| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, vector velocity form |
| Difficulty | Moderate -0.8 This is a straightforward M2 collision question requiring standard application of momentum conservation and impulse formulas with vector notation. Part (a) is direct KE calculation, part (b) is impulse = change in momentum, and part (c) applies Newton's third law. All steps are routine textbook exercises with no problem-solving insight required, making it easier than average A-level questions. |
| Spec | 6.03a Linear momentum: p = mv6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Expression for total KE before collision | M1 | Dimensionally correct. Condone confusion between before and after for \(A\). Allow if vectors seen in the working but the modulus is used correctly. The two parts must be added together. Allow confusion of 2 kg and 3 kg |
| \(\frac{1}{2} \times 2 \times 5^2 + \frac{1}{2} \times 3 \times (3^2 + (-1)^2)\) | A1 | Correct unsimplified expression |
| \(= 40\) (J) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2((3\mathbf{i} + 2\mathbf{j}) - 5\mathbf{j})\) | M1 | Change in momentum of \(A\), must be a difference but allow subtraction in either order. Must be using the correct mass, 2 kg |
| \(= (6\mathbf{i} - 6\mathbf{j})\) (N s) | A1 | cao. The final answer should be in terms of \(\mathbf{i}\) and \(\mathbf{j}\). Accept \(2(3\mathbf{i} - 3\mathbf{j})\). ISW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Impulse-momentum equation for \(B\) | M1 | Must use negative of their answer to (b) and the initial velocity of \(B\). Must be using the correct mass, 3 kg. Or CLM with correct terms (allow slip) and plus signs |
| \(3(\mathbf{v}_B - (3\mathbf{i} - \mathbf{j})) = (-6\mathbf{i} + 6\mathbf{j})\) or \(2 \times 5\mathbf{j} + 3(3\mathbf{i} - \mathbf{j}) = 2(3\mathbf{i} + 2\mathbf{j}) + 3\mathbf{v}_B\) | A1ft | Correct unsimplified equation. ft on their impulse from (b) |
| \(\mathbf{v}_B = (\mathbf{i} + \mathbf{j})\) (m s\(^{-1}\)) | A1 | cao. Accept column vector. ISW |
# Question 1:
## Part 1(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Expression for total KE before collision | M1 | Dimensionally correct. Condone confusion between before and after for $A$. Allow if vectors seen in the working but the modulus is used correctly. The two parts must be added together. Allow confusion of 2 kg and 3 kg |
| $\frac{1}{2} \times 2 \times 5^2 + \frac{1}{2} \times 3 \times (3^2 + (-1)^2)$ | A1 | Correct unsimplified expression |
| $= 40$ (J) | A1 | cao |
## Part 1(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2((3\mathbf{i} + 2\mathbf{j}) - 5\mathbf{j})$ | M1 | Change in momentum of $A$, must be a difference but allow subtraction in either order. Must be using the correct mass, 2 kg |
| $= (6\mathbf{i} - 6\mathbf{j})$ (N s) | A1 | cao. The final answer should be in terms of $\mathbf{i}$ and $\mathbf{j}$. Accept $2(3\mathbf{i} - 3\mathbf{j})$. ISW |
## Part 1(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Impulse-momentum equation for $B$ | M1 | Must use negative of their answer to (b) and the initial velocity of $B$. Must be using the correct mass, 3 kg. Or CLM with correct terms (allow slip) and plus signs |
| $3(\mathbf{v}_B - (3\mathbf{i} - \mathbf{j})) = (-6\mathbf{i} + 6\mathbf{j})$ or $2 \times 5\mathbf{j} + 3(3\mathbf{i} - \mathbf{j}) = 2(3\mathbf{i} + 2\mathbf{j}) + 3\mathbf{v}_B$ | A1ft | Correct unsimplified equation. ft on their impulse from (b) |
| $\mathbf{v}_B = (\mathbf{i} + \mathbf{j})$ (m s$^{-1}$) | A1 | cao. Accept column vector. ISW |
---\begin{enumerate}
\item \hspace{0pt} [In this question, $\mathbf { i }$ and $\mathbf { j }$ are horizontal perpendicular unit vectors.]
\end{enumerate}
A particle $A$ has mass 2 kg and a particle $B$ has mass 3 kg . The particles are moving on a smooth horizontal plane when they collide.
Immediately before the collision, the velocity of $A$ is $5 \mathbf { j } \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the velocity of $B$ is $( 3 \mathbf { i } - \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$
Immediately after the collision, the velocity of $A$ is $( 3 \mathbf { i } + 2 \mathbf { j } ) \mathrm { ms } ^ { - 1 }$\\
(a) Find the total kinetic energy of the two particles before the collision.\\
(b) Find, in terms of $\mathbf { i }$ and $\mathbf { j }$, the impulse received by $A$ in the collision.
Given that, in the collision, the impulse of $A$ on $B$ is equal and opposite to the impulse of $B$ on $A$,\\
(c) find the velocity of $B$ immediately after the collision.
\hfill \mbox{\textit{Edexcel M2 2024 Q1 [8]}}