| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with energy loss |
| Difficulty | Standard +0.8 This is a two-part Further Mechanics collision problem requiring careful application of conservation of momentum and restitution equations, with part (b) involving energy loss calculations. While the setup is standard FM1 material, finding the range of e values requires insight into the physical constraints (both particles must reverse direction), and the energy loss condition in part (b) leads to a quadratic equation requiring algebraic manipulation. This is more demanding than typical A-level mechanics but represents a solid FM1 question rather than an exceptionally difficult one. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Use of CLM | M1 | Use of CLM. All terms required. Must be dimensionally correct. Condone sign errors |
| \(3mu - 4mu = 2mw - 3mv \quad (-u = -3v + 2w)\) | A1 | Correct unsimplified equation |
| Use of impact law | M1 | Use of impact law. Must be dimensionally correct and used correctly. Condone sign errors |
| \(w + v = 3ue\) | A1 | Correct unsimplified equation. Signs consistent with CLM equation |
| Correct strategy to form equation in \(w\) and find critical value of \(e \in (0,1)\); \(\left(5w = u(9e-1)\right)\) | M1 | Correct overall strategy to find the critical value of \(e\) in \((0,1)\) in \(e\) eg by using CLM and impact law to form equation or inequality in \(w\) and solve for \(e\) |
| \(w > 0 : e > \frac{1}{9}\) | A1 | One inequality for \(e\) correct. Condone \(e \geq \frac{1}{9}\) |
| Complete strategy to justify the range of values of \(e\); \(\left(5v = u(1+6e)\right) \quad v > 0\): true for all \(e\) | M1 | Correct strategy to find the range of possible value of \(e\), i.e find second speed and form second inequality |
| Therefore \(\frac{1}{9} < e \leq 1\) | A1 | Correct final conclusion |
| Total: 8 marks | (8) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Final KE \(= 25\%\) of initial KE | M1 | Use KE to form equation in \(e\). 25% should be used correctly. Condone if mass cancelled throughout |
| \(\frac{1}{2} \times 2m \times \frac{u^2(9e-1)^2}{25} = \frac{1}{4} \times \frac{1}{2} \times 2m \times 4u^2 \quad \left(\text{or } w = \frac{1}{2} \times 2u\right)\) | A1ft | Correct unsimplified equation – follow their \(w\) |
| \(\Rightarrow (9e-1)^2 = 25, \quad e = \frac{2}{3}\) only | A1 | Or equivalent. Correct conclusion. ISW after correct answer. |
| Total: 3 marks | (3) | |
| Question Total: 11 marks | (11) |
## Question 5(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of CLM | M1 | Use of CLM. All terms required. Must be dimensionally correct. Condone sign errors |
| $3mu - 4mu = 2mw - 3mv \quad (-u = -3v + 2w)$ | A1 | Correct unsimplified equation |
| Use of impact law | M1 | Use of impact law. Must be dimensionally correct and used correctly. Condone sign errors |
| $w + v = 3ue$ | A1 | Correct unsimplified equation. Signs consistent with CLM equation |
| Correct strategy to form equation in $w$ and find critical value of $e \in (0,1)$; $\left(5w = u(9e-1)\right)$ | M1 | Correct overall strategy to find the critical value of $e$ in $(0,1)$ in $e$ eg by using CLM and impact law to form equation or inequality in $w$ and solve for $e$ |
| $w > 0 : e > \frac{1}{9}$ | A1 | One inequality for $e$ correct. Condone $e \geq \frac{1}{9}$ |
| Complete strategy to justify the range of values of $e$; $\left(5v = u(1+6e)\right) \quad v > 0$: true for all $e$ | M1 | Correct strategy to find the range of possible value of $e$, i.e find second speed and form second inequality |
| Therefore $\frac{1}{9} < e \leq 1$ | A1 | Correct final conclusion |
| **Total: 8 marks** | **(8)** | |
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## Question 5(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Final KE $= 25\%$ of initial KE | M1 | Use KE to form equation in $e$. 25% should be used correctly. Condone if mass cancelled throughout |
| $\frac{1}{2} \times 2m \times \frac{u^2(9e-1)^2}{25} = \frac{1}{4} \times \frac{1}{2} \times 2m \times 4u^2 \quad \left(\text{or } w = \frac{1}{2} \times 2u\right)$ | A1ft | Correct unsimplified equation – follow their $w$ |
| $\Rightarrow (9e-1)^2 = 25, \quad e = \frac{2}{3}$ only | A1 | Or equivalent. Correct conclusion. ISW after correct answer. |
| **Total: 3 marks** | **(3)** | |
| **Question Total: 11 marks** | **(11)** | |
---\begin{enumerate}
\item A particle $P$ of mass $3 m$ and a particle $Q$ of mass $2 m$ are moving along the same straight line on a smooth horizontal plane. The particles are moving in opposite directions towards each other and collide directly.
\end{enumerate}
Immediately before the collision the speed of $P$ is $u$ and the speed of $Q$ is $2 u$.\\
Immediately after the collision $P$ and $Q$ are moving in opposite directions.\\
The coefficient of restitution between $P$ and $Q$ is $e$.\\
(a) Find the range of possible values of $e$, justifying your answer.
Given that $Q$ loses 75\% of its kinetic energy as a result of the collision,\\
(b) find the value of $e$.
\hfill \mbox{\textit{Edexcel FM1 2019 Q5 [11]}}