well-practiced procedure with clear steps and standard techniques, though it does require careful component resolution and multi-step working. |
| Spec | 6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Use of \(m\mathbf{v} = \mathbf{I} + m\mathbf{u}\) | M1 | Need to consider both components |
| Component of momentum parallel to original direction \(= 6\times 0.75 + \sqrt{24}\cos 60 \left(= 4.5 + \sqrt{6}\right)\) | A1 | Or equivalent |
| A1 | Or equivalent | |
| Use of Pythagoras: \(\left(\frac{3}{4}v =\right)\sqrt{\left(4.5+\sqrt{6}\right)^2 + 18}\) | M1 | |
| \(v = 10.9\ (\text{m s}^{-1}),\ 11\ (\text{m s}^{-1})\) | A1 | Or better |
| Alternative (first 5 marks): | ||
| Vector triangle for impulses or velocities | M1 | Must be using correct triangle — need \(120°\) seen or implied |
| Use of cosine rule | M1 | |
| \(\left(\frac{3}{4}v\right)^2 = 4.5^2 + 24 - 2\times 4.5\times\sqrt{24}\times\cos 120°\) | A1 | Correct unsimplified |
| A1 | ||
| \(v = 10.9\ (\text{m s}^{-1}),\ 11\ (\text{m s}^{-1})\) | A1 | Or better |
| Change in direction \(= \tan^{-1}\dfrac{3\sqrt{2}}{4.5+\sqrt{6}}\) | M1 | Or equivalent use of trig with components to find the required angle. E.g. \(\text{angle} = \cos^{-1}\!\left(\dfrac{4.5^2+(mv)^2-24}{2\times4.5\times(mv)}\right)\) or from scalar product \(\cos^{-1}\!\left(\dfrac{6\times9.27...}{6\times10.9...}\right)\) |
| \(= 31.4°\ (31°)\) | A1 (7) | 0.548 radians (0.55 radians) or better. Do not ISW |
## Question 3:
| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of $m\mathbf{v} = \mathbf{I} + m\mathbf{u}$ | M1 | Need to consider both components |
| Component of momentum parallel to original direction $= 6\times 0.75 + \sqrt{24}\cos 60 \left(= 4.5 + \sqrt{6}\right)$ | A1 | Or equivalent |
| | A1 | Or equivalent |
| Use of Pythagoras: $\left(\frac{3}{4}v =\right)\sqrt{\left(4.5+\sqrt{6}\right)^2 + 18}$ | M1 | |
| $v = 10.9\ (\text{m s}^{-1}),\ 11\ (\text{m s}^{-1})$ | A1 | Or better |
| **Alternative (first 5 marks):** | | |
| Vector triangle for impulses or velocities | M1 | Must be using correct triangle — need $120°$ seen or implied |
| Use of cosine rule | M1 | |
| $\left(\frac{3}{4}v\right)^2 = 4.5^2 + 24 - 2\times 4.5\times\sqrt{24}\times\cos 120°$ | A1 | Correct unsimplified |
| | A1 | |
| $v = 10.9\ (\text{m s}^{-1}),\ 11\ (\text{m s}^{-1})$ | A1 | Or better |
| Change in direction $= \tan^{-1}\dfrac{3\sqrt{2}}{4.5+\sqrt{6}}$ | M1 | Or equivalent use of trig with components to find the required angle. E.g. $\text{angle} = \cos^{-1}\!\left(\dfrac{4.5^2+(mv)^2-24}{2\times4.5\times(mv)}\right)$ or from scalar product $\cos^{-1}\!\left(\dfrac{6\times9.27...}{6\times10.9...}\right)$ |
| $= 31.4°\ (31°)$ | A1 (7) | 0.548 radians (0.55 radians) or better. Do not ISW |
**Total: [7]**
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c16c17b6-2c24-4939-b3b5-63cd63646b76-06_291_481_255_733}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A particle $P$ of mass 0.75 kg is moving along a straight line on a horizontal surface. At the instant when the speed of $P$ is $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, it receives an impulse of magnitude $\sqrt { 24 } \mathrm { Ns }$. The impulse acts in the plane of the horizontal surface. At the instant when $P$ receives the impulse, the line of action of the impulse makes an angle of $60 ^ { \circ }$ with the direction of motion of $P$, as shown in Figure 2.
Find\\
(i) the speed of $P$ immediately after receiving the impulse,\\
(ii) the size of the angle between the direction of motion of $P$ immediately before receiving the impulse and the direction of motion of $P$ immediately after receiving the impulse.\\
\includegraphics[max width=\textwidth, alt={}, center]{c16c17b6-2c24-4939-b3b5-63cd63646b76-06_2252_51_311_1980}\\
\includegraphics[max width=\textwidth, alt={}, center]{c16c17b6-2c24-4939-b3b5-63cd63646b76-07_36_65_2722_109}\\
\hfill \mbox{\textit{Edexcel M2 2020 Q3 [7]}}