| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | June |
| Marks | 17 |
| 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 question involving conservation of momentum, coefficient of restitution, and energy loss. Parts (i)-(iii) are routine applications of standard formulas with straightforward algebra. Part (iv) requires a second collision calculation but follows the same methodology. The multi-step nature and bookwork add some complexity, but no novel insight is required—this is slightly easier than average for M2. |
| 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 |
|---|---|---|
| \(2mu = 2mv + 3mv\) | M1 | Conservation of momentum |
| \(v = \frac{2}{5}u\) | A1 3 | Must be \(v =\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(e = \frac{(3v - v)}{u}\) | M1 | Using restitution |
| \(e = \frac{4}{5}\) | A1 2 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Initial K.E. \(= \frac{9mu^2}{2} = 18mu^2/25\) | B1 FT | FT on their \(v\) from (i) |
| Final K.E. \(= \frac{9mv^2}{8} = 9mu^2/50\) | B1 FT | FT on their \(v\) from (i) |
| \(\frac{1}{2}m(V)^2 = \text{Final K.E.}\) | M1 | |
| \(V = 3u/5\) | A1 4 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(4mu/5 - 3mu/5 = 2mx + my\) | M1 | Conservation of momentum |
| \(u/5 = 2x + y\) | A1 FT | FT on their \(v\) from (i); aef |
| \(e = \frac{4}{5} = \frac{(y-x)}{u}\) | M1 FT | Using restitution |
| \(4u = 5y - 5x\) | A1 | FT on their \(v\) from (i); aef |
| solving 2 relevant equations | M1 | |
| \(x = -u/5\) \(y = 3u/5\) | A1 A1 | |
| away from wall (\(x\)) + towards wall (\(y\)) | A1 8 | both |
Total for Question 6: 17
## (i)
$2mu = 2mv + 3mv$ | M1 | Conservation of momentum
$v = \frac{2}{5}u$ | A1 3 | Must be $v =$
## (ii)
$e = \frac{(3v - v)}{u}$ | M1 | Using restitution
$e = \frac{4}{5}$ | A1 2 | AG
## (iii)
Initial K.E. $= \frac{9mu^2}{2} = 18mu^2/25$ | B1 FT | FT on their $v$ from (i)
Final K.E. $= \frac{9mv^2}{8} = 9mu^2/50$ | B1 FT | FT on their $v$ from (i)
$\frac{1}{2}m(V)^2 = \text{Final K.E.}$ | M1 |
$V = 3u/5$ | A1 4 | AG
## (iv)
$4mu/5 - 3mu/5 = 2mx + my$ | M1 | Conservation of momentum
$u/5 = 2x + y$ | A1 FT | FT on their $v$ from (i); aef
$e = \frac{4}{5} = \frac{(y-x)}{u}$ | M1 FT | Using restitution
$4u = 5y - 5x$ | A1 | FT on their $v$ from (i); aef
solving 2 relevant equations | M1 |
$x = -u/5$ $y = 3u/5$ | A1 A1 |
away from wall ($x$) + towards wall ($y$) | A1 8 | both
Total for Question 6: **17**
---A particle $A$ of mass $2m$ is moving with speed $u$ on a smooth horizontal surface when it collides with a stationary particle $B$ of mass $m$. After the collision the speed of $A$ is $v$, the speed of $B$ is $3v$ and the particles move in the same direction.
\begin{enumerate}[label=(\roman*)]
\item Find $v$ in terms of $u$. [3]
\item Show that the coefficient of restitution between $A$ and $B$ is $\frac{1}{3}$. [2]
\end{enumerate}
$B$ subsequently hits a vertical wall which is perpendicular to the direction of motion. As a result of the impact, $B$ loses $\frac{3}{4}$ of its kinetic energy.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Show that the speed of $B$ after hitting the wall is $\frac{3}{4}u$. [4]
\item $B$ then hits $A$. Calculate the speeds of $A$ and $B$, in terms of $u$, after this collision and state their directions of motion. [8]
\end{enumerate}
\hfill \mbox{\textit{OCR M2 2010 Q6 [17]}}