| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Particle brought to rest by collision |
| Difficulty | Standard +0.3 This is a standard M2 momentum-collision question requiring systematic application of conservation of momentum and Newton's restitution law across two collisions. Part (a) is routine algebraic manipulation, part (b) requires interpreting a physical condition (reversed motion means v < 0), and part (c) involves straightforward kinetic energy calculation. While multi-part with several steps, it follows predictable M2 patterns without requiring novel insight or particularly complex algebra. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
6. Three particles $P , Q$ and $R$ have masses $3 m , k m$ and 7.5m respectively. The three particles lie at rest in a straight line on a smooth horizontal table with $Q$ between $P$ and $R$. Particle $P$ is projected towards $Q$ with speed $u$ and collides directly with $Q$. The coefficient of restitution between $P$ and $Q$ is $\frac { 1 } { 9 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the speed of $Q$ immediately after the collision is $\frac { 10 u } { 3 ( 3 + k ) }$.
\item Find the range of values of $k$ for which the direction of motion of $P$ is reversed as a result of the collision.
Following the collision between $P$ and $Q$ there is a collision between $Q$ and $R$. Given that $k = 7$ and that $Q$ is brought to rest by the collision with $R$,
\item find the total kinetic energy lost in the collision between $Q$ and $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2014 Q6 [14]}}