| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2020 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, vector velocity form |
| Difficulty | Challenging +1.2 This is a standard Further Mechanics 1 oblique collision question requiring conservation of momentum along the line of centres, perpendicular component preservation, the given energy condition, and Newton's experimental law. While it involves multiple steps and vector components, the techniques are routine for FM1 students with no novel insight required—slightly above average difficulty due to the algebraic manipulation and multi-part nature. |
| Spec | 1.10d Vector operations: addition and scalar multiplication6.02d Mechanical energy: KE and PE concepts6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03i Coefficient of restitution: e |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Components perpendicular to line of centres after collision: \(\mathbf{v}_{Pj} = 2\mathbf{j}\ (\text{ms}^{-1})\), \(\mathbf{v}_{Qj} = \mathbf{j}\ (\text{ms}^{-1})\) | B1 | Seen or implied. Correct only |
| Equation for KE of Q: \(\frac{1}{2} \times 0.2 \times (v^2 + 1) = \frac{1}{2} \times \frac{1}{2} \times 0.2 \times (9+1)\) | M1 | Equation for KE of Q. Dimensionally correct. Condone \(\frac{1}{2}\) on wrong side |
| Correct unsimplified equation in \(v^2\) | A1 | Correct unsimplified equation in \(v^2\) |
| CLM parallel to line of centres: \(0.3 \times 4 - 0.2 \times 3 = 0.2v - 0.3u \quad (6 = 2v - 3u)\) | M1 | Equation for CLM. Correct terms required. Condone sign errors. Dimensionally correct |
| Correct unsimplified equation | A1 | Correct unsimplified equation |
| Impact law parallel to line of centres: \(v + u = e(4+3)\) | M1 | Correct use of impact law. Condone sign errors |
| Correct unsimplified equation | A1 | Correct unsimplified equation |
| Complete method to solve for \(\mathbf{v}_P\), \(\mathbf{v}_Q\) or \(e\) | M1 | Working in \(e\) gives \(v = \frac{1}{5}(6+21e)\) and \(441e^2 + 252e - 64 = 0\) |
| \(\mathbf{v}_P = \frac{2}{3}\mathbf{i} + 2\mathbf{j}\ (\text{ms}^{-1})\) and \(\mathbf{v}_Q = 2\mathbf{i} + \mathbf{j}\ (\text{ms}^{-1})\) | A1 | Both velocities correct. Need answers in form \(a\mathbf{i} + b\mathbf{j}\) or equivalent |
| \(e = \frac{4}{21}\) | A1 | Correct only. 0.19 or better (0.19047…) |
| \(v = -2 \Rightarrow u = -\frac{10}{3} \Rightarrow P\) and \(Q\) have passed through each other: impossible, so solution is unique | A1* | Or equivalent justification. e.g. a negative value for \(e\) is not possible |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use trig to find angle between velocities | M1 | Use of trig or equivalent to find relevant angle between two velocities, e.g. scalar product or difference between angles |
| \(\cos\theta = \left(\dfrac{\frac{8}{3}+4}{\sqrt{20}\sqrt{4\frac{4}{9}}}\right)\) or \(\theta = \tan^{-1}\dfrac{2}{\frac{2}{3}} - \tan^{-1}\dfrac{1}{2}\) | A1ft | Correct unsimplified equation in \(\theta\). Follow their \(\mathbf{v}_P\) |
| \(\theta = 45°\ \left(\dfrac{\pi}{4}\ \text{rads}\right)\) | A1 | Correct only. (0.785… radians) Do not ISW |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Components perpendicular to line of centres after collision: $\mathbf{v}_{Pj} = 2\mathbf{j}\ (\text{ms}^{-1})$, $\mathbf{v}_{Qj} = \mathbf{j}\ (\text{ms}^{-1})$ | B1 | Seen or implied. Correct only |
| Equation for KE of Q: $\frac{1}{2} \times 0.2 \times (v^2 + 1) = \frac{1}{2} \times \frac{1}{2} \times 0.2 \times (9+1)$ | M1 | Equation for KE of Q. Dimensionally correct. Condone $\frac{1}{2}$ on wrong side |
| Correct unsimplified equation in $v^2$ | A1 | Correct unsimplified equation in $v^2$ |
| CLM parallel to line of centres: $0.3 \times 4 - 0.2 \times 3 = 0.2v - 0.3u \quad (6 = 2v - 3u)$ | M1 | Equation for CLM. Correct terms required. Condone sign errors. Dimensionally correct |
| Correct unsimplified equation | A1 | Correct unsimplified equation |
| Impact law parallel to line of centres: $v + u = e(4+3)$ | M1 | Correct use of impact law. Condone sign errors |
| Correct unsimplified equation | A1 | Correct unsimplified equation |
| Complete method to solve for $\mathbf{v}_P$, $\mathbf{v}_Q$ or $e$ | M1 | Working in $e$ gives $v = \frac{1}{5}(6+21e)$ and $441e^2 + 252e - 64 = 0$ |
| $\mathbf{v}_P = \frac{2}{3}\mathbf{i} + 2\mathbf{j}\ (\text{ms}^{-1})$ and $\mathbf{v}_Q = 2\mathbf{i} + \mathbf{j}\ (\text{ms}^{-1})$ | A1 | Both velocities correct. Need answers in form $a\mathbf{i} + b\mathbf{j}$ or equivalent |
| $e = \frac{4}{21}$ | A1 | Correct only. 0.19 or better (0.19047…) |
| $v = -2 \Rightarrow u = -\frac{10}{3} \Rightarrow P$ and $Q$ have passed through each other: impossible, so solution is unique | A1* | Or equivalent justification. e.g. a negative value for $e$ is not possible |
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## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use trig to find angle between velocities | M1 | Use of trig or equivalent to find relevant angle between two velocities, e.g. scalar product or difference between angles |
| $\cos\theta = \left(\dfrac{\frac{8}{3}+4}{\sqrt{20}\sqrt{4\frac{4}{9}}}\right)$ or $\theta = \tan^{-1}\dfrac{2}{\frac{2}{3}} - \tan^{-1}\dfrac{1}{2}$ | A1ft | Correct unsimplified equation in $\theta$. Follow their $\mathbf{v}_P$ |
| $\theta = 45°\ \left(\dfrac{\pi}{4}\ \text{rads}\right)$ | A1 | Correct only. (0.785… radians) Do not ISW |
---\begin{enumerate}
\item A smooth uniform sphere $P$ has mass 0.3 kg . Another smooth uniform sphere $Q$, with the same radius as $P$, has mass 0.2 kg .
\end{enumerate}
The spheres are moving on a smooth horizontal surface when they collide obliquely. Immediately before the collision the velocity of $P$ is $( 4 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ and the velocity of $Q$ is $( - 3 \mathbf { i } + \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.
At the instant of collision, the line joining the centres of the spheres is parallel to $\mathbf { i }$.\\
The kinetic energy of $Q$ immediately after the collision is half the kinetic energy of $Q$ immediately before the collision.\\
(a) Find\\
(i) the velocity of $P$ immediately after the collision,\\
(ii) the velocity of $Q$ immediately after the collision,\\
(iii) the coefficient of restitution between $P$ and $Q$,\\
carefully justifying your answers.\\
(b) Find the size of the angle through which the direction of motion of $P$ is deflected by the collision.
\hfill \mbox{\textit{Edexcel FM1 2020 Q5 [14]}}