| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2019 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with direction reversal |
| Difficulty | Standard +0.3 This is a standard M2 collision problem requiring conservation of momentum and Newton's restitution law. Part (a) is a straightforward 'show that' requiring basic algebra, part (b) involves finding a range by solving an inequality, and part (c) is routine kinetic energy calculation. All techniques are standard textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle6.03b Conservation of momentum: 1D two particles6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
7. A particle $P$ of mass $3 m$ is moving in a straight line with speed $u$ on a smooth horizontal table. A second particle $Q$ of mass $2 m$ is moving with speed $2 u$ in the opposite direction to $P$ along the same straight line. Particle $P$ collides directly with $Q$. The coefficient of restitution between $P$ and $Q$ is $e$.
\begin{enumerate}[label=(\alph*)]
\item Show that the direction of motion of $P$ is reversed as a result of the collision with $Q$.
\item Find the range of values of $e$ for which the direction of motion of $Q$ is also reversed as a result of the collision.
Given that $e = \frac { 1 } { 2 }$
\item find, in terms of $m$ and $u$, the kinetic energy lost in the collision between $P$ and $Q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2019 Q7 [13]}}