Question 4 - A-Level Maths

Edexcel M2 2021 October — Question 4 6 marks

Exam Board	Edexcel
Module	M2 (Mechanics 2)
Year	2021
Session	October
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Impulse and momentum (advanced)
Type	Velocity after impulse (find unknown constant)
Difficulty	Standard +0.3 This is a standard M2 impulse-momentum question requiring application of the impulse-momentum theorem in vector form, followed by solving a quadratic equation. The setup is straightforward with clear given information, and while it involves multiple steps (vector addition, magnitude calculation, quadratic formula), these are all routine techniques for M2 students with no novel insight required.
Spec	1.10d Vector operations: addition and scalar multiplication 6.03f Impulse-momentum: relation 6.03g Impulse in 2D: vector form

4. A particle \(P\) of mass 0.75 kg is moving with velocity \(4 \mathbf { i } \mathrm {~ms} ^ { - 1 }\) when it receives an impulse \(\mathbf { J }\) Ns. Immediately after \(P\) receives the impulse, the speed of \(P\) is \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) Given that \(\mathbf { J } = c ( - \mathbf { i } + 2 \mathbf { j } )\), where \(c\) is a constant, find the two possible values of \(c\).
(6)

Show mark scheme Show mark scheme source

Question 4:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Use of \(\mathbf{I} = m(\mathbf{v} - \mathbf{u})\)	M1	As a single vector equation or two separate equations
\(c\begin{pmatrix}-1\\2\end{pmatrix} = \frac{3}{4}\mathbf{v} - \begin{pmatrix}3\\0\end{pmatrix}\)	A1	Any equivalent substituted form
\(\mathbf{v} = \frac{4}{3}\begin{pmatrix}3-c\\2c\end{pmatrix}\)
Use of Pythagoras	M1
\(64 = \frac{16}{9}\left((3-c)^2 + 4c^2\right)\)	A1	Correct unsimplified equation in \(c\) or a component of \(\mathbf{v}\). \(\left(5a^2 - 32a = 0 \text{ or } 5b^2 - 16b - 192 = 0\right)\)
Simplify to 3 term quadratic and solve for \(c\)	M1	\(5c^2 - 6c - 27 = 0\)
\(c = 3\) or \(c = -\frac{9}{5}\ (-1.8)\)	A1	Correct only
Total: (6) [6]

Alternative Method (Q4):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Form vector triangle (diagram with sides 6, \(\sqrt{5}c\), 3)	M1	Form vector triangle. Dimensionally correct
Three correct lengths and \(\	\cos\theta\	= \frac{1}{\sqrt{5}}\) seen or implied
Use of cosine rule	M1
\(36 = 9 + 5c^2 - 2 \times 3\sqrt{5}c\cos\theta\)	A1	Correct unsimplified equation in \(c\) with \(\cos\theta\) or their \(\cos\theta\)
Rearrange as 3 term quadratic and solve for \(c\)	M1	\(5c^2 - 6c - 27 = 0\)
\(c = 3\) or \(c = -\frac{9}{5}\ (-1.8)\)	A1	Correct only
Total: (6) [6]

# Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $\mathbf{I} = m(\mathbf{v} - \mathbf{u})$ | M1 | As a single vector equation or two separate equations |
| $c\begin{pmatrix}-1\\2\end{pmatrix} = \frac{3}{4}\mathbf{v} - \begin{pmatrix}3\\0\end{pmatrix}$ | A1 | Any equivalent substituted form |
| $\mathbf{v} = \frac{4}{3}\begin{pmatrix}3-c\\2c\end{pmatrix}$ | | |
| Use of Pythagoras | M1 | |
| $64 = \frac{16}{9}\left((3-c)^2 + 4c^2\right)$ | A1 | Correct unsimplified equation in $c$ or a component of $\mathbf{v}$. $\left(5a^2 - 32a = 0 \text{ or } 5b^2 - 16b - 192 = 0\right)$ |
| Simplify to 3 term quadratic and solve for $c$ | M1 | $5c^2 - 6c - 27 = 0$ |
| $c = 3$ or $c = -\frac{9}{5}\ (-1.8)$ | A1 | Correct only |
| **Total: (6) [6]** | | |

### Alternative Method (Q4):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Form vector triangle (diagram with sides 6, $\sqrt{5}c$, 3) | M1 | Form vector triangle. Dimensionally correct |
| Three correct lengths and $\|\cos\theta\| = \frac{1}{\sqrt{5}}$ seen or implied | A1 | |
| Use of cosine rule | M1 | |
| $36 = 9 + 5c^2 - 2 \times 3\sqrt{5}c\cos\theta$ | A1 | Correct unsimplified equation in $c$ with $\cos\theta$ or their $\cos\theta$ |
| Rearrange as 3 term quadratic and solve for $c$ | M1 | $5c^2 - 6c - 27 = 0$ |
| $c = 3$ or $c = -\frac{9}{5}\ (-1.8)$ | A1 | Correct only |
| **Total: (6) [6]** | | |

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Show LaTeX source

4. A particle $P$ of mass 0.75 kg is moving with velocity $4 \mathbf { i } \mathrm {~ms} ^ { - 1 }$ when it receives an impulse $\mathbf { J }$ Ns. Immediately after $P$ receives the impulse, the speed of $P$ is $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$

Given that $\mathbf { J } = c ( - \mathbf { i } + 2 \mathbf { j } )$, where $c$ is a constant, find the two possible values of $c$.\\
(6)\\

\hfill \mbox{\textit{Edexcel M2 2021 Q4 [6]}}

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