| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, vector velocity form |
| Difficulty | Standard +0.3 This is a standard Further Mechanics 1 oblique collision question requiring conservation of momentum along the line of centres, Newton's law of restitution, and recognition that perpendicular components are unchanged. While it involves multiple steps and vector calculations, it follows a well-established procedure taught explicitly in FM1 with no novel insight required—making it slightly easier than average for A-level overall but typical for FM1. |
| Spec | 1.10d Vector operations: addition and scalar multiplication6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Perpendicular to line of centres: \(2\mathbf{j}\) | B1 | Use the model to find the component perpendicular to the line of centres. Correct value seen or implied |
| CLM parallel to the line of centres | M1 | Use of CLM parallel to line of centres. Need all terms and dimensionally correct. Condone sign errors |
| \(0.2 \times 3 - 0.4 \times 4 = 0.4w - 0.2v \quad (-5 = 2w - v)\) | A1 | Correct unsimplified equation |
| Impact law parallel to the line of centres | M1 | Correct use of impact law parallel to the line of centres. Condone sign errors |
| \(7e = v + w \Rightarrow 3 = v + w\) | A1 | Correct equation with \(\frac{3}{7}\) used |
| Complete strategy to find \(\mathbf{v}_A\) | M1 | Complete strategy to find components parallel and perpendicular to line of centres, eg by using CLM and impact law |
| \(\mathbf{v}_A = -\frac{11}{3}\mathbf{i} + 2\mathbf{j}\) (ms\(^{-1}\)) follow their \(2\mathbf{j}\) | A1ft | \(\mathbf{v}_A\) correct, follow their \(a\mathbf{j}\) for \(2\mathbf{j}\) \((a \neq 0)\) |
| Total: 7 marks | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Magnitude of impulse on \(A\): \(0.2\left(\frac{11}{3} - (-3)\right)\) | M1 | Evidence of use of \(m(v-u)\) parallel to the line of centres |
| \(= 0.2\left(\frac{11}{3} + 3\right) = \frac{4}{3}\) (Ns) | A1 | 1.3 (Ns) or better |
| Total: 2 marks | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of scalar product to find the angle | M1 | Complete method for finding the required angle. Allow for \(\tan^{-1}\frac{3}{2}\) or \(\tan^{-1}\frac{2}{3}\) and \(\tan^{-1}\frac{6}{11}\) or \(\tan^{-1}\frac{11}{6}\) |
| \(\cos\theta = \dfrac{(3\mathbf{i}+2\mathbf{j})\cdot\left(-\frac{11}{3}\mathbf{i}+2\mathbf{j}\right)}{\sqrt{13}\times\sqrt{\frac{157}{9}}}\) | A1ft | A correct unsimplified expression. Follow their \(\mathbf{V}_A\). Do not ISW |
| \(\theta = 118°\) | A1 | Correct answer only. (Q asks for the nearest degree) Do not ISW |
| Alternative: \(180° - \tan^{-1}\frac{2}{3} - \tan^{-1}\frac{6}{11}\) or \(\tan^{-1}\frac{3}{2} + \tan^{-1}\frac{11}{6}\) | \(62°\) probably scores M1A0A0 |
## Question 6(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Perpendicular to line of centres: $2\mathbf{j}$ | B1 | Use the model to find the component perpendicular to the line of centres. Correct value seen or implied |
| CLM parallel to the line of centres | M1 | Use of CLM parallel to line of centres. Need all terms and dimensionally correct. Condone sign errors |
| $0.2 \times 3 - 0.4 \times 4 = 0.4w - 0.2v \quad (-5 = 2w - v)$ | A1 | Correct unsimplified equation |
| Impact law parallel to the line of centres | M1 | Correct use of impact law parallel to the line of centres. Condone sign errors |
| $7e = v + w \Rightarrow 3 = v + w$ | A1 | Correct equation with $\frac{3}{7}$ used |
| Complete strategy to find $\mathbf{v}_A$ | M1 | Complete strategy to find components parallel and perpendicular to line of centres, eg by using CLM and impact law |
| $\mathbf{v}_A = -\frac{11}{3}\mathbf{i} + 2\mathbf{j}$ (ms$^{-1}$) follow their $2\mathbf{j}$ | A1ft | $\mathbf{v}_A$ correct, follow their $a\mathbf{j}$ for $2\mathbf{j}$ $(a \neq 0)$ |
| **Total: 7 marks** | **(7)** | |
---
## Question 6(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Magnitude of impulse on $A$: $0.2\left(\frac{11}{3} - (-3)\right)$ | M1 | Evidence of use of $m(v-u)$ parallel to the line of centres |
| $= 0.2\left(\frac{11}{3} + 3\right) = \frac{4}{3}$ (Ns) | A1 | 1.3 (Ns) or better |
| **Total: 2 marks** | **(2)** | |
## Question 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of scalar product to find the angle | M1 | Complete method for finding the required angle. Allow for $\tan^{-1}\frac{3}{2}$ or $\tan^{-1}\frac{2}{3}$ and $\tan^{-1}\frac{6}{11}$ or $\tan^{-1}\frac{11}{6}$ |
| $\cos\theta = \dfrac{(3\mathbf{i}+2\mathbf{j})\cdot\left(-\frac{11}{3}\mathbf{i}+2\mathbf{j}\right)}{\sqrt{13}\times\sqrt{\frac{157}{9}}}$ | A1ft | A correct unsimplified expression. Follow their $\mathbf{V}_A$. Do not ISW |
| $\theta = 118°$ | A1 | Correct answer only. (Q asks for the nearest degree) Do not ISW |
| Alternative: $180° - \tan^{-1}\frac{2}{3} - \tan^{-1}\frac{6}{11}$ or $\tan^{-1}\frac{3}{2} + \tan^{-1}\frac{11}{6}$ | | $62°$ probably scores M1A0A0 |
---\begin{enumerate}
\item \hspace{0pt} [In this question $\mathbf { i }$ and $\mathbf { j }$ are perpendicular unit vectors in a horizontal plane.]
\end{enumerate}
A smooth uniform sphere $A$ has mass 0.2 kg and another smooth uniform sphere $B$, with the same radius as $A$, has mass 0.4 kg .
The spheres are moving on a smooth horizontal surface when they collide obliquely. Immediately before the collision, the velocity of $A$ is $( 3 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ and the velocity of $B$ is $( - 4 \mathbf { i } - \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$
At the instant of collision, the line joining the centres of the spheres is parallel to $\mathbf { i }$\\
The coefficient of restitution between the spheres is $\frac { 3 } { 7 }$\\
(a) Find the velocity of $A$ immediately after the collision.\\
(b) Find the magnitude of the impulse received by $A$ in the collision.\\
(c) Find, to the nearest degree, the size of the angle through which the direction of motion of $A$ is deflected as a result of the collision.
\hfill \mbox{\textit{Edexcel FM1 2019 Q6 [12]}}