| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with energy loss |
| Difficulty | Standard +0.3 This is a standard M2 collision problem requiring conservation of momentum and the given energy condition to find two unknowns, followed by applying the restitution formula. While it involves simultaneous equations and careful algebra with the energy condition, it follows a well-practiced procedure with no novel insight required. The 10 marks for part (a) indicate substantial working, but this is a textbook-style question slightly easier than average A-level standard. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| \(2mu = 2mv_P + mv_O\) | M1 | CLM. Needs all 3 terms of correct form but condone sign slips |
| A1 | Correct equation | |
| \(\frac{3}{2}mu^2 = \frac{1}{2}2mv_P^2 + \frac{1}{2}mv_O^2\) | M1 | KE after impact. 3 terms of correct form |
| A1 | Correct equation | |
| \(3v_O^2 - 4uv_O + u^2 = 0\) or \(12v_P^2 - 16uv_P + 5u^2 = 0\) | M1 | Use CLM equation to form quadratic in \(v_P\) or \(v_O\) |
| A1 | Correct equation | |
| \(v_O = \frac{u}{3}, v_P = \frac{5u}{6}\) or \(v_O = u, v_P = \frac{u}{2}\) | DM1 A1 | Solve for a value of \(v_O\). Dependent on the previous M1. A \(v_O, v_P\) pair correct or two correct values for \(v_O\) |
| \(v_O = u\) | DM1 | Select solution from a choice of two. Dependent on all 4 M marks. |
| \(\ldots\ldots\ldots\) since \(v_O > v_P\) | A1 | Correct justification |
| (10) |
| Answer | Marks | Guidance |
|---|---|---|
| \(e = \frac{u - \frac{u}{2}}{\frac{v_O - v_P}{u}}\) | M1 | Impact law. Must be used correctly. Condone ±e Follow their speeds from (a). |
| A1 ft | Correct for their speeds | |
| \(= \frac{1}{2}\) | A1 | |
| (3) | ||
| [13] |
**Part (a):**
| $2mu = 2mv_P + mv_O$ | M1 | CLM. Needs all 3 terms of correct form but condone sign slips |
| | A1 | Correct equation |
| $\frac{3}{2}mu^2 = \frac{1}{2}2mv_P^2 + \frac{1}{2}mv_O^2$ | M1 | KE after impact. 3 terms of correct form |
| | A1 | Correct equation |
| $3v_O^2 - 4uv_O + u^2 = 0$ or $12v_P^2 - 16uv_P + 5u^2 = 0$ | M1 | Use CLM equation to form quadratic in $v_P$ or $v_O$ |
| | A1 | Correct equation |
| $v_O = \frac{u}{3}, v_P = \frac{5u}{6}$ or $v_O = u, v_P = \frac{u}{2}$ | DM1 A1 | Solve for a value of $v_O$. Dependent on the previous M1. A $v_O, v_P$ pair correct or two correct values for $v_O$ |
| $v_O = u$ | DM1 | Select solution from a choice of two. Dependent on all 4 M marks. |
| $\ldots\ldots\ldots$ since $v_O > v_P$ | A1 | Correct justification |
| | (10) | |
**Part (b):**
| $e = \frac{u - \frac{u}{2}}{\frac{v_O - v_P}{u}}$ | M1 | Impact law. Must be used correctly. Condone ±e Follow their speeds from (a). |
| | A1 ft | Correct for their speeds |
| $= \frac{1}{2}$ | A1 | |
| | (3) |
| | [13] | |
---Two particles $P$ and $Q$, of masses $2m$ and $m$ respectively, are on a smooth horizontal table. Particle $Q$ is at rest and particle $P$ collides directly with it when moving with speed $u$. After the collision the total kinetic energy of the two particles is $\frac{3}{4}mu^2$. Find
\begin{enumerate}[label=(\alph*)]
\item the speed of $Q$ immediately after the collision, [10]
\item the coefficient of restitution between the particles. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2013 Q5 [13]}}