Edexcel FM1 2020 June — Question 4 9 marks

Exam BoardEdexcel
ModuleFM1 (Further Mechanics 1)
Year2020
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicOblique and successive collisions
TypeSphere rebounds off fixed wall obliquely
DifficultyStandard +0.3 This is a standard FM1 collision question requiring knowledge that velocity parallel to the wall is unchanged and application of Newton's law of restitution perpendicular to the wall. Part (a) uses the standard result that the component parallel to the wall is conserved, requiring resolution of velocities. Part (b) is a direct application of e = (separation speed)/(approach speed) in the perpendicular direction. Both parts follow textbook methods with no novel insight required, making this slightly easier than average for Further Maths.
Spec1.10d Vector operations: addition and scalar multiplication6.03i Coefficient of restitution: e6.03l Newton's law: oblique impacts
  1. \hspace{0pt} [In this question, \(\mathbf { i }\) and \(\mathbf { j }\) are perpendicular unit vectors in a horizontal plane.]
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{361d263e-0ee1-47e9-8fc2-0f127f1c2d7e-12_588_633_301_724} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 represents the plan view of part of a smooth horizontal floor, where \(A B\) represents a fixed smooth vertical wall. A small ball of mass 0.5 kg is moving on the floor when it strikes the wall.
Immediately before the impact the velocity of the ball is \(( 7 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\).
Immediately after the impact the velocity of the ball is \(( \mathbf { i } + 6 \mathbf { j } ) \mathrm { ms } ^ { - 1 }\).
The coefficient of restitution between the ball and the wall is \(e\).
  1. Show that \(A B\) is parallel to \(( 2 \mathbf { i } + 3 \mathbf { j } )\).
  2. Find the value of \(e\).
Question 4:
Part (a):
AnswerMarks Guidance
Working/AnswerMark Guidance
Use of \(\mathbf{I}=m\mathbf{v}-m\mathbf{u}\) or \(\mathbf{v}-\mathbf{u}\)M1 Must be finding the difference between two momenta or two velocities
\(\mathbf{I}=0.5((\mathbf{i}+6\mathbf{j})-(7\mathbf{i}+2\mathbf{j}))\) \((=(-3\mathbf{i}+2\mathbf{j}))\)A1 Correct unsimplified equation for the impulse or for change in velocity
Use of scalar product: \((-3\mathbf{i}+2\mathbf{j}).(2\mathbf{i}+3\mathbf{j})=-6+6=0\)M1 Use of scalar product or equivalent. In alt method allow full marks if \(\sqrt{13}\) not used
Hence impulse perpendicular to \((2\mathbf{i}+3\mathbf{j})\), so \(AB\) must be parallel to \((2\mathbf{i}+3\mathbf{j})\)A1* Reach given conclusion from correct working
Total: (4)
Alternative method 1:
AnswerMarks Guidance
Working/AnswerMark Guidance
Components of velocities parallel to \((2\mathbf{i}+3\mathbf{j})\)M1
\(\left(\frac{1}{\sqrt{13}}\right)(7\mathbf{i}+2\mathbf{j}).(2\mathbf{i}+3\mathbf{j})=\left(\frac{1}{\sqrt{13}}\right)(14+6)\) and \(\left(\frac{1}{\sqrt{13}}\right)(\mathbf{i}+6\mathbf{j}).(2\mathbf{i}+3\mathbf{j})=\left(\frac{1}{\sqrt{13}}\right)(2+18)\)A1
Simplify and compare valuesM1
Hence component of velocity parallel to \((2\mathbf{i}+3\mathbf{j})\) is unchanged, so \(AB\) must be parallel to \((2\mathbf{i}+3\mathbf{j})\)A1*
Alternative method 2:
AnswerMarks Guidance
Working/AnswerMark Guidance
Use conservation of velocity parallel to \(a\mathbf{i}+b\mathbf{j}\)M1
\((7\mathbf{i}+2\mathbf{j}).(a\mathbf{i}+b\mathbf{j})=(\mathbf{i}+6\mathbf{j}).(a\mathbf{i}+b\mathbf{j})\) \((\Rightarrow 7a+2b=a+6b)\)A1
Find ratio of \(a\) and \(b\) to obtain direction: \(\left(b=\frac{2}{3}a\right)\)M1
Hence \(AB\) must be parallel to \((2\mathbf{i}+3\mathbf{j})\)A1*
Part (b):
AnswerMarks Guidance
Working/AnswerMark Guidance
Use scalar product to find components of velocities perpendicular to the wallM1 Condone if not using a unit vector
\(\left(\frac{1}{\sqrt{13}}\right)(-3\mathbf{i}+2\mathbf{j})(7\mathbf{i}+2\mathbf{j})=\left(\frac{1}{\sqrt{13}}\right)(-21+4)\) \(\left(=\frac{-17}{\sqrt{13}}\right)\)A1 One correct value
\(\left(\frac{1}{\sqrt{13}}\right)(-3\mathbf{i}+2\mathbf{j})(\mathbf{i}+6\mathbf{j})=\left(\frac{1}{\sqrt{13}}\right)(-3+12)\) \(\left(=\frac{9}{\sqrt{13}}\right)\)A1 Second correct value. (If working with angles: M1A1A1 for \(\sqrt{53}\sin40.36..°\) and \(\sqrt{37}\sin24.23..°\))
Use of impact lawM1 Use their components the right way round in the impact law. Condone sign error
\(e=\frac{9}{17}\)A1 0.53 or better \((0.52941\ldots)\)
Total: (5) 
## Question 4:

### Part (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of $\mathbf{I}=m\mathbf{v}-m\mathbf{u}$ or $\mathbf{v}-\mathbf{u}$ | M1 | Must be finding the difference between two momenta or two velocities |
| $\mathbf{I}=0.5((\mathbf{i}+6\mathbf{j})-(7\mathbf{i}+2\mathbf{j}))$ $(=(-3\mathbf{i}+2\mathbf{j}))$ | A1 | Correct unsimplified equation for the impulse or for change in velocity |
| Use of scalar product: $(-3\mathbf{i}+2\mathbf{j}).(2\mathbf{i}+3\mathbf{j})=-6+6=0$ | M1 | Use of scalar product or equivalent. In alt method allow full marks if $\sqrt{13}$ not used |
| Hence impulse perpendicular to $(2\mathbf{i}+3\mathbf{j})$, so $AB$ must be parallel to $(2\mathbf{i}+3\mathbf{j})$ | A1* | Reach given conclusion from correct working |
| **Total: (4)** | | |

**Alternative method 1:**

| Working/Answer | Mark | Guidance |
|---|---|---|
| Components of velocities parallel to $(2\mathbf{i}+3\mathbf{j})$ | M1 | |
| $\left(\frac{1}{\sqrt{13}}\right)(7\mathbf{i}+2\mathbf{j}).(2\mathbf{i}+3\mathbf{j})=\left(\frac{1}{\sqrt{13}}\right)(14+6)$ and $\left(\frac{1}{\sqrt{13}}\right)(\mathbf{i}+6\mathbf{j}).(2\mathbf{i}+3\mathbf{j})=\left(\frac{1}{\sqrt{13}}\right)(2+18)$ | A1 | |
| Simplify and compare values | M1 | |
| Hence component of velocity parallel to $(2\mathbf{i}+3\mathbf{j})$ is unchanged, so $AB$ must be parallel to $(2\mathbf{i}+3\mathbf{j})$ | A1* | |

**Alternative method 2:**

| Working/Answer | Mark | Guidance |
|---|---|---|
| Use conservation of velocity parallel to $a\mathbf{i}+b\mathbf{j}$ | M1 | |
| $(7\mathbf{i}+2\mathbf{j}).(a\mathbf{i}+b\mathbf{j})=(\mathbf{i}+6\mathbf{j}).(a\mathbf{i}+b\mathbf{j})$ $(\Rightarrow 7a+2b=a+6b)$ | A1 | |
| Find ratio of $a$ and $b$ to obtain direction: $\left(b=\frac{2}{3}a\right)$ | M1 | |
| Hence $AB$ must be parallel to $(2\mathbf{i}+3\mathbf{j})$ | A1* | |

### Part (b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Use scalar product to find components of velocities perpendicular to the wall | M1 | Condone if not using a unit vector |
| $\left(\frac{1}{\sqrt{13}}\right)(-3\mathbf{i}+2\mathbf{j})(7\mathbf{i}+2\mathbf{j})=\left(\frac{1}{\sqrt{13}}\right)(-21+4)$ $\left(=\frac{-17}{\sqrt{13}}\right)$ | A1 | One correct value |
| $\left(\frac{1}{\sqrt{13}}\right)(-3\mathbf{i}+2\mathbf{j})(\mathbf{i}+6\mathbf{j})=\left(\frac{1}{\sqrt{13}}\right)(-3+12)$ $\left(=\frac{9}{\sqrt{13}}\right)$ | A1 | Second correct value. (If working with angles: M1A1A1 for $\sqrt{53}\sin40.36..°$ and $\sqrt{37}\sin24.23..°$) |
| Use of impact law | M1 | Use their components the right way round in the impact law. Condone sign error |
| $e=\frac{9}{17}$ | A1 | 0.53 or better $(0.52941\ldots)$ |
| **Total: (5)** | | |
\begin{enumerate}
  \item \hspace{0pt} [In this question, $\mathbf { i }$ and $\mathbf { j }$ are perpendicular unit vectors in a horizontal plane.]
\end{enumerate}

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{361d263e-0ee1-47e9-8fc2-0f127f1c2d7e-12_588_633_301_724}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Figure 1 represents the plan view of part of a smooth horizontal floor, where $A B$ represents a fixed smooth vertical wall.

A small ball of mass 0.5 kg is moving on the floor when it strikes the wall.\\
Immediately before the impact the velocity of the ball is $( 7 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.\\
Immediately after the impact the velocity of the ball is $( \mathbf { i } + 6 \mathbf { j } ) \mathrm { ms } ^ { - 1 }$.\\
The coefficient of restitution between the ball and the wall is $e$.\\
(a) Show that $A B$ is parallel to $( 2 \mathbf { i } + 3 \mathbf { j } )$.\\
(b) Find the value of $e$.

\begin{center}

\end{center}

\hfill \mbox{\textit{Edexcel FM1 2020 Q4 [9]}}
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