| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Vector impulse: find velocity or speed after impulse |
| Difficulty | Moderate -0.8 This is a straightforward application of the impulse-momentum theorem with vector components. Part (a) requires using impulse = change in momentum to find final velocity, then calculating speed using Pythagoras. Parts (b) and (c) are direct calculations using standard formulas (arctan for angle, KE = ½mv²). All steps are routine M2 techniques with no problem-solving insight required, making it easier than average but not trivial due to the vector manipulation and multi-part structure. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{I} = m\mathbf{v} - m\mathbf{u}\) | M1 | |
| \(-4\mathbf{i} + 7\mathbf{j} = 0.5(\mathbf{v} - 12\mathbf{i})\) | ||
| \(4i + 14j = v\) | A1 | |
| Speed \(= \sqrt{16 + 196} = \sqrt{212}\) m s\(^{-1}\) (14.6 or better) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\tan \theta = \frac{7}{2}\) | M1 | |
| \(\theta = 74.0\ldots\) | ||
| \(\theta = 74°\) | A1 ft |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Gain in K.E. \(= \frac{1}{2} \times 0.5(212 - 12^2) = 17\) J | M1 A1 |
## Part (a):
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| $\mathbf{I} = m\mathbf{v} - m\mathbf{u}$ | M1 | |
| $-4\mathbf{i} + 7\mathbf{j} = 0.5(\mathbf{v} - 12\mathbf{i})$ | | |
| $4i + 14j = v$ | A1 | |
| Speed $= \sqrt{16 + 196} = \sqrt{212}$ m s$^{-1}$ (14.6 or better) | M1 A1 | |
## Part (b):
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| $\tan \theta = \frac{7}{2}$ | M1 | |
| $\theta = 74.0\ldots$ | | |
| $\theta = 74°$ | A1 ft | |
## Part (c):
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| Gain in K.E. $= \frac{1}{2} \times 0.5(212 - 12^2) = 17$ J | M1 A1 | |A ball of mass 0.5 kg is moving with velocity $12\mathbf{i}$ m s$^{-1}$ when it is struck by a bat. The impulse received by the ball is $(-4\mathbf{i} + 7\mathbf{j})$ N s. By modelling the ball as a particle, find
\begin{enumerate}[label=(\alph*)]
\item the speed of the ball immediately after the impact, [4]
\item the angle, in degrees, between the velocity of the ball immediately after the impact and the vector $\mathbf{i}$, [2]
\item the kinetic energy gained by the ball as a result of the impact. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2011 Q3 [8]}}