| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2005 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with direction reversal |
| Difficulty | Standard +0.3 This is a standard M2 collision problem requiring conservation of momentum and Newton's restitution law. Part (a) is a routine 'show that' derivation, part (b) requires interpreting the physical constraint that P reverses direction, and part (c) applies the impulse-momentum theorem. All techniques are textbook-standard with no novel insight required, making it slightly easier than average for M2. |
| Spec | 6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| LM: \(6mu - 2mu = 3mx + 2my\) | M1 A1 | |
| NEL: \(y - x = 3eu\) | B1 | |
| Solving to \(y = \frac{1}{5}u(9e + 4)\) | M1 A1 | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Solving to \(x = \frac{2}{5}u(2 - 3e)\) | M1 A1 | oe |
| \(x < 0 \Rightarrow e > \frac{2}{3}\) | M1 A1 | |
| \(\frac{2}{3} < e \leq 1\) | A1ft | ft their \(e\) for glb |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2m\left[\frac{1}{5}u(9e+4) + u\right] = \frac{32}{5}mu\) | M1 A1 | |
| Solving to \(e = \frac{7}{9}\) | M1 A1 | awrt 0.78 |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| LM: $6mu - 2mu = 3mx + 2my$ | M1 A1 | |
| NEL: $y - x = 3eu$ | B1 | |
| Solving to $y = \frac{1}{5}u(9e + 4)$ | M1 A1 | cso |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Solving to $x = \frac{2}{5}u(2 - 3e)$ | M1 A1 | oe |
| $x < 0 \Rightarrow e > \frac{2}{3}$ | M1 A1 | |
| $\frac{2}{3} < e \leq 1$ | A1ft | ft their $e$ for glb |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2m\left[\frac{1}{5}u(9e+4) + u\right] = \frac{32}{5}mu$ | M1 A1 | |
| Solving to $e = \frac{7}{9}$ | M1 A1 | awrt 0.78 |
---6. A particle $P$ of mass $3 m$ is moving with speed $2 u$ in a straight line on a smooth horizontal table. The particle $P$ collides with a particle $Q$ of mass $2 m$ moving with speed $u$ in the opposite direction to $P$. The coefficient of restitution between $P$ and $Q$ is $e$.
\begin{enumerate}[label=(\alph*)]
\item Show that the speed of $Q$ after the collision is $\frac { 1 } { 5 } u ( 9 e + 4 )$.
As a result of the collision, the direction of motion of $P$ is reversed.
\item Find the range of possible values of $e$.
Given that the magnitude of the impulse of $P$ on $Q$ is $\frac { 32 } { 5 } m u$,
\item find the value of $e$.\\
(4)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2005 Q6 [14]}}