| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Session | Specimen |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Collision followed by wall impact |
| Difficulty | Standard +0.3 This is a standard M2 collision problem requiring conservation of momentum and Newton's restitution law, followed by a wall collision and a meeting-time calculation. All techniques are routine for M2 students, though the algebra in part (c) requires careful tracking of multiple steps. Slightly easier than average due to the structured parts guiding students through the solution. |
| Spec | 6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Conservation of Momentum: \(3mu - mu = 3mv + mw\) → \(2u = 3v + w\) (1) | M1# A1 | |
| N.L.R.: \(\frac{1}{2}(u+u) = w - v\) → \(u = w - v\) (2) | M1# A1 | |
| \((1)-(2)\): \(u = 4v\) → \(v = \frac{1}{4}u\) | DM1# A1 | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| In (2): \(u = w - \frac{1}{4}u\) → \(w = \frac{5}{4}u\) | A1 | (7 total) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(B\) to wall N.L.R.: \(\dfrac{5}{4}u \times \dfrac{2}{5} = V\) | M1 | |
| \(V = \dfrac{1}{2}u\) | A1ft | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(B\) to wall: time \(= 4a \div \dfrac{5}{4}u = \dfrac{16a}{5u}\) | B1ft | |
| Distance travelled by \(A = \dfrac{1}{4}u \times \dfrac{16a}{5u} = \dfrac{4}{5}a\) | B1ft | |
| Collide when speed of approach \(= \dfrac{1}{2}ut + \dfrac{1}{4}ut\), distance to cover \(= 4a - \dfrac{4}{5}a\) | M1$ | |
| \(\therefore t = \dfrac{4a - \frac{4}{5}a}{\frac{3}{4}u} = \dfrac{16a}{5} \times \dfrac{4}{3u} = \dfrac{64a}{15u}\) | DM1$ A1 | |
| Total time \(= \dfrac{16a}{5u} + \dfrac{64a}{15u} = \dfrac{112a}{15u}\) * | A1 | (6) [15] |
# Question 8:
## Part (a)(i):
| Working | Marks | Notes |
|---------|-------|-------|
| Conservation of Momentum: $3mu - mu = 3mv + mw$ → $2u = 3v + w$ (1) | M1# A1 | |
| N.L.R.: $\frac{1}{2}(u+u) = w - v$ → $u = w - v$ (2) | M1# A1 | |
| $(1)-(2)$: $u = 4v$ → $v = \frac{1}{4}u$ | DM1# A1 | (7) |
## Part (a)(ii):
| Working | Marks | Notes |
|---------|-------|-------|
| In (2): $u = w - \frac{1}{4}u$ → $w = \frac{5}{4}u$ | A1 | (7 total) |
## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $B$ to wall N.L.R.: $\dfrac{5}{4}u \times \dfrac{2}{5} = V$ | M1 | |
| $V = \dfrac{1}{2}u$ | A1ft | (2) |
## Part (c):
| Working | Marks | Notes |
|---------|-------|-------|
| $B$ to wall: time $= 4a \div \dfrac{5}{4}u = \dfrac{16a}{5u}$ | B1ft | |
| Distance travelled by $A = \dfrac{1}{4}u \times \dfrac{16a}{5u} = \dfrac{4}{5}a$ | B1ft | |
| Collide when speed of approach $= \dfrac{1}{2}ut + \dfrac{1}{4}ut$, distance to cover $= 4a - \dfrac{4}{5}a$ | M1$ | |
| $\therefore t = \dfrac{4a - \frac{4}{5}a}{\frac{3}{4}u} = \dfrac{16a}{5} \times \dfrac{4}{3u} = \dfrac{64a}{15u}$ | DM1$ A1 | |
| Total time $= \dfrac{16a}{5u} + \dfrac{64a}{15u} = \dfrac{112a}{15u}$ * | A1 | (6) **[15]** |8. A small ball A of mass 3 m is moving with speed u in a straight line on a smooth horizontal table. The ball collides directly with another small ball B of mass m moving with speed $u$ towards $A$ along the same straight line. The coefficient of restitution between $A$ and $B$ is $\frac { 1 } { 2 }$. The balls have the same radius and can be modelled as particles.
\begin{enumerate}[label=(\alph*)]
\item Find
\begin{enumerate}[label=(\roman*)]
\item the speed of A immediately after the collision,
\item the speed of B immediately after the collision.
A fter the collision $B$ hits a smooth vertical wall which is perpendicular to the direction of motion of $B$. The coefficient of restitution between $B$ and the wall is $\frac { 2 } { 5 }$.
\end{enumerate}\item Find the speed of B immediately after hitting the wall.\\
(2)
The first collision between A and B occurred at a distance 4a from the wall. The balls collide again $T$ seconds after the first collision.
\item Show that $T = \frac { 112 a } { 15 u }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q8 [15]}}