| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Direct collision, find mass |
| Difficulty | Moderate -0.3 This is a straightforward application of conservation of momentum to a collision problem. Part (a) requires setting up and solving a single linear equation using momentum conservation, while part (b) applies the impulse-momentum theorem directly. The problem involves no conceptual subtleties—the collision conditions are clearly stated, and the solution follows standard M1 procedures with algebraic manipulation of given quantities. |
| Spec | 6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03e Impulse: by a force6.03f Impulse-momentum: relation |
Particle $P$ has mass $m$ kg and particle $Q$ has mass $3m$ kg. The particles are moving in opposite directions along a smooth horizontal plane when they collide directly. Immediately before the collision $P$ has speed $4u$ m s$^{-1}$ and $Q$ has speed $ku$ m s$^{-1}$, where $k$ is a constant. As a result of the collision the direction of motion of each particle is reversed and the speed of each particle is halved.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$.
[4]
\item Find, in terms of $m$ and $u$, the magnitude of the impulse exerted on $P$ by $Q$.
[3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q2 [7]}}