Standard +0.3 This is a standard M2 impulse-momentum question requiring vector manipulation and solving a quadratic equation. Students must apply the impulse-momentum theorem in 2D, use the magnitude constraint, and solve the resulting quadratic. While it involves multiple steps, the techniques are routine for M2 students with no novel insight required, making it slightly easier than average.
3. A particle \(P\) of mass 0.5 kg is moving with velocity \(\lambda ( \mathbf { i } + \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\) when \(P\) receives an impulse of magnitude \(\sqrt { \frac { 5 } { 2 } } \mathrm { Ns }\)
Immediately after \(P\) receives the impulse, the velocity of \(P\) is \(4 \mathbf { i } \mathrm {~ms} ^ { - 1 }\) Given that \(\lambda\) is a constant, find the two possible values of \(\lambda\)
Form a 3 term quadratic (seen or implied). Not necessarily stated "\(= 0\)". From \(\mathbf{I} = a\mathbf{i}+b\mathbf{j}\) can obtain \(4a^2-8a+3=0\) or \(4b^2+8b+3=0\). Dependent on preceding M1. Solving not required for the M1
\(\lambda = 3\) and \(\lambda = 1\)
A1cso
From correct solution only. Do not need to see method of solution
Total [6]
Part 3alt:
Answer
Marks
Guidance
Answer/Working
Mark
Guidance
Use of \(\mathbf{I} = m\mathbf{v} - m\mathbf{u}\) to form a vector triangle
M1
Triangle with sides of length \(\sqrt{\frac{5}{2}},\ \
2\mathbf{i}\
\) and \(\left
Use of cosine rule with \(45°\ \left(\frac{\pi}{4}\right)\)
Form a 3 term quadratic (seen or implied). Dependent on preceding M1
\(\lambda = 3\) and \(\lambda = 1\)
A1
Correct solution only
Total [6]
## Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $\mathbf{I} = m\mathbf{v} - m\mathbf{u}$ | M1 | Must be subtracting |
| $(\mathbf{I} =) \pm 0.5((4-\lambda)\mathbf{i} + (-\lambda)\mathbf{j})$ | A1 | Accept $\pm$ correct unsimplified expression on right hand side. Allow $2\mathbf{i} - \frac{\lambda}{2}(\mathbf{i}+\mathbf{j})$ or equivalent |
| Use of magnitude to form an equation in one variable | M1 | Correct use of Pythagoras |
| $\frac{5}{2} = \frac{1}{4}\left((4-\lambda)^2 + (-\lambda)^2\right)$ | A1ft | Follow their $\mathbf{I}$ |
| $0 = 2\lambda^2 - 8\lambda + 6\ \left(= (2\lambda-6)(\lambda-1)\right)$ | DM1 | Form a 3 term quadratic (seen or implied). Not necessarily stated "$= 0$". From $\mathbf{I} = a\mathbf{i}+b\mathbf{j}$ can obtain $4a^2-8a+3=0$ or $4b^2+8b+3=0$. Dependent on preceding M1. Solving not required for the M1 |
| $\lambda = 3$ and $\lambda = 1$ | A1cso | From correct solution only. Do not need to see method of solution |
| **Total [6]** | | |
**Part 3alt:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $\mathbf{I} = m\mathbf{v} - m\mathbf{u}$ to form a vector triangle | M1 | |
| Triangle with sides of length $\sqrt{\frac{5}{2}},\ \|2\mathbf{i}\|$ and $\left|\frac{\lambda}{2}(\mathbf{i}+\mathbf{j})\right|$ | A1 | |
| Use of cosine rule with $45°\ \left(\frac{\pi}{4}\right)$ | M1 | |
| $\frac{5}{2} = 2^2 + \left(\frac{\lambda}{2}\right)^2 \times 2 - 2 \times 2 \times \frac{\lambda}{2}\sqrt{2}\cos 45°$ | A1ft | Correct unsimplified equation. Follow their magnitudes |
| $0 = \lambda^2 - 4\lambda + 3\ \left(= (\lambda-3)(\lambda-1)\right)$ | DM1 | Form a 3 term quadratic (seen or implied). Dependent on preceding M1 |
| $\lambda = 3$ and $\lambda = 1$ | A1 | Correct solution only |
| **Total [6]** | | |
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3. A particle $P$ of mass 0.5 kg is moving with velocity $\lambda ( \mathbf { i } + \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ when $P$ receives an impulse of magnitude $\sqrt { \frac { 5 } { 2 } } \mathrm { Ns }$
Immediately after $P$ receives the impulse, the velocity of $P$ is $4 \mathbf { i } \mathrm {~ms} ^ { - 1 }$ Given that $\lambda$ is a constant, find the two possible values of $\lambda$\\
\hfill \mbox{\textit{Edexcel M2 2022 Q3 [6]}}