| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Three-particle sequential collisions |
| Difficulty | Standard +0.3 This is a standard M2 collision question requiring systematic application of conservation of momentum and Newton's restitution law across two collisions. While it involves multiple steps and algebraic manipulation, the techniques are routine and well-practiced at this level, with no novel problem-solving insight required. The 'show that' part in (b) provides the answer to verify against, reducing difficulty further. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03b Conservation of momentum: 1D two particles6.03k Newton's experimental law: direct impact |
| Answer | Marks |
|---|---|
| Momentum: \(u = u' + v\) | M1 A1 |
| NEL: \(v - u' = eu\) | M1 A1 |
| \(2v = u(1 + \frac{2}{3})\), \(v = \frac{1}{2}u \times \frac{5}{3} = \frac{5u}{6}\) | DM1 A1 |
| \(u' = u - v = \frac{u}{6}\) | A1 |
| Answer | Marks |
|---|---|
| KE lost \(= \frac{1}{2}mu^2 - \left(\frac{1}{2}m \times \frac{25}{36}u^2 + \frac{1}{2}m \times \frac{1}{36}u^2\right)\) their speeds | M1 |
| \(= \frac{1}{2}mu^2 - \left(\frac{1}{2}m \times \frac{26}{36}u^2\right)\) | A2 - lee |
| \(= \frac{1}{2}mu^2 \times \frac{10}{35} = \frac{5}{36}mu^2\) AG | A1 |
| Answer | Marks |
|---|---|
| Speed of C \(= \frac{1}{2}\left(\frac{u}{3}\right)\left(\frac{5}{3}\right) = \frac{1}{2}\frac{5u}{6}\frac{5}{3} = \frac{25}{36}u\) | M1 A1 DM1 A1 |
## (a)
| Momentum: $u = u' + v$ | M1 A1 |
| NEL: $v - u' = eu$ | M1 A1 |
| $2v = u(1 + \frac{2}{3})$, $v = \frac{1}{2}u \times \frac{5}{3} = \frac{5u}{6}$ | DM1 A1 |
| $u' = u - v = \frac{u}{6}$ | A1 |
**Total: 7 marks**
## (b)
| KE lost $= \frac{1}{2}mu^2 - \left(\frac{1}{2}m \times \frac{25}{36}u^2 + \frac{1}{2}m \times \frac{1}{36}u^2\right)$ their speeds | M1 |
| $= \frac{1}{2}mu^2 - \left(\frac{1}{2}m \times \frac{26}{36}u^2\right)$ | A2 - lee |
| $= \frac{1}{2}mu^2 \times \frac{10}{35} = \frac{5}{36}mu^2$ AG | A1 |
**Total: 4 marks**
## (c)
| Speed of C $= \frac{1}{2}\left(\frac{u}{3}\right)\left(\frac{5}{3}\right) = \frac{1}{2}\frac{5u}{6}\frac{5}{3} = \frac{25}{36}u$ | M1 A1 DM1 A1 |
**Total: 4 marks**
**Question 6 Total: 15 marks**
---Three identical particles, $A$, $B$ and $C$, lie at rest in a straight line on a smooth horizontal table with $B$ between $A$ and $C$. The mass of each particle is $m$. Particle $A$ is projected towards $B$ with speed $u$ and collides directly with $B$. The coefficient of restitution between each pair of particles is $\frac{2}{3}$.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $u$,
\begin{enumerate}[label=(\roman*)]
\item the speed of $A$ after this collision,
\item the speed of $B$ after this collision.
\end{enumerate}
[7]
\item Show that the kinetic energy lost in this collision is $\frac{5}{36}mu^2$
[4]
\end{enumerate}
After the collision between $A$ and $B$, particle $B$ collides directly with $C$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find, in terms of $u$, the speed of $C$ immediately after this collision between $B$ and $C$.
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2012 Q6 [15]}}