| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 14 |
| 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. While it involves multiple steps and algebraic manipulation, the techniques are routine for this module. Part (a) is a 'show that' requiring straightforward application of two equations, part (b) requires physical reasoning about direction reversal, and part (c) is a direct energy calculation. The problem is slightly easier than average because the method is completely standard with no novel insight required. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(6mu - 3mu = 3my - 2mx\) | M1 | CLM — needs all terms. Condone sign errors |
| \(3y - 2x = 3u\) | A1 | Correct equation |
| \(4ue = x + y\) | M1 | Impact law. Must be used the right way round |
| A1 | Correct equation. Signs with \(x\), \(y\) must be consistent with CLM equation | |
| \(y = \dfrac{u}{5}(8e+3)\) ** | DM1 | Dependent on the two preceding M marks |
| A1 (6) | Obtain the given result correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 4ue - y\) | M1 | Use one of their equations from (a) and \(\pm\) the given \(y\) to find \(x\) |
| \(x = \dfrac{1}{5}u(20e - 8e - 3) = \dfrac{3}{5}u(4e-1)\) | A1 | Any equivalent form. Accept \(\pm\) |
| \(x > 0 \Rightarrow e > \dfrac{1}{4}\) | M1 | Use \(x > 0\) to find values of \(e\). Inequality must match their \(x\) |
| \(\therefore \dfrac{1}{4} < e \leqslant 1\) | A1 (4) | Need both limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(e = \frac{1}{2}\), \(\quad x = \dfrac{3u}{5}\), \(\quad y = \dfrac{u}{5}(4+3) = \dfrac{7u}{5}\) | B1 | Allow \(x = -\dfrac{3u}{5}\) |
| \(T = \dfrac{1}{2} \times 2m \times 9u^2 + \dfrac{1}{2} \times 3m \times u^2 \left(= \dfrac{21mu^2}{2}\right)\) | M1 | KE before or KE after, in terms of \(u\) |
| \(kT = \dfrac{1}{2} \times 2m \times \dfrac{9u^2}{25} + \dfrac{1}{2} \times 3m \times \dfrac{49u^2}{25} \left(= \dfrac{165mu^2}{50}\right)\) | M1 | Second KE in terms of \(u\) and use to find \(k\) |
| \(k = \dfrac{\frac{1}{2}\times 2\times\frac{9}{25}+\frac{1}{2}\times 3\times\frac{49}{25}}{\frac{1}{2}\times 2\times 9+\frac{1}{2}\times 3} = \dfrac{165}{50} \div \dfrac{21}{2} = \dfrac{11}{35}\) | A1 (4) | Or equivalent. 0.314 or better |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $6mu - 3mu = 3my - 2mx$ | M1 | CLM — needs all terms. Condone sign errors |
| $3y - 2x = 3u$ | A1 | Correct equation |
| $4ue = x + y$ | M1 | Impact law. Must be used the right way round |
| | A1 | Correct equation. Signs with $x$, $y$ must be consistent with CLM equation |
| $y = \dfrac{u}{5}(8e+3)$ ** | DM1 | Dependent on the two preceding M marks |
| | A1 (6) | Obtain the **given result** correctly |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 4ue - y$ | M1 | Use one of their equations from (a) and $\pm$ the given $y$ to find $x$ |
| $x = \dfrac{1}{5}u(20e - 8e - 3) = \dfrac{3}{5}u(4e-1)$ | A1 | Any equivalent form. Accept $\pm$ |
| $x > 0 \Rightarrow e > \dfrac{1}{4}$ | M1 | Use $x > 0$ to find values of $e$. Inequality must match their $x$ |
| $\therefore \dfrac{1}{4} < e \leqslant 1$ | A1 (4) | Need both limits |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $e = \frac{1}{2}$, $\quad x = \dfrac{3u}{5}$, $\quad y = \dfrac{u}{5}(4+3) = \dfrac{7u}{5}$ | B1 | Allow $x = -\dfrac{3u}{5}$ |
| $T = \dfrac{1}{2} \times 2m \times 9u^2 + \dfrac{1}{2} \times 3m \times u^2 \left(= \dfrac{21mu^2}{2}\right)$ | M1 | KE before or KE after, in terms of $u$ |
| $kT = \dfrac{1}{2} \times 2m \times \dfrac{9u^2}{25} + \dfrac{1}{2} \times 3m \times \dfrac{49u^2}{25} \left(= \dfrac{165mu^2}{50}\right)$ | M1 | Second KE in terms of $u$ and use to find $k$ |
| $k = \dfrac{\frac{1}{2}\times 2\times\frac{9}{25}+\frac{1}{2}\times 3\times\frac{49}{25}}{\frac{1}{2}\times 2\times 9+\frac{1}{2}\times 3} = \dfrac{165}{50} \div \dfrac{21}{2} = \dfrac{11}{35}$ | A1 (4) | Or equivalent. 0.314 or better |7. A particle $P$ of mass $2 m$ is moving in a straight line with speed $3 u$ on a smooth horizontal table. A second particle $Q$ of mass $3 m$ is moving in the opposite direction to $P$ along the same straight line with speed $u$. The particle $P$ collides directly with $Q$. The direction of motion of $P$ is reversed by the collision. The coefficient of restitution between $P$ and $Q$ is $e$.
\begin{enumerate}[label=(\alph*)]
\item Show that the speed of $Q$ immediately after the collision is $\frac { u } { 5 } ( 8 e + 3 )$
\item Find the range of possible values of $e$.
The total kinetic energy of the particles before the collision is $T$. The total kinetic energy of the particles after the collision is $k T$. Given that $e = \frac { 1 } { 2 }$
\item find the value of $k$.\\
\includegraphics[max width=\textwidth, alt={}, center]{82cadc37-4cb0-455e-9531-e09ec0c19533-14_104_61_2407_1836}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2014 Q7 [14]}}