| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Collision followed by wall impact |
| Difficulty | Standard +0.3 This is a standard M2 momentum/collision question requiring systematic application of Newton's experimental law (e = 1/3) and conservation of momentum. Part (a) uses KE loss and restitution to find mass (routine algebra). Part (b) requires checking velocities after collision to verify P moves back toward the wall—straightforward logic but multi-step. Slightly easier than average due to clear structure and standard techniques. |
| Spec | 6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| KE lost: \(\frac{1}{2} \times m \times 36 - \frac{1}{2} \times m \times v^2 = 64\) | M1A1 | |
| Restitution: \(v = \frac{1}{3} \times 6 = 2\) | M1A1 | |
| Substitute and solve for \(m\): \(\frac{1}{2} \times m \times 36 - \frac{1}{2} \times m \times 4 = 64 = 16m\) | DM1 | |
| \(m = 4\), answer given | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Conservation of momentum: \(6 - 8 = 4w - 2v\) | M1A1ft | their "2" |
| Restitution: \(v + w = \frac{1}{3}(2 + 3)\) | M1A1ft | their "2" |
| \(v = \frac{5}{3} - w\) | ||
| Solve for \(w\): \(-2 = 4w - 2\left(\frac{5}{3} - w\right) = 6w - \frac{10}{3}\) | DM1 | |
| \(\frac{4}{3} = 6w\) | A1 | |
| \(w = \frac{4}{18} = \frac{2}{9}\) m s\(^{-1}\) | ||
| \(w > 0 \Rightarrow\) will collide with the wall again | A1 |
# Question 8:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| KE lost: $\frac{1}{2} \times m \times 36 - \frac{1}{2} \times m \times v^2 = 64$ | M1A1 | |
| Restitution: $v = \frac{1}{3} \times 6 = 2$ | M1A1 | |
| Substitute and solve for $m$: $\frac{1}{2} \times m \times 36 - \frac{1}{2} \times m \times 4 = 64 = 16m$ | DM1 | |
| $m = 4$, answer given | A1 | |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Conservation of momentum: $6 - 8 = 4w - 2v$ | M1A1ft | their "2" |
| Restitution: $v + w = \frac{1}{3}(2 + 3)$ | M1A1ft | their "2" |
| $v = \frac{5}{3} - w$ | | |
| Solve for $w$: $-2 = 4w - 2\left(\frac{5}{3} - w\right) = 6w - \frac{10}{3}$ | DM1 | |
| $\frac{4}{3} = 6w$ | A1 | |
| $w = \frac{4}{18} = \frac{2}{9}$ m s$^{-1}$ | | |
| $w > 0 \Rightarrow$ will collide with the wall again | A1 | |\begin{enumerate}
\item A particle $P$ of mass $m \mathrm {~kg}$ is moving with speed $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a straight line on a smooth horizontal floor. The particle strikes a fixed smooth vertical wall at right angles and rebounds. The kinetic energy lost in the impact is 64 J . The coefficient of restitution between $P$ and the wall is $\frac { 1 } { 3 }$.\\
(a) Show that $m = 4$.\\
(6)
\end{enumerate}
After rebounding from the wall, $P$ collides directly with a particle $Q$ which is moving towards $P$ with speed $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The mass of $Q$ is 2 kg and the coefficient of restitution between $P$ and $Q$ is $\frac { 1 } { 3 }$.\\
(b) Show that there will be a second collision between $P$ and the wall.
\hfill \mbox{\textit{Edexcel M2 2011 Q8 [13]}}