Edexcel FM1 2021 June — Question 5 10 marks

Exam BoardEdexcel
ModuleFM1 (Further Mechanics 1)
Year2021
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicOblique and successive collisions
TypeBall between two walls, successive rebounds
DifficultyChallenging +1.2 This is a two-collision momentum problem requiring application of coefficient of restitution and energy loss conditions. While it involves multiple steps and careful component resolution, the techniques are standard for FM1: resolving velocities parallel and perpendicular to walls, applying e = v_separation/v_approach, and using KE = ½mv². The novel aspect is using energy loss to find the angle, but this is a straightforward algebraic manipulation once the framework is set up. Harder than typical A-level mechanics but standard for Further Maths.
Spec6.02d Mechanical energy: KE and PE concepts6.03i Coefficient of restitution: e6.03l Newton's law: oblique impacts
5. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{7f077b82-6b39-4cb5-8574-bfa308c88df3-16_575_665_246_699} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 represents the plan view of part of a horizontal floor, where \(A B\) and \(B C\) represent fixed vertical walls, with \(A B\) perpendicular to \(B C\). A small ball is projected along the floor towards the wall \(A B\). Immediately before hitting the wall \(A B\) the ball is moving with speed \(v \mathrm {~ms} ^ { - 1 }\) at an angle \(\theta\) to \(A B\). The ball hits the wall \(A B\) and then hits the wall \(B C\).
The coefficient of restitution between the ball and the wall \(A B\) is \(\frac { 1 } { 3 }\) The coefficient of restitution between the ball and the wall \(B C\) is \(e\).
The floor and the walls are modelled as being smooth.
The ball is modelled as a particle.
The ball loses half of its kinetic energy in the impact with the wall \(A B\).
  1. Find the exact value of \(\cos \theta\). The ball loses half of its remaining kinetic energy in the impact with the wall \(B C\).
  2. Find the exact value of \(e\).
Question 5:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Components after first impact: \(\frac{1}{3}v\sin\theta\) (perpendicular), \(v\cos\theta\) (parallel)B1, B1 Diagram showing correct components
KE: \(\frac{1}{2}\times\frac{1}{2}mv^2 = \frac{1}{2}m\left(v^2\cos^2\theta + \frac{1}{9}v^2\sin^2\theta\right)\)M1
\(\frac{1}{2} = \frac{1}{9} + \frac{8}{9}\cos^2\theta\)M1
\(\frac{7}{16} = \cos^2\theta,\quad \cos\theta = \frac{\sqrt{7}}{4}\)A1
Part (a) alt:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Initial velocity \(\mathbf{v} = x\mathbf{i} - y\mathbf{j}\), after impact \(\mathbf{v} = x\mathbf{i} + \frac{1}{3}y\mathbf{j}\)B1, B1
KE: \(\frac{1}{2}\times\frac{1}{2}m(x^2+y^2) = \frac{1}{2}m\left(x^2 + \frac{1}{9}y^2\right)\)M1
\(y^2 = \frac{9}{7}x^2,\quad \frac{y}{x} = \tan\theta = \frac{3}{\sqrt{7}}\)M1
\(\cos\theta = \frac{\sqrt{7}}{4}\)A1
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Components after second impact: \(ev\cos\theta\) (horizontal), \(\frac{1}{3}v\sin\theta\) (vertical)B1, B1 Diagram showing correct components
KE: \(\frac{1}{4}\times\frac{1}{2}mv^2 = \frac{1}{2}m\left(e^2v^2\cos^2\theta + \frac{1}{9}v^2\sin^2\theta\right)\)M1
\(\frac{1}{4} = \frac{7}{16}e^2 + \frac{1}{9}\times\frac{9}{16}\)M1
\(\Rightarrow e^2 = \frac{3}{7},\quad e = \sqrt{\frac{3}{7}}\)A1
Part (b) alt:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
After second impact \(\mathbf{v} = -ex\mathbf{i} + \frac{1}{3}y\mathbf{j}\)B1, B1
Question 5 (Collision/Kinetic Energy):
AnswerMarks Guidance
AnswerMark Guidance
KE: \(\frac{1}{4} \times \frac{1}{2}m(x^2+y^2) = \frac{1}{2}m\left(e^2x^2+\frac{1}{9}y^2\right)\)M1 Equation for KE in \(v, \theta\). Dimensionally correct. Includes all components. Condone \(\frac{1}{2}\) used on wrong side
\(4e^2x^2 + \frac{4}{9}y^2 = x^2+y^2\), \(\quad 4e^2 = 1+\frac{5}{9}\left(\frac{y}{x}\right)^2\)M1 Form and solve equation in \(\cos\theta\)
\(e^2 = \frac{3}{7}\), \(\quad e = \sqrt{\frac{3}{7}}\)A1 Or exact equivalent 
# Question 5:

## Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Components after first impact: $\frac{1}{3}v\sin\theta$ (perpendicular), $v\cos\theta$ (parallel) | B1, B1 | Diagram showing correct components |
| KE: $\frac{1}{2}\times\frac{1}{2}mv^2 = \frac{1}{2}m\left(v^2\cos^2\theta + \frac{1}{9}v^2\sin^2\theta\right)$ | M1 | |
| $\frac{1}{2} = \frac{1}{9} + \frac{8}{9}\cos^2\theta$ | M1 | |
| $\frac{7}{16} = \cos^2\theta,\quad \cos\theta = \frac{\sqrt{7}}{4}$ | A1 | |

### Part (a) alt:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Initial velocity $\mathbf{v} = x\mathbf{i} - y\mathbf{j}$, after impact $\mathbf{v} = x\mathbf{i} + \frac{1}{3}y\mathbf{j}$ | B1, B1 | |
| KE: $\frac{1}{2}\times\frac{1}{2}m(x^2+y^2) = \frac{1}{2}m\left(x^2 + \frac{1}{9}y^2\right)$ | M1 | |
| $y^2 = \frac{9}{7}x^2,\quad \frac{y}{x} = \tan\theta = \frac{3}{\sqrt{7}}$ | M1 | |
| $\cos\theta = \frac{\sqrt{7}}{4}$ | A1 | |

## Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Components after second impact: $ev\cos\theta$ (horizontal), $\frac{1}{3}v\sin\theta$ (vertical) | B1, B1 | Diagram showing correct components |
| KE: $\frac{1}{4}\times\frac{1}{2}mv^2 = \frac{1}{2}m\left(e^2v^2\cos^2\theta + \frac{1}{9}v^2\sin^2\theta\right)$ | M1 | |
| $\frac{1}{4} = \frac{7}{16}e^2 + \frac{1}{9}\times\frac{9}{16}$ | M1 | |
| $\Rightarrow e^2 = \frac{3}{7},\quad e = \sqrt{\frac{3}{7}}$ | A1 | |

### Part (b) alt:

| Answer/Working | Marks | Guidance |
|---|---|---|
| After second impact $\mathbf{v} = -ex\mathbf{i} + \frac{1}{3}y\mathbf{j}$ | B1, B1 | |

## Question 5 (Collision/Kinetic Energy):

| Answer | Mark | Guidance |
|--------|------|----------|
| KE: $\frac{1}{4} \times \frac{1}{2}m(x^2+y^2) = \frac{1}{2}m\left(e^2x^2+\frac{1}{9}y^2\right)$ | M1 | Equation for KE in $v, \theta$. Dimensionally correct. Includes all components. Condone $\frac{1}{2}$ used on wrong side |
| $4e^2x^2 + \frac{4}{9}y^2 = x^2+y^2$, $\quad 4e^2 = 1+\frac{5}{9}\left(\frac{y}{x}\right)^2$ | M1 | Form and solve equation in $\cos\theta$ |
| $e^2 = \frac{3}{7}$, $\quad e = \sqrt{\frac{3}{7}}$ | A1 | Or exact equivalent |

---
5.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{7f077b82-6b39-4cb5-8574-bfa308c88df3-16_575_665_246_699}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Figure 1 represents the plan view of part of a horizontal floor, where $A B$ and $B C$ represent fixed vertical walls, with $A B$ perpendicular to $B C$.

A small ball is projected along the floor towards the wall $A B$. Immediately before hitting the wall $A B$ the ball is moving with speed $v \mathrm {~ms} ^ { - 1 }$ at an angle $\theta$ to $A B$.

The ball hits the wall $A B$ and then hits the wall $B C$.\\
The coefficient of restitution between the ball and the wall $A B$ is $\frac { 1 } { 3 }$\\
The coefficient of restitution between the ball and the wall $B C$ is $e$.\\
The floor and the walls are modelled as being smooth.\\
The ball is modelled as a particle.\\
The ball loses half of its kinetic energy in the impact with the wall $A B$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact value of $\cos \theta$.

The ball loses half of its remaining kinetic energy in the impact with the wall $B C$.
\item Find the exact value of $e$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel FM1 2021 Q5 [10]}}
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