| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | January |
| Marks | 13 |
| 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 routine application of conservation of momentum and the restitution formula. All steps are algorithmic: apply momentum conservation to find A's final velocity, use the restitution definition, calculate KE loss by subtraction, and apply impulse-momentum for the wall collision. No novel insight or problem-solving is required—just systematic application of memorized formulas. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks |
|---|---|
| LM: \(6mu - 2mu = 2mx + \frac{8}{3}mu\) | M1 A1 |
| \(x = \frac{2}{3}u\) | M1 A1 |
| NEL: \(\frac{8}{3}u - x = 5ue\) | M1 A1 |
| Solving to \(e = \frac{2}{5}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Initial K.E. \(= \frac{1}{2} \times 2m(3u)^2 + \frac{1}{2} \times m(2u)^2 = 11mu^2\) | M1 | |
| Final K.E. \(= \frac{1}{2} \times 2m\left(\frac{2}{3}u\right)^2 + \frac{1}{2} \times m\left(\frac{8}{3}u\right)^2 = 4mu^2\) | both M1 | |
| Change in K.E. \(= 7mu^2\) | M1 Subtracting and simplifying to \(kmu^2\) M1 A1 Also | (3) |
| Answer | Marks |
|---|---|
| \(m\left(\frac{8}{3} + v\right) = \frac{14}{3}mu\) | M1 A1 |
| \(v = 2u\) | M1 A1 |
| \(e = \frac{2}{\frac{8}{3}} = \frac{3}{4}\) | M1 A1 |
## (a)
LM: $6mu - 2mu = 2mx + \frac{8}{3}mu$ | M1 A1 |
$x = \frac{2}{3}u$ | M1 A1 |
NEL: $\frac{8}{3}u - x = 5ue$ | M1 A1 |
Solving to $e = \frac{2}{5}$ | M1 A1 |
## (b)
Initial K.E. $= \frac{1}{2} \times 2m(3u)^2 + \frac{1}{2} \times m(2u)^2 = 11mu^2$ | M1 |
Final K.E. $= \frac{1}{2} \times 2m\left(\frac{2}{3}u\right)^2 + \frac{1}{2} \times m\left(\frac{8}{3}u\right)^2 = 4mu^2$ | both M1 |
Change in K.E. $= 7mu^2$ | M1 Subtracting and simplifying to $kmu^2$ M1 A1 Also | (3)
## (c)
$m\left(\frac{8}{3} + v\right) = \frac{14}{3}mu$ | M1 A1 |
$v = 2u$ | M1 A1 |
$e = \frac{2}{\frac{8}{3}} = \frac{3}{4}$ | M1 A1 |
---A particle $A$ of mass $2m$ is moving with speed $3u$ in a straight line on a smooth horizontal table. The particle collides directly with a particle $B$ of mass $m$ moving with speed $2u$ in the opposite direction to $A$. Immediately after the collision the speed of $B$ is $\frac{8}{3}u$ and the direction of motion of $B$ is reversed.
\begin{enumerate}[label=(\alph*)]
\item Calculate the coefficient of restitution between $A$ and $B$. [6]
\item Show that the kinetic energy lost in the collision is $7mu^2$. [3]
\end{enumerate}
After the collision $B$ strikes a fixed vertical wall that is perpendicular to the direction of motion of $B$. The magnitude of the impulse of the wall on $B$ is $\frac{14}{3}mu$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Calculate the coefficient of restitution between $B$ and the wall. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2006 Q4 [13]}}