| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Multiple wall bounces or returns |
| Difficulty | Standard +0.3 This is a straightforward mechanics problem requiring systematic application of coefficient of restitution and constant velocity motion. Students must track speeds after each bounce (multiplying by 2/3) and calculate times for each segment, but the structure is routine and requires no geometric insight or novel problem-solving—just careful bookkeeping across multiple bounces. |
| Spec | 6.03j Perfectly elastic/inelastic: collisions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Speed after first collision \(= \frac{2}{3}u\) | B1 | Seen or implied (possibly on diagram) |
| Speed after second collision \(= \frac{4}{9}u\) | B1 | Seen or implied (possibly on diagram) |
| Correct method for total time | M1 | Correct formula, dimensionally correct and including all 3 elements |
| \(T_1 = \frac{d}{u} + \frac{3d}{\frac{2}{3}u} + \frac{2d}{\frac{4}{9}u} \left(= \frac{d}{u} + \frac{9d}{2u} + \frac{18d}{4u}\right)\) | A1 | Correct unsimplified expression for \(T_1\) |
| \(T_1 = \frac{10d}{u}\) | A1 | Correct single term. Allow unsimplified fraction e.g. \(T_1 = \frac{40d}{4u}\) |
| Total [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T_2 = \frac{10d}{\frac{4}{9}u} = \frac{45d}{2u} \quad \left(T_2 = \frac{9}{4}T_1\right)\) | B1ft | Follow through on their \(T_1\) and/or their \(\frac{4}{9}u\). Any equivalent form e.g. \(\frac{90d}{4u}\) |
| Total [1] |
## Question 2:
**Part 2a:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Speed after first collision $= \frac{2}{3}u$ | B1 | Seen or implied (possibly on diagram) |
| Speed after second collision $= \frac{4}{9}u$ | B1 | Seen or implied (possibly on diagram) |
| Correct method for total time | M1 | Correct formula, dimensionally correct and including all 3 elements |
| $T_1 = \frac{d}{u} + \frac{3d}{\frac{2}{3}u} + \frac{2d}{\frac{4}{9}u} \left(= \frac{d}{u} + \frac{9d}{2u} + \frac{18d}{4u}\right)$ | A1 | Correct unsimplified expression for $T_1$ |
| $T_1 = \frac{10d}{u}$ | A1 | Correct single term. Allow unsimplified fraction e.g. $T_1 = \frac{40d}{4u}$ |
| **Total [5]** | | |
**Part 2b:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_2 = \frac{10d}{\frac{4}{9}u} = \frac{45d}{2u} \quad \left(T_2 = \frac{9}{4}T_1\right)$ | B1ft | Follow through on their $T_1$ and/or their $\frac{4}{9}u$. Any equivalent form e.g. $\frac{90d}{4u}$ |
| **Total [1]** | | |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7eedd755-0dfd-4506-b7fd-23b9def4ebc8-04_508_780_258_584}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The point $A$ lies on a smooth horizontal floor between two fixed smooth parallel vertical walls $W X$ and $Y Z$, as shown in the plan view in Figure 1.\\
The distance between $W X$ and $Y Z$ is $3 d$.\\
The distance of $A$ from $Y Z$ is $d$.\\
A particle is projected from $A$ along the floor with speed $u$ towards $Y Z$ in a direction perpendicular to $Y Z$.
The coefficient of restitution between the particle and each wall is $\frac { 2 } { 3 }$\\
The time taken for the particle to move from $A$, bounce off each wall once and return to A for the first time is $T _ { 1 }$
\begin{enumerate}[label=(\alph*)]
\item Find $T _ { 1 }$ in terms of $d$ and $u$.
The ball returns to $A$ for the first time after bouncing off each wall once. The further time taken for the particle to move from $A$, bounce off each wall once and return to $A$ for the second time is $T _ { 2 }$
\item Find $T _ { 2 }$ in terms of $d$ and $u$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2022 Q2 [6]}}