Question 5 - A-Level Maths

Edexcel M2 2018 June — Question 5 13 marks

Exam Board	Edexcel
Module	M2 (Mechanics 2)
Year	2018
Session	June
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Momentum and Collisions 1
Type	Collision followed by wall impact
Difficulty	Standard +0.3 This is a standard M2 collision problem requiring conservation of momentum and Newton's restitution law, followed by tracking particle positions. The 'show that' part guides students through the first collision, and part (b) requires setting up distance-time equations for both particles to find when they meet again. While multi-step, it follows a well-practiced template with no novel insights required, making it slightly easier than average.
Spec	6.03b Conservation of momentum: 1D two particles 6.03j Perfectly elastic/inelastic: collisions 6.03k Newton's experimental law: direct impact 6.03l Newton's law: oblique impacts

5. A particle $A$ of mass $3 m$ is moving in a straight line with speed $2 u$ on a smooth horizontal floor. Particle $A$ collides directly with another particle $B$ of mass $2 m$ which is moving along the same straight line with speed $u$ but in the opposite direction to $A$. The coefficient of restitution between $A$ and $B$ is $\frac { 1 } { 3 }$.

1. Show that the speed of $B$ immediately after the collision is $\frac { 7 } { 5 } u$
2. Find the speed of $A$ immediately after the collision. After the collision, $B$ hits a smooth vertical wall which is perpendicular to the direction of motion of $B$. The coefficient of restitution between $B$ and the wall is $\frac { 1 } { 2 }$. The first collision between $A$ and $B$ occurred at a distance $x$ from the wall. The particles collide again at a distance $y$ from the wall.
Find $y$ in terms of $x$.

Show mark scheme Show mark scheme source

Part 5a

CLM: $3m \cdot 2u + 2m \cdot u = 3mv + 2mw$

$4u = 3v + 2w$

Answer	Marks
M1A1

Impact law: $w - v = \frac{1}{2}u - u = -\frac{1}{2}u$

Answer	Marks
M1A1

Solve for simultaneous equations for $w$ or $v$:

$3w - 3v = 3u$, $2w - 3v = 4u$

$5w = 7u$, $w = \frac{7}{5}u$

Answer	Marks
DM1	Dependent on both previous M marks

$v = \frac{2}{5}u$

Answer	Marks
A1	Or equivalent. Must be positive

(7)

Part 5b

Speed of B after collision with wall:

$\frac{1}{2} \cdot \frac{7}{5}u = \frac{7}{10}u$

Answer	Marks
B1	Accept +/-

Total time for either particle

Answer	Marks
B1

Equate the time travelled for each particle:

$\frac{x}{y} = \frac{\frac{7}{5}u}{\frac{7}{10}u}$

Answer	Marks
M1

$\frac{x}{y} = \frac{5x + 10y}{7u} \cdot \frac{5x + 5y}{2u}$

Answer	Marks
A1	Correct unsimplified

$5x + 10y = 35x + 35y$, $10x + 20y = 35x + 35y$

$7u \cdot 7u = 2u \cdot 2u$

Answer	Marks
DM1	Dependent on previous M1

$55y = 25x$, $y = \frac{5}{11}x$

Answer	Marks
A1	Or equivalent. 0.45x or better

(6)

Alternative 1

Speed of B after collision with wall:

$\frac{1}{2} \cdot \frac{7}{5}u = \frac{7}{10}u$

Answer	Marks
B1	Accept +/-

Time of travel for B:

$\frac{x}{y} = \frac{\frac{7}{5}u}{\frac{7}{10}u} = \frac{5x + 10y}{7u}$

Answer	Marks
B1

Distance moved by A:

$\frac{2}{5}u$

Answer	Marks
M1	Correct method for distance

$\frac{2x + 4y}{7} = y \cdot x$

Answer	Marks
A1	Correct unsimplified

$2x + 4y = 7y - 7x$

Answer	Marks
DM1	Dependent on previous M1. Form equation in $x$ and $y$

$y = \frac{5}{11}x$

Answer	Marks
A1	Or equivalent. 0.45x or better

(6)

Alternative 2

Speed of B after collision with wall:

$\frac{1}{2} \cdot \frac{7}{5}u = \frac{7}{10}u$

Answer	Marks
B1	Accept +/-

Distance moved by A when B hits wall:

$x \cdot \frac{2}{5}u \cdot x = \frac{2x}{7u} \cdot \frac{5}{7}$

Answer	Marks
B1	Distance apart when B hits the wall

Gap closing at:

$\frac{7}{10}u - \frac{2}{5}u = \frac{11

## Part 5a

CLM: $3m \cdot 2u + 2m \cdot u = 3mv + 2mw$

$4u = 3v + 2w$

| M1A1 |

Impact law: $w - v = \frac{1}{2}u - u = -\frac{1}{2}u$

| M1A1 |

Solve for simultaneous equations for $w$ or $v$:

$3w - 3v = 3u$, $2w - 3v = 4u$

$5w = 7u$, $w = \frac{7}{5}u$

| DM1 | Dependent on both previous M marks

$v = \frac{2}{5}u$

| A1 | Or equivalent. Must be positive

(7)

## Part 5b

Speed of B after collision with wall:

$\frac{1}{2} \cdot \frac{7}{5}u = \frac{7}{10}u$

| B1 | Accept +/-

Total time for either particle

| B1 |

Equate the time travelled for each particle:

$\frac{x}{y} = \frac{\frac{7}{5}u}{\frac{7}{10}u}$

| M1 |

$\frac{x}{y} = \frac{5x + 10y}{7u} \cdot \frac{5x + 5y}{2u}$

| A1 | Correct unsimplified

$5x + 10y = 35x + 35y$, $10x + 20y = 35x + 35y$

$7u \cdot 7u = 2u \cdot 2u$

| DM1 | Dependent on previous M1

$55y = 25x$, $y = \frac{5}{11}x$

| A1 | Or equivalent. 0.45x or better

(6)

---

## Alternative 1

Speed of B after collision with wall:

$\frac{1}{2} \cdot \frac{7}{5}u = \frac{7}{10}u$

| B1 | Accept +/-

Time of travel for B:

$\frac{x}{y} = \frac{\frac{7}{5}u}{\frac{7}{10}u} = \frac{5x + 10y}{7u}$

| B1 |

Distance moved by A:

$\frac{2}{5}u$

| M1 | Correct method for distance

$\frac{2x + 4y}{7} = y \cdot x$

| A1 | Correct unsimplified

$2x + 4y = 7y - 7x$

| DM1 | Dependent on previous M1. Form equation in $x$ and $y$

$y = \frac{5}{11}x$

| A1 | Or equivalent. 0.45x or better

(6)

---

## Alternative 2

Speed of B after collision with wall:

$\frac{1}{2} \cdot \frac{7}{5}u = \frac{7}{10}u$

| B1 | Accept +/-

Distance moved by A when B hits wall:

$x \cdot \frac{2}{5}u \cdot x = \frac{2x}{7u} \cdot \frac{5}{7}$

| B1 | Distance apart when B hits the wall

Gap closing at:

$\frac{7}{10}u - \frac{2}{5}u = \frac{11

Show LaTeX source

5. A particle $A$ of mass $3 m$ is moving in a straight line with speed $2 u$ on a smooth horizontal floor. Particle $A$ collides directly with another particle $B$ of mass $2 m$ which is moving along the same straight line with speed $u$ but in the opposite direction to $A$. The coefficient of restitution between $A$ and $B$ is $\frac { 1 } { 3 }$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the speed of $B$ immediately after the collision is $\frac { 7 } { 5 } u$
\item Find the speed of $A$ immediately after the collision.

After the collision, $B$ hits a smooth vertical wall which is perpendicular to the direction of motion of $B$. The coefficient of restitution between $B$ and the wall is $\frac { 1 } { 2 }$. The first collision between $A$ and $B$ occurred at a distance $x$ from the wall. The particles collide again at a distance $y$ from the wall.
\end{enumerate}\item Find $y$ in terms of $x$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2 2018 Q5 [13]}}

This paper (7 questions)

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Q1 8 Q2 7 Q3 8 Q4 10 Q5 13 Q6 14 Q7 15

Edexcel M2 2018 June — Question 5 13 marks

Part 5a

CLM: \(3m \cdot 2u + 2m \cdot u = 3mv + 2mw\)

\(4u = 3v + 2w\)

Impact law: \(w - v = \frac{1}{2}u - u = -\frac{1}{2}u\)

Solve for simultaneous equations for \(w\) or \(v\):

\(3w - 3v = 3u\), \(2w - 3v = 4u\)

\(5w = 7u\), \(w = \frac{7}{5}u\)

\(v = \frac{2}{5}u\)

(7)

Part 5b

Speed of B after collision with wall:

\(\frac{1}{2} \cdot \frac{7}{5}u = \frac{7}{10}u\)

Equate the time travelled for each particle:

\(\frac{x}{y} = \frac{\frac{7}{5}u}{\frac{7}{10}u}\)

\(\frac{x}{y} = \frac{5x + 10y}{7u} \cdot \frac{5x + 5y}{2u}\)

\(5x + 10y = 35x + 35y\), \(10x + 20y = 35x + 35y\)

\(7u \cdot 7u = 2u \cdot 2u\)

\(55y = 25x\), \(y = \frac{5}{11}x\)

(6)

Alternative 1

Speed of B after collision with wall:

\(\frac{1}{2} \cdot \frac{7}{5}u = \frac{7}{10}u\)

Time of travel for B:

\(\frac{x}{y} = \frac{\frac{7}{5}u}{\frac{7}{10}u} = \frac{5x + 10y}{7u}\)

Distance moved by A:

\(\frac{2}{5}u\)

\(\frac{2x + 4y}{7} = y \cdot x\)

\(2x + 4y = 7y - 7x\)

\(y = \frac{5}{11}x\)

(6)

Alternative 2

Speed of B after collision with wall:

\(\frac{1}{2} \cdot \frac{7}{5}u = \frac{7}{10}u\)

Distance moved by A when B hits wall:

\(x \cdot \frac{2}{5}u \cdot x = \frac{2x}{7u} \cdot \frac{5}{7}\)

Gap closing at:

$\frac{7}{10}u - \frac{2}{5}u = \frac{11

This paper (7 questions)