Edexcel FM1 AS 2023 June — Question 4 14 marks

Exam BoardEdexcel
ModuleFM1 AS (Further Mechanics 1 AS)
Year2023
SessionJune
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions 1
TypeThree-particle sequential collisions
DifficultyStandard +0.3 This is a standard Further Mechanics 1 collision problem with straightforward application of conservation of momentum and Newton's restitution law. Part (a) is a routine 'show that' requiring two equations, parts (b)-(c) involve simple inequality/sign checks, part (d) is standard KE loss calculation, and part (e) tests conceptual understanding. While it's a multi-part question requiring careful bookkeeping, all techniques are direct applications of standard FM1 methods with no novel insight required. It's slightly easier than average A-level due to being a typical textbook-style sequential collision problem.
Spec6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact
4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{0cec16c3-23a0-4620-a80f-b5d4e014e2fc-12_81_1383_255_342} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Three particles, \(P , Q\) and \(R\), lie at rest on a smooth horizontal plane. The particles are in a straight line with \(Q\) between \(P\) and \(R\), as shown in Figure 1 . Particle \(P\) is projected towards \(Q\) with speed \(u\). At the same time, \(R\) is projected with speed \(\frac { 1 } { 2 } u\) away from \(Q\), in the direction \(Q R\). Particle \(P\) has mass \(m\) and particle \(Q\) has mass \(2 m\).
The coefficient of restitution between \(P\) and \(Q\) is \(e\).
  1. Show that the speed of \(Q\) immediately after the collision between \(P\) and \(Q\) is $$\frac { u ( 1 + e ) } { 3 }$$ It is given that \(e > \frac { 1 } { 2 }\)
  2. Determine whether there is a collision between \(Q\) and \(R\).
  3. Determine the direction of motion of \(P\) immediately after the collision between \(P\) and \(Q\).
  4. Find, in terms of \(m , u\) and \(e\), the total kinetic energy lost in the collision between \(P\) and \(Q\), simplifying your answer.
  5. Explain how using \(e = 1\) could be used to check your answer to part (d).
Question 4(a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Use of CLMM1
\(mu = -mv_P + 2mv_Q\)A1
Use of NELM1
\(eu = v_P + v_Q\)A1
Solve for \(v_Q\)M1
\(v_Q = \frac{u(1+e)}{3}\)A1* Allow \(\frac{u(e+1)}{3}\)
(6 marks)
Question 4(b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(e > \frac{1}{2} \Rightarrow \frac{u(1+e)}{3} > \frac{1}{2}u\). Allow this argument reversed.M1
Hence collision between \(Q\) and \(R\) will occurA1
(2 marks)
Question 4(c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(v_P = \frac{u(2e-1)}{3}\)M1 Correct expression using their \(v_P\)
\(e > \frac{1}{2} \Rightarrow v_P > 0\), so \(P\) moves in opposite direction to its original directionA1 A0 for any of: direction changes, \(P\) moves away from \(Q\), goes left, \(P\) will travel in negative direction, \(P\) moves backwards
ALT: \(e > \frac{1}{2} \Rightarrow v_Q > \frac{1}{2}u\); Since \(v_P = 2v_Q - u, v_P > 0\)M1
So \(P\) moves in opposite direction to its original directionA1
(2 marks)
Question 4:
Part 4(d):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\frac{1}{2}mu^2 - \frac{1}{2}m\left[\frac{u(2e-1)}{3}\right]^2 - \frac{1}{2}2m\left[\frac{u(e+1)}{3}\right]^2\)A1 1.1b
\(= \frac{1}{3}mu^2 - \frac{1}{3}mu^2e^2 = \frac{1}{3}mu^2(1-e^2)\)A1 1.1b
(3)
Notes for 4d:
AnswerMarks
- M1Correct no. of terms, dimensionally correct. Allow if \(m\) and \(2m\) are swapped or \(m\) used for mass of \(Q\) but otherwise correct. Allow an expression for the KE gain.
- A1Correct unsimplified expression
- A1Any correct simplified two term expression, not necessarily factorised.
Part 4(e):
AnswerMarks Guidance
Answer/WorkingMark Guidance
The answer for part (d) should equal 0 when \(e = 1\).B1 2.4
(1)
Notes for 4e:
AnswerMarks
- B1Correct explanation. This can be scored after an incorrect answer to (d). Need to say that when \(e = 1\), KE change \(= 0\). Not enough to say KE Before \(=\) KE After.
(14 marks total)
## Question 4(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of CLM | M1 | |
| $mu = -mv_P + 2mv_Q$ | A1 | |
| Use of NEL | M1 | |
| $eu = v_P + v_Q$ | A1 | |
| Solve for $v_Q$ | M1 | |
| $v_Q = \frac{u(1+e)}{3}$ | A1* | Allow $\frac{u(e+1)}{3}$ |

**(6 marks)**

---

## Question 4(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $e > \frac{1}{2} \Rightarrow \frac{u(1+e)}{3} > \frac{1}{2}u$. Allow this argument reversed. | M1 | |
| Hence collision between $Q$ and $R$ will occur | A1 | |

**(2 marks)**

---

## Question 4(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $v_P = \frac{u(2e-1)}{3}$ | M1 | Correct expression using their $v_P$ |
| $e > \frac{1}{2} \Rightarrow v_P > 0$, so $P$ moves in opposite direction to its original direction | A1 | A0 for any of: direction changes, $P$ moves away from $Q$, goes left, $P$ will travel in negative direction, $P$ moves backwards |
| **ALT:** $e > \frac{1}{2} \Rightarrow v_Q > \frac{1}{2}u$; Since $v_P = 2v_Q - u, v_P > 0$ | M1 | |
| So $P$ moves in opposite direction to its original direction | A1 | |

**(2 marks)**

## Question 4:

### Part 4(d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}mu^2 - \frac{1}{2}m\left[\frac{u(2e-1)}{3}\right]^2 - \frac{1}{2}2m\left[\frac{u(e+1)}{3}\right]^2$ | A1 | 1.1b |
| $= \frac{1}{3}mu^2 - \frac{1}{3}mu^2e^2 = \frac{1}{3}mu^2(1-e^2)$ | A1 | 1.1b |
| | **(3)** | |

**Notes for 4d:**
- M1 | Correct no. of terms, dimensionally correct. Allow if $m$ and $2m$ are swapped or $m$ used for mass of $Q$ but otherwise correct. Allow an expression for the KE gain.
- A1 | Correct unsimplified expression
- A1 | Any correct simplified two term expression, not necessarily factorised.

---

### Part 4(e):

| Answer/Working | Mark | Guidance |
|---|---|---|
| The answer for part (d) should equal 0 when $e = 1$. | B1 | 2.4 |
| | **(1)** | |

**Notes for 4e:**
- B1 | Correct explanation. This can be scored after an incorrect answer to (d). Need to say that when $e = 1$, KE change $= 0$. Not enough to say KE Before $=$ KE After.

---

**(14 marks total)**
4.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{0cec16c3-23a0-4620-a80f-b5d4e014e2fc-12_81_1383_255_342}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Three particles, $P , Q$ and $R$, lie at rest on a smooth horizontal plane. The particles are in a straight line with $Q$ between $P$ and $R$, as shown in Figure 1 .

Particle $P$ is projected towards $Q$ with speed $u$. At the same time, $R$ is projected with speed $\frac { 1 } { 2 } u$ away from $Q$, in the direction $Q R$.

Particle $P$ has mass $m$ and particle $Q$ has mass $2 m$.\\
The coefficient of restitution between $P$ and $Q$ is $e$.
\begin{enumerate}[label=(\alph*)]
\item Show that the speed of $Q$ immediately after the collision between $P$ and $Q$ is

$$\frac { u ( 1 + e ) } { 3 }$$

It is given that $e > \frac { 1 } { 2 }$
\item Determine whether there is a collision between $Q$ and $R$.
\item Determine the direction of motion of $P$ immediately after the collision between $P$ and $Q$.
\item Find, in terms of $m , u$ and $e$, the total kinetic energy lost in the collision between $P$ and $Q$, simplifying your answer.
\item Explain how using $e = 1$ could be used to check your answer to part (d).
\end{enumerate}

\hfill \mbox{\textit{Edexcel FM1 AS 2023 Q4 [14]}}
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