| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with direction reversal |
| Difficulty | Moderate -0.3 This is a standard M2 collision problem requiring application of conservation of momentum and Newton's law of restitution with identical masses. The setup is straightforward with given values, requiring two simultaneous equations to solve. Slightly easier than average due to identical particles simplifying the algebra and being a routine textbook-style question. |
| Spec | 6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Conservation of momentum: \(m(5) - m(3) = mv_1 + mv_2\) | M1 | |
| \(v_1 + v_2 = 2\) | A1 | |
| \(\frac{v_2-v_1}{5-(-3)} = \frac{1}{2}\) \(\therefore v_2 - v_1 = 4\) | M1 A1 | |
| Solve simultaneously, giving \(v_1 = -1 \text{ ms}^{-1}\) so speed is \(1 \text{ ms}^{-1}\), \(v_2 = 3 \text{ ms}^{-1}\) | M1 A1 | (6 marks) |
Conservation of momentum: $m(5) - m(3) = mv_1 + mv_2$ | M1 |
$v_1 + v_2 = 2$ | A1 |
$\frac{v_2-v_1}{5-(-3)} = \frac{1}{2}$ $\therefore v_2 - v_1 = 4$ | M1 A1 |
Solve simultaneously, giving $v_1 = -1 \text{ ms}^{-1}$ so speed is $1 \text{ ms}^{-1}$, $v_2 = 3 \text{ ms}^{-1}$ | M1 A1 | (6 marks)\begin{enumerate}
\item Two identical particles are approaching each other along a straight horizontal track. Just before they collide, they are moving with speeds $5 \mathrm {~ms} ^ { - 1 }$ and $3 \mathrm {~ms} ^ { - 1 }$ respectively. The coefficient of restitution between the particles is $\frac { 1 } { 2 }$.
\end{enumerate}
Find the speeds of the particles immediately after the impact.\\
\hfill \mbox{\textit{Edexcel M2 Q1 [6]}}