| 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 systematic application of conservation of momentum and Newton's law of restitution across two collision events. While it involves multiple steps and careful bookkeeping of velocities/directions, the techniques are routine for M2 students with no novel insight required. The 'show that' in part (a) guides students toward the key relationship, and part (b) follows a standard energy-coefficient approach. Slightly easier than average due to its structured, methodical nature. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Q collides with wall: \(v' = \frac{1}{3} \times 3u = u\) towards P | M1 | Correct use of impact law |
| A1 | ||
| P and Q collide. | M1 | Requires all 4 terms. Condone sign errors |
| Conservation of momentum: | ||
| \(m \times 2u - 2m \times u = 2m \times w - m \times v\) | ||
| \(0 = 2m \times w - m \times v, \quad 2w = v\) *AG* | A1 | |
| A1 | Watch out | |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Impact law: \(v + w = e(2u + u)\) | M1 | Correct use of impact law to find w |
| \(3w = 3eu, \quad w = eu\) | A1 | |
| Kinetic energy: | M1 | Requires all terms. Each term of correct structure |
| \(\frac{1}{2}\left[\frac{1}{2}m.4u^2 + \frac{1}{2}2m.u^2\right] = \frac{1}{2}m.v^2 + \frac{1}{2}2m.w^2\) | A2 | -1 each error |
| \(\frac{1}{2}3u^2 = \frac{1}{2}4w^2 + \frac{1}{2}2w^2 = 3w^2\) | M1 | Solve for w |
| \(w = \frac{1}{\sqrt{2}}u\) | A1 | |
| \(e = \frac{1}{\sqrt{2}}\) | A1 | 0.71 or better |
| (8) | ||
| [13] |
**(a)**
| Q collides with wall: $v' = \frac{1}{3} \times 3u = u$ towards P | M1 | Correct use of impact law |
| | A1 | |
| P and Q collide. | M1 | Requires all 4 terms. Condone sign errors |
| Conservation of momentum: | | |
| $m \times 2u - 2m \times u = 2m \times w - m \times v$ | | |
| $0 = 2m \times w - m \times v, \quad 2w = v$ *AG* | A1 | |
| | A1 | **Watch out** |
| | (5) | |
**(b)**
| Impact law: $v + w = e(2u + u)$ | M1 | Correct use of impact law to find w |
| $3w = 3eu, \quad w = eu$ | A1 | |
| Kinetic energy: | M1 | Requires all terms. Each term of correct structure |
| $\frac{1}{2}\left[\frac{1}{2}m.4u^2 + \frac{1}{2}2m.u^2\right] = \frac{1}{2}m.v^2 + \frac{1}{2}2m.w^2$ | A2 | -1 each error |
| $\frac{1}{2}3u^2 = \frac{1}{2}4w^2 + \frac{1}{2}2w^2 = 3w^2$ | M1 | Solve for w |
| $w = \frac{1}{\sqrt{2}}u$ | A1 | |
| $e = \frac{1}{\sqrt{2}}$ | A1 | 0.71 or better |
| | (8) |
| | [13] | |\includegraphics{figure_4}
Two smooth particles $P$ and $Q$ have masses $m$ and $2m$ respectively. The particles are moving in the same direction in the same straight line, on a smooth horizontal plane, with $Q$ in front of $P$. The particles are moving towards a fixed smooth vertical wall which is perpendicular to the direction of motion of the particles, as shown in Figure 4. The speed of $P$ is $2u$ and the speed of $Q$ is $3u$. The coefficient of restitution between $Q$ and the wall is $\frac{1}{3}$. Particle $Q$ strikes the wall, rebounds and then collides directly with $P$. The direction of motion of each particle is reversed by this collision. Immediately after this collision the speed of $P$ is $v$ and the speed of $Q$ is $w$.
\begin{enumerate}[label=(\alph*)]
\item Show that $v = 2w$. [5]
\end{enumerate}
The total kinetic energy of $P$ and $Q$ immediately after they collide is half the total kinetic energy of $P$ and $Q$ immediately before they collide.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the coefficient of restitution between $P$ and $Q$. [8]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2013 Q7 [13]}}