| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Successive collisions, three particles in line |
| Difficulty | Standard +0.3 This is a standard M2 momentum/collisions question with multiple parts requiring conservation of momentum, coefficient of restitution, and checking collision conditions. While it involves successive collisions and algebraic manipulation, the techniques are routine for M2 students and the question provides significant scaffolding through 'show that' parts, making it slightly easier than average. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Conservation of momentum: \(4mu - 3mv = 3mkv\) | M1A1 | |
| Impact law: \(kv = \frac{3}{4}(u + v)\) | M1A1 | |
| Eliminate \(k\): \(4mu - 3mv = 3m \times \frac{3}{4}(u+v)\) | DM1 | |
| \(u = 3v\) (answer given) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(kv = \frac{3}{4}(3v + v),\ k = 3\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Impact law: \((kv + 2v)e = v_C - v_B\ (5ve = v_C - v_B)\) | B1 | |
| Conservation of momentum: \(3 \times kv - 1 \times 2v = 3v_B + v_c\ (7v = 3v_B + v_c)\) | B1 | |
| Eliminate \(v_C\): \(v_B = \frac{v}{4}(7 - 5e) > 0\) hence no further collision with \(A\) | M1A1 |
## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Conservation of momentum: $4mu - 3mv = 3mkv$ | M1A1 | |
| Impact law: $kv = \frac{3}{4}(u + v)$ | M1A1 | |
| Eliminate $k$: $4mu - 3mv = 3m \times \frac{3}{4}(u+v)$ | DM1 | |
| $u = 3v$ (answer given) | A1 | |
**Subtotal: (6)**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $kv = \frac{3}{4}(3v + v),\ k = 3$ | M1, A1 | |
**Subtotal: (2)**
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Impact law: $(kv + 2v)e = v_C - v_B\ (5ve = v_C - v_B)$ | B1 | |
| Conservation of momentum: $3 \times kv - 1 \times 2v = 3v_B + v_c\ (7v = 3v_B + v_c)$ | B1 | |
| Eliminate $v_C$: $v_B = \frac{v}{4}(7 - 5e) > 0$ hence no further collision with $A$ | M1A1 | |
**Subtotal: (4) Total: [12]**\begin{enumerate}
\item Particles $A , B$ and $C$ of masses $4 m , 3 m$ and $m$ respectively, lie at rest in a straight line on a smooth horizontal plane with $B$ between $A$ and $C$. Particles $A$ and $B$ are projected towards each other with speeds $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ respectively, and collide directly.
\end{enumerate}
As a result of the collision, $A$ is brought to rest and $B$ rebounds with speed $k v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The coefficient of restitution between $A$ and $B$ is $\frac { 3 } { 4 }$.\\
(a) Show that $u = 3 v$.\\
(b) Find the value of $k$.
Immediately after the collision between $A$ and $B$, particle $C$ is projected with speed $2 v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ towards $B$ so that $B$ and $C$ collide directly.\\
(c) Show that there is no further collision between $A$ and $B$.\\
\hfill \mbox{\textit{Edexcel M2 2009 Q8 [12]}}