| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2002 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Collision with two possible outcomes |
| Difficulty | Standard +0.3 This is a standard M1 collision problem requiring impulse-momentum theorem and conservation of momentum. Part (a) is straightforward application of impulse = change in momentum. Part (b) requires setting up a momentum equation with two cases (B's direction after collision), leading to a simple linear equation. The multi-step nature and sign considerations make it slightly above average, but it follows a well-practiced template with no novel insight required. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation |
The masses of two particles $A$ and $B$ are $0.5 \text{ kg}$ and $m \text{ kg}$ respectively. The particles are moving on a smooth horizontal table in opposite directions and collide directly. Immediately before the collision the speed of $A$ is $5 \text{ m s}^{-1}$ and the speed of $B$ is $3 \text{ m s}^{-1}$. In the collision, the magnitude of the impulse exerted by $B$ on $A$ is $3.6 \text{ Ns}$. As a result of the collision the direction of motion of $A$ is reversed.
\begin{enumerate}[label=(\alph*)]
\item Find the speed of $A$ immediately after the collision. [3]
\end{enumerate}
The speed of $B$ immediately after the collision is $1 \text{ m s}^{-1}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the two possible values of $m$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2002 Q2 [7]}}