| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2021 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Velocity after impulse (find unknown constant) |
| Difficulty | Moderate -0.3 This is a straightforward M2 impulse-momentum question requiring application of the impulse-momentum theorem with vector components. Students must recognize that impulse magnitude gives two possible directions (±), then solve using Pythagoras. Standard bookwork with clear method, slightly easier than average A-level due to clean numbers and direct application of a single principle. |
| Spec | 6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance Notes |
| \((\mathbf{I}=)1.5\{v\mathbf{i}-(4\mathbf{i}+6\mathbf{j})\}\) | M1 | Use of \(\mathbf{I}=m\mathbf{v}-m\mathbf{u}\). Must be using \(v\mathbf{i}\). Condone u, v confusion. Ignore the left hand side |
| \(=1.5\{(v-4)\mathbf{i}-6\mathbf{j}\}\) | A1 | Or equivalent seen or implied. Condone subtraction the wrong way round. Ignore the left hand side |
| \(\Rightarrow 15^2=1.5^2\{(v-4)^2+6^2\}\) | M1 | Use of modulus. Allow for \(p^2+q^2=100\) |
| \(100=(v-4)^2+36\) | A1 | Correct unsimplified equation in \(v\) |
| \(v^2-8v-48=0\) | A1 | Correct simplified equation in \(v\) seen or implied |
| \(\Rightarrow v=12\) | A1 | One correct value |
| or \(v=-4\) | A1 | Both correct values |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance Notes |
| Initial momentum \(=(6\mathbf{i}+9\mathbf{j})\) Ns | M1 | Impulse momentum triangle. Accept \(\sqrt{117}\) Ns |
| \(\cos\alpha=\dfrac{6}{\sqrt{117}}\left(=\dfrac{2}{\sqrt{13}}\right)\) | A1 | Or equivalent |
| \(m^2+117-2m\sqrt{117}\cos\alpha=225\) | M1 | Use of cosine formula (final momentum \(m\)) |
| \(m^2-12m-108=0\) | A1 | Or equivalent |
| \(\Rightarrow m=-6\) or \(m=18\) | A1 | |
| \(\Rightarrow v=12\) | A1 | One correct value |
| or \(v=-4\) | A1 | Both correct values |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance Notes |
| Initial momentum \(=(6\mathbf{i}+9\mathbf{j})\) Ns | M1 | Impulse momentum triangle. Accept \(\sqrt{117}\) Ns |
| \(\sin\alpha=\dfrac{3}{\sqrt{13}}\) | A1 | Or equivalent |
| \(\dfrac{15}{\sin\alpha}=\dfrac{\sqrt{117}}{\sin\theta}\) | M1 | Use of sine formula |
| \(\Rightarrow\sin\theta=\dfrac{3}{5}\), \(\theta=36.9°\) or \(\theta=143.1°\) | A1 | |
| \(\dfrac{m}{\sin 86.8}=\dfrac{15}{\sin\alpha}\) or \(\dfrac{m}{\sin 19.4}=\dfrac{15}{\sin(180-\alpha)}\) | A1 | Correct equation in \(m\) |
| \(\Rightarrow v=12\) | A1 | One correct value |
| or \(v=-4\) | A1 | Both correct values |
# Question 1:
| Working/Answer | Mark | Guidance Notes |
|---|---|---|
| $(\mathbf{I}=)1.5\{v\mathbf{i}-(4\mathbf{i}+6\mathbf{j})\}$ | M1 | Use of $\mathbf{I}=m\mathbf{v}-m\mathbf{u}$. Must be using $v\mathbf{i}$. Condone **u**, **v** confusion. Ignore the left hand side |
| $=1.5\{(v-4)\mathbf{i}-6\mathbf{j}\}$ | A1 | Or equivalent seen or implied. Condone subtraction the wrong way round. Ignore the left hand side |
| $\Rightarrow 15^2=1.5^2\{(v-4)^2+6^2\}$ | M1 | Use of modulus. Allow for $p^2+q^2=100$ |
| $100=(v-4)^2+36$ | A1 | Correct unsimplified equation in $v$ |
| $v^2-8v-48=0$ | A1 | Correct simplified equation in $v$ seen or implied |
| $\Rightarrow v=12$ | A1 | One correct value |
| or $v=-4$ | A1 | Both correct values |
## Question 1 alt1:
| Working/Answer | Mark | Guidance Notes |
|---|---|---|
| Initial momentum $=(6\mathbf{i}+9\mathbf{j})$ Ns | M1 | Impulse momentum triangle. Accept $\sqrt{117}$ Ns |
| $\cos\alpha=\dfrac{6}{\sqrt{117}}\left(=\dfrac{2}{\sqrt{13}}\right)$ | A1 | Or equivalent |
| $m^2+117-2m\sqrt{117}\cos\alpha=225$ | M1 | Use of cosine formula (final momentum $m$) |
| $m^2-12m-108=0$ | A1 | Or equivalent |
| $\Rightarrow m=-6$ or $m=18$ | A1 | |
| $\Rightarrow v=12$ | A1 | One correct value |
| or $v=-4$ | A1 | Both correct values |
## Question 1 alt2:
| Working/Answer | Mark | Guidance Notes |
|---|---|---|
| Initial momentum $=(6\mathbf{i}+9\mathbf{j})$ Ns | M1 | Impulse momentum triangle. Accept $\sqrt{117}$ Ns |
| $\sin\alpha=\dfrac{3}{\sqrt{13}}$ | A1 | Or equivalent |
| $\dfrac{15}{\sin\alpha}=\dfrac{\sqrt{117}}{\sin\theta}$ | M1 | Use of sine formula |
| $\Rightarrow\sin\theta=\dfrac{3}{5}$, $\theta=36.9°$ or $\theta=143.1°$ | A1 | |
| $\dfrac{m}{\sin 86.8}=\dfrac{15}{\sin\alpha}$ or $\dfrac{m}{\sin 19.4}=\dfrac{15}{\sin(180-\alpha)}$ | A1 | Correct equation in $m$ |
| $\Rightarrow v=12$ | A1 | One correct value |
| or $v=-4$ | A1 | Both correct values |
---\begin{enumerate}
\item A particle $P$ of mass 1.5 kg is moving with velocity $( 4 \mathbf { i } + 6 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ when it receives an impulse of magnitude 15Ns. Immediately after $P$ receives the impulse, the velocity of $P$ is $\boldsymbol { v } \mathrm { m } \mathrm { s } ^ { - 1 }$.
\end{enumerate}
Find the two possible values of $v$.\\
\hfill \mbox{\textit{Edexcel M2 2021 Q1 [7]}}