Edexcel FM1 2022 June — Question 8 10 marks

Exam BoardEdexcel
ModuleFM1 (Further Mechanics 1)
Year2022
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicOblique and successive collisions
TypeSphere rebounds off fixed wall obliquely
DifficultyChallenging +1.2 This is a standard Further Mechanics 1 collision problem requiring understanding of coefficient of restitution and vector resolution. Part (a) needs geometric reasoning about which values of e allow the ball to reach ST, while part (b) involves routine application of collision formulas with vectors. The vector geometry adds some complexity beyond basic mechanics, but the techniques are well-practiced in FM1.
Spec1.10a Vectors in 2D: i,j notation and column vectors6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts
8. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{86a37170-046f-46e5-9c8c-06d5f98ca4fe-28_567_1406_244_333} \captionsetup{labelformat=empty} \caption{Figure 5}
\end{figure} Figure 5 represents the plan view of part of a smooth horizontal floor, where \(R S\) and \(S T\) are smooth fixed vertical walls. The vector \(\overrightarrow { R S }\) is in the direction of \(\mathbf { i }\) and the vector \(\overrightarrow { S T }\) is in the direction of \(( 2 \mathbf { i } + \mathbf { j } )\). A small ball \(B\) is projected across the floor towards \(R S\). Immediately before the impact with \(R S\), the velocity of \(B\) is \(( 6 \mathbf { i } - 8 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\). The ball bounces off \(R S\) and then hits \(S T\). The ball is modelled as a particle.
Given that the coefficient of restitution between \(B\) and \(R S\) is \(e\),
  1. find the full range of possible values of \(e\). It is now given that \(e = \frac { 1 } { 4 }\) and that the coefficient of restitution between \(B\) and \(S T\) is \(\frac { 1 } { 2 }\)
  2. Find, in terms of \(\mathbf { i }\) and \(\mathbf { j }\), the velocity of \(B\) immediately after its impact with \(S T\).
8(a)
AnswerMarks
B1Velocity before impact: \(\mathbf{v} = 6\mathbf{i} + \ldots\)
B1\(\ldots - 8e\mathbf{j}\)
B1Impact with ST: \(\frac{8e}{6} = \frac{3}{2}\), so \(0 < e \leq \frac{3}{8}\)
8(b)
AnswerMarks
B1Perpendicular to ST: direction \((-\mathbf{i} + 2\mathbf{j})\) (or \(\sqrt{2}\) times this)
M1Component parallel to ST: \(\frac{(6\mathbf{i} + 2\mathbf{j}) \cdot (2\mathbf{i} + \mathbf{j})}{5}\)
A1\(= \frac{1}{5}(6\mathbf{i} + 2\mathbf{j}) \cdot (2\mathbf{i} + \mathbf{j}) = \frac{1}{5}(12 + 2) = \frac{14}{5}\)
M1Component perpendicular to ST: \(\frac{1}{2}(6\mathbf{i} + 2\mathbf{j}) \cdot (-\mathbf{i} + 2\mathbf{j}) \cdot \sqrt{5}\)
A1\(= \frac{1}{2}(-6 + 4) = -\frac{1}{\sqrt{5}}\)
M1\(\mathbf{w} = \frac{1}{5}(2\mathbf{i} + \mathbf{j}) + \frac{1}{5}(-\mathbf{i} + 2\mathbf{j})\)
A1\(\mathbf{w} = \frac{27}{5}\mathbf{i} + \frac{16}{5}\mathbf{j}\) or \((5.4\mathbf{i} + 3.2\mathbf{j})\) (m s\(^{-1}\))
8(b) Alternative 1
AnswerMarks
B1Perpendicular to ST: direction \((-\mathbf{i} + 2\mathbf{j})\)
M1\(\mathbf{w} = a\mathbf{i} + b\mathbf{j}\); \((6\mathbf{i} + 2\mathbf{j}) \cdot (2\mathbf{i} + \mathbf{j}) = (a\mathbf{i} + b\mathbf{j}) \cdot (2\mathbf{i} + \mathbf{j})\)
A1\(14 = 2a + b\)
M1\(\frac{1}{2}(6\mathbf{i} + 2\mathbf{j}) \cdot (-\mathbf{i} + 2\mathbf{j}) = (a\mathbf{i} + b\mathbf{j}) \cdot (-\mathbf{i} + 2\mathbf{j})\)
A1\(2b - a = 1\)
M1Solve simultaneous equations for \(a\) and \(b\)
A1\(\mathbf{w} = \frac{27}{5}\mathbf{i} + \frac{16}{5}\mathbf{j}\) or \((5.4\mathbf{i} + 3.2\mathbf{j})\) (m s\(^{-1}\))
8(b) Alternative 2
AnswerMarks
B1Perpendicular to ST: direction \((-\mathbf{i} + 2\mathbf{j})\)
M1\(6\mathbf{i} + 2\mathbf{j} = p(2\mathbf{i} + \mathbf{j}) + q(-\mathbf{i} + 2\mathbf{j})\)
A1\(6 = 2p - q\), \(2 = p + 2q\); \(p = \frac{14}{5}\), \(q = -\frac{2}{5}\)
M1Component perpendicular to ST: \(\frac{1}{2}q(-\mathbf{i} + 2\mathbf{j})\)
A1\(\frac{1}{2}q(-\mathbf{i} + 2\mathbf{j})\)
M1Solve for \(p\) and \(q\) to obtain velocity: \(\mathbf{w} = \frac{14}{5}(2\mathbf{i} + \mathbf{j}) + \frac{2}{5}(-\mathbf{i} + 2\mathbf{j})\)
A1\(\mathbf{w} = \frac{27}{5}\mathbf{i} + \frac{16}{5}\mathbf{j}\) or \((5.4\mathbf{i} + 3.2\mathbf{j})\) (m s\(^{-1}\))
8(b) Alternative 3
AnswerMarks
B1\(\sin(\alpha - \beta) = \frac{8}{50}\), \(\cos(\alpha - \beta) = \frac{7}{50}\); $\ 
## 8(a)

| B1 | Velocity before impact: $\mathbf{v} = 6\mathbf{i} + \ldots$ |
| B1 | $\ldots - 8e\mathbf{j}$ |
| B1 | Impact with ST: $\frac{8e}{6} = \frac{3}{2}$, so $0 < e \leq \frac{3}{8}$ |

## 8(b)

| B1 | Perpendicular to ST: direction $(-\mathbf{i} + 2\mathbf{j})$ (or $\sqrt{2}$ times this) |
| M1 | Component parallel to ST: $\frac{(6\mathbf{i} + 2\mathbf{j}) \cdot (2\mathbf{i} + \mathbf{j})}{5}$ |
| A1 | $= \frac{1}{5}(6\mathbf{i} + 2\mathbf{j}) \cdot (2\mathbf{i} + \mathbf{j}) = \frac{1}{5}(12 + 2) = \frac{14}{5}$ |
| M1 | Component perpendicular to ST: $\frac{1}{2}(6\mathbf{i} + 2\mathbf{j}) \cdot (-\mathbf{i} + 2\mathbf{j}) \cdot \sqrt{5}$ |
| A1 | $= \frac{1}{2}(-6 + 4) = -\frac{1}{\sqrt{5}}$ |
| M1 | $\mathbf{w} = \frac{1}{5}(2\mathbf{i} + \mathbf{j}) + \frac{1}{5}(-\mathbf{i} + 2\mathbf{j})$ |
| A1 | $\mathbf{w} = \frac{27}{5}\mathbf{i} + \frac{16}{5}\mathbf{j}$ or $(5.4\mathbf{i} + 3.2\mathbf{j})$ (m s$^{-1}$) |

### 8(b) Alternative 1

| B1 | Perpendicular to ST: direction $(-\mathbf{i} + 2\mathbf{j})$ |
| M1 | $\mathbf{w} = a\mathbf{i} + b\mathbf{j}$; $(6\mathbf{i} + 2\mathbf{j}) \cdot (2\mathbf{i} + \mathbf{j}) = (a\mathbf{i} + b\mathbf{j}) \cdot (2\mathbf{i} + \mathbf{j})$ |
| A1 | $14 = 2a + b$ |
| M1 | $\frac{1}{2}(6\mathbf{i} + 2\mathbf{j}) \cdot (-\mathbf{i} + 2\mathbf{j}) = (a\mathbf{i} + b\mathbf{j}) \cdot (-\mathbf{i} + 2\mathbf{j})$ |
| A1 | $2b - a = 1$ |
| M1 | Solve simultaneous equations for $a$ and $b$ |
| A1 | $\mathbf{w} = \frac{27}{5}\mathbf{i} + \frac{16}{5}\mathbf{j}$ or $(5.4\mathbf{i} + 3.2\mathbf{j})$ (m s$^{-1}$) |

### 8(b) Alternative 2

| B1 | Perpendicular to ST: direction $(-\mathbf{i} + 2\mathbf{j})$ |
| M1 | $6\mathbf{i} + 2\mathbf{j} = p(2\mathbf{i} + \mathbf{j}) + q(-\mathbf{i} + 2\mathbf{j})$ |
| A1 | $6 = 2p - q$, $2 = p + 2q$; $p = \frac{14}{5}$, $q = -\frac{2}{5}$ |
| M1 | Component perpendicular to ST: $\frac{1}{2}q(-\mathbf{i} + 2\mathbf{j})$ |
| A1 | $\frac{1}{2}q(-\mathbf{i} + 2\mathbf{j})$ |
| M1 | Solve for $p$ and $q$ to obtain velocity: $\mathbf{w} = \frac{14}{5}(2\mathbf{i} + \mathbf{j}) + \frac{2}{5}(-\mathbf{i} + 2\mathbf{j})$ |
| A1 | $\mathbf{w} = \frac{27}{5}\mathbf{i} + \frac{16}{5}\mathbf{j}$ or $(5.4\mathbf{i} + 3.2\mathbf{j})$ (m s$^{-1}$) |

### 8(b) Alternative 3

| B1 | $\sin(\alpha - \beta) = \frac{8}{50}$, $\cos(\alpha - \beta) = \frac{7}{50}$; $\
8.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{86a37170-046f-46e5-9c8c-06d5f98ca4fe-28_567_1406_244_333}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}

Figure 5 represents the plan view of part of a smooth horizontal floor, where $R S$ and $S T$ are smooth fixed vertical walls. The vector $\overrightarrow { R S }$ is in the direction of $\mathbf { i }$ and the vector $\overrightarrow { S T }$ is in the direction of $( 2 \mathbf { i } + \mathbf { j } )$.

A small ball $B$ is projected across the floor towards $R S$. Immediately before the impact with $R S$, the velocity of $B$ is $( 6 \mathbf { i } - 8 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$. The ball bounces off $R S$ and then hits $S T$.

The ball is modelled as a particle.\\
Given that the coefficient of restitution between $B$ and $R S$ is $e$,
\begin{enumerate}[label=(\alph*)]
\item find the full range of possible values of $e$.

It is now given that $e = \frac { 1 } { 4 }$ and that the coefficient of restitution between $B$ and $S T$ is $\frac { 1 } { 2 }$
\item Find, in terms of $\mathbf { i }$ and $\mathbf { j }$, the velocity of $B$ immediately after its impact with $S T$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel FM1 2022 Q8 [10]}}
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