| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2009 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, vector velocity form |
| Difficulty | Standard +0.3 This is a standard M4 oblique collision problem requiring conservation of momentum (straightforward vector calculation), impulse-momentum theorem (routine application), and Newton's law of restitution along the line of centres. All steps follow well-established procedures with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03f Impulse-momentum: relation6.03k Newton's experimental law: direct impact |
| Answer | Marks |
|---|---|
| M1 A1 A1 |
| Answer | Marks |
|---|---|
| M1 A1 B1 |
| Answer | Marks |
|---|---|
| M1 A3 DM1 A1 |
**Part (a)**
CLM: $2(i + 2j) + (-2i) = 2j + \mathbf{v}$
$\mathbf{v} = 2j$ m s$^{-1}$
| M1 A1 A1 |
**Part (b)**
$\mathbf{I} = 2(j - (i + 2j))$
$= (-2i - 2j)$ Ns
Since $\mathbf{I}$ acts along l.o.c.c., l.o.c.c is parallel to $i + j$
| M1 A1 B1 |
**Part (c)**
Before $A$: $(i+2j) \cdot \frac{1}{\sqrt{2}}(i+j) = \frac{3}{\sqrt{2}}$
$B$: $-2j \cdot \frac{1}{\sqrt{2}}(i+j) = \frac{-2}{\sqrt{2}}$
After $A$: $j \cdot \frac{1}{\sqrt{2}}(i+j) = \frac{1}{\sqrt{2}}$
$B$: $2j \cdot \frac{1}{\sqrt{2}}(i+j) = \frac{2}{\sqrt{2}}$
NIL: $e = \frac{\frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}}}{\frac{3}{\sqrt{2}} - \frac{-2}{\sqrt{2}}} = \frac{1}{5}$
| M1 A3 DM1 A1 |
**Total: [13]**
---5. Two small smooth spheres $A$ and $B$, of mass 2 kg and 1 kg respectively, are moving on a smooth horizontal plane when they collide. Immediately before the collision the velocity of $A$ is $( \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ and the velocity of $B$ is $- 2 \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Immediately after the collision the velocity of $A$ is $\mathbf { j } \mathrm { m } \mathrm { s } ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the velocity of $B$ immediately after the collision is $2 \mathbf { j } \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item Find the impulse of $B$ on $A$ in the collision, giving your answer as a vector, and hence show that the line of centres is parallel to $\mathbf { i } + \mathbf { j }$.
\item Find the coefficient of restitution between $A$ and $B$.\\
\section*{June 2009}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2009 Q5 [13]}}