| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2020 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Impulse from velocity change |
| Difficulty | Moderate -0.3 This is a straightforward M2 impulse-momentum question requiring application of the impulse-momentum equation in vector form, solving simultaneous equations (one linear from direction constraint, one from magnitude), and using Pythagoras. It's slightly easier than average because the method is standard and the algebra is manageable, though the two-solution aspect adds minor complexity. |
| Spec | 6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{I} = 2[\lambda\mathbf{i} + \lambda\mathbf{j} - 5\mathbf{i} - 3\mathbf{j}]\) | M1 | Use of \(\mathbf{I} = m(\mathbf{v} - \mathbf{u})\) |
| \(= 2(\lambda-5)\mathbf{i} + 2(\lambda-3)\mathbf{j}\) | A1 | Any equivalent form |
| \( | I | = \sqrt{40} \Rightarrow (\lambda-5)^2 + (\lambda-3)^2 = 10\) |
| \(\lambda^2 - 8\lambda + 12 = 0 \Rightarrow \lambda = 2\) or \(\lambda = 6\) | DM1 | Solve to find both values for \(\lambda\). Dependent on the 2 preceding M marks |
| \(\mathbf{I} = -6\mathbf{i} - 2\mathbf{j}\) or \(\mathbf{I} = 2\mathbf{i} + 6\mathbf{j}\), \((a=-6, b=-2\) or \(a=2, b=6)\) | A1 | And no others |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{I}(= a\mathbf{i}+b\mathbf{j}) = 2(\mathbf{v}-(5\mathbf{i}+3\mathbf{j}))\) | M1A1 | |
| \(\mathbf{v} = \frac{a+10}{2}\mathbf{i} + \frac{b+6}{2}\mathbf{j} \Rightarrow (a+10 = b+6)\) | ||
| \(a^2 + b^2 = 40 \Rightarrow b^2 - 4b - 12 = 0\) or \(a^2 + 4a - 12 = 0\) | M1 | Correct use of Pythagoras and impulse to form an equation in \(a\) or \(b\). Any equivalent form |
| \(b^2 - 4b - 12 = 0 \Rightarrow b = 6\) or \(b = -2\) | DM1 | |
| \(\mathbf{I} = -6\mathbf{i} - 2\mathbf{j}\) or \(\mathbf{I} = 2\mathbf{i} + 6\mathbf{j}\) | A1 | Or simplified equivalent |
# Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{I} = 2[\lambda\mathbf{i} + \lambda\mathbf{j} - 5\mathbf{i} - 3\mathbf{j}]$ | M1 | Use of $\mathbf{I} = m(\mathbf{v} - \mathbf{u})$ |
| $= 2(\lambda-5)\mathbf{i} + 2(\lambda-3)\mathbf{j}$ | A1 | Any equivalent form |
| $|I| = \sqrt{40} \Rightarrow (\lambda-5)^2 + (\lambda-3)^2 = 10$ | M1 | Correct use of Pythagoras and their impulse to form an equation in $\lambda$ |
| $\lambda^2 - 8\lambda + 12 = 0 \Rightarrow \lambda = 2$ or $\lambda = 6$ | DM1 | Solve to find both values for $\lambda$. Dependent on the 2 preceding M marks |
| $\mathbf{I} = -6\mathbf{i} - 2\mathbf{j}$ or $\mathbf{I} = 2\mathbf{i} + 6\mathbf{j}$, $(a=-6, b=-2$ or $a=2, b=6)$ | A1 | And no others |
**Alternative working:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{I}(= a\mathbf{i}+b\mathbf{j}) = 2(\mathbf{v}-(5\mathbf{i}+3\mathbf{j}))$ | M1A1 | |
| $\mathbf{v} = \frac{a+10}{2}\mathbf{i} + \frac{b+6}{2}\mathbf{j} \Rightarrow (a+10 = b+6)$ | | |
| $a^2 + b^2 = 40 \Rightarrow b^2 - 4b - 12 = 0$ or $a^2 + 4a - 12 = 0$ | M1 | Correct use of Pythagoras and impulse to form an equation in $a$ or $b$. Any equivalent form |
| $b^2 - 4b - 12 = 0 \Rightarrow b = 6$ or $b = -2$ | DM1 | |
| $\mathbf{I} = -6\mathbf{i} - 2\mathbf{j}$ or $\mathbf{I} = 2\mathbf{i} + 6\mathbf{j}$ | A1 | Or simplified equivalent |
---\begin{enumerate}
\item A particle of mass 2 kg is moving with velocity $( 5 \mathbf { i } + 3 \mathbf { j } ) \mathrm { ms } ^ { - 1 }$ when it receives an impulse $\mathbf { I N }$ s, such that $\mathbf { I } = a \mathbf { i } + b \mathbf { j }$
\end{enumerate}
Immediately after receiving the impulse, the particle is moving with velocity $\lambda ( \mathbf { i } + \mathbf { j } ) \mathrm { ms } ^ { - 1 }$, where $\lambda$ is a constant.
Given that the magnitude of $\mathbf { I }$ is $\sqrt { 40 }$, find the two possible impulses.\\
(5)\\
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\hfill \mbox{\textit{Edexcel M2 2020 Q1 [5]}}