| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Collision with vector velocities |
| Difficulty | Moderate -0.3 This is a standard FM1 collision problem requiring routine application of conservation of momentum and impulse-momentum theorem with vector notation. Parts (a) and (b) are direct calculations, while (c) requires straightforward use of momentum conservation. The vector component work is mechanical rather than conceptually challenging, making this slightly easier than an average A-level question overall. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Form an expression for total KE | M1 | Must add 2 KE terms of correct structure. Allow missing minus signs on components when squaring. M0 for use of \((3\mathbf{i}-\mathbf{j})^2\) unless recovered. M0 if KE terms never added. M0 for using \(\frac{1}{2}mv\) unless correct KE formula also stated. |
| \(\frac{1}{2}\times3\times(3^2+(-1)^2)+\frac{1}{2}\times2\times((-6)^2+2^2)\) | A1 | Correct unsimplified expression for total KE. If the 2 KE terms are incorrect when total is formed, this is A0. |
| \(= 55\) (J) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of change in momentum | M1 | N.B. Only penalise use of column vectors in final answers once across (b) and (c). Penalise first occurrence. Allow column vectors throughout working in (b) and (c). |
| \(3\left((-2\mathbf{i}+\frac{2}{3}\mathbf{j})-(3\mathbf{i}-\mathbf{j})\right)\) | A1 | |
| \(=(-15\mathbf{i}+5\mathbf{j})\) (N s) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Either: Use impulse on \(B\) to form equation | M1 | |
| \((15\mathbf{i}-5\mathbf{j})=2\left(\mathbf{v}_B-(-6\mathbf{i}+2\mathbf{j})\right)\) | A1 | |
| \(\mathbf{v}_B=(1.5\mathbf{i}-0.5\mathbf{j})\) (m s\(^{-1}\)) | A1 | |
| Or: Use CLM to form equation | M1 | |
| \(3(3\mathbf{i}-\mathbf{j})+2(-6\mathbf{i}+2\mathbf{j})=3(-2\mathbf{i}+\frac{2}{3}\mathbf{j})+2\mathbf{v}_B\) | A1 | |
| \(\mathbf{v}_B=(1.5\mathbf{i}-0.5\mathbf{j})\) (m s\(^{-1}\)) | A1 |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Form an expression for **total KE** | M1 | Must add 2 KE terms of correct structure. Allow missing minus signs on components when squaring. M0 for use of $(3\mathbf{i}-\mathbf{j})^2$ unless recovered. M0 if KE terms never added. M0 for using $\frac{1}{2}mv$ unless correct KE formula also stated. |
| $\frac{1}{2}\times3\times(3^2+(-1)^2)+\frac{1}{2}\times2\times((-6)^2+2^2)$ | A1 | Correct unsimplified expression for total KE. If the 2 KE terms are incorrect when total is formed, this is A0. |
| $= 55$ (J) | A1 | cao |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of change in momentum | M1 | N.B. Only penalise use of column vectors in **final answers** once across (b) and (c). Penalise first occurrence. Allow column vectors throughout working in (b) and (c). |
| $3\left((-2\mathbf{i}+\frac{2}{3}\mathbf{j})-(3\mathbf{i}-\mathbf{j})\right)$ | A1 | |
| $=(-15\mathbf{i}+5\mathbf{j})$ (N s) | A1 | |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| **Either:** Use impulse on $B$ to form equation | M1 | |
| $(15\mathbf{i}-5\mathbf{j})=2\left(\mathbf{v}_B-(-6\mathbf{i}+2\mathbf{j})\right)$ | A1 | |
| $\mathbf{v}_B=(1.5\mathbf{i}-0.5\mathbf{j})$ (m s$^{-1}$) | A1 | |
| **Or:** Use CLM to form equation | M1 | |
| $3(3\mathbf{i}-\mathbf{j})+2(-6\mathbf{i}+2\mathbf{j})=3(-2\mathbf{i}+\frac{2}{3}\mathbf{j})+2\mathbf{v}_B$ | A1 | |
| $\mathbf{v}_B=(1.5\mathbf{i}-0.5\mathbf{j})$ (m s$^{-1}$) | A1 | |
**Total: 9 marks**\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 3 kg and a particle $B$ has mass 2 kg .\\
The particles are moving on a smooth horizontal plane when they collide directly.\\
Immediately before the collision, the velocity of $A$ is $( 3 \mathbf { i } - \mathbf { j } ) \mathrm { ms } ^ { - 1 }$ and the velocity of $B$ is $( - 6 \mathbf { i } + 2 \mathbf { j } ) \mathrm { ms } ^ { - 1 }$
Immediately after the collision the velocity of $A$ is $\left( - 2 \mathbf { i } + \frac { 2 } { 3 } \mathbf { j } \right) \mathrm { ms } ^ { - 1 }$\\
(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 exerted on $A$ by $B$ in the collision.\\
(c) Find, in terms of $\mathbf { i }$ and $\mathbf { j }$, the velocity of $B$ immediately after the collision.
\hfill \mbox{\textit{Edexcel FM1 2024 Q1 [9]}}