| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2022 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Find position by integrating velocity |
| Difficulty | Standard +0.3 This is a straightforward M2 mechanics question requiring standard techniques: (a) finding when j-component of velocity equals zero, then differentiating to find acceleration; (b) integrating velocity components and applying initial conditions. While it involves integration of √(5-t) and a cubic, these are routine A-level techniques with no novel problem-solving required. Slightly easier than average due to clear structure and standard methods. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.10b Vectors in 3D: i,j,k notation3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\lambda^2 + 2\lambda - 3 = 0\ (=(\lambda+3)(\lambda-1))\) | M1 | Set \(\mathbf{j}\) component \(= 0\) and solve for \(\lambda\) |
| \(\Rightarrow \lambda = 1\) | A1 | Only. Seen or implied. Accept \(t=1\) |
| Use \(\mathbf{a} = \dfrac{d\mathbf{v}}{dt}\) | M1 | Attempt derivative of both components with respect to \(t\). Powers going down. Condone errors in dealing with signs/indices for the square root. Answer must be a vector |
| \(= \dfrac{-1}{2\sqrt{5-t}}\mathbf{i} + (2t+2)\mathbf{j}\) | A1 | Any equivalent form |
| \(= -\dfrac{1}{4}\mathbf{i} + 4\mathbf{j}\) | A1 | Only. Any equivalent form. ISW if they go on to find the magnitude |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use \(\mathbf{s} = \int \mathbf{v}\, dt\) | M1 | Attempt integral of both components. (M0 if they have assumed that one component is zero.) Powers going up. Condone errors in dealing with signs/indices for the square root |
| \(\mathbf{s} = \left(-\dfrac{2}{3}(5-t)^{\frac{3}{2}}(+A)\right)\mathbf{i} + \left(\dfrac{1}{3}t^3 + t^2 - 3t(+B)\right)\mathbf{j}\) | A1 | Unsimplified expression with error in at most one term |
| A1 | Correct unsimplified expression. Allow with no constant(s) of integration | |
| Use \(t=1\), \(\mathbf{s} = -2\mathbf{i}+\mathbf{j}\) | DM1 | Use of initial condition to find constant(s) of integration. Dependent on the previous M1 |
| \(\mathbf{s} = \left(-\dfrac{2}{3}(5-T)^{\frac{3}{2}}+\dfrac{10}{3}\right)\mathbf{i}+\left(\dfrac{1}{3}T^3+T^2-3T+\dfrac{8}{3}\right)\mathbf{j}\) | A1 | Any equivalent form for the position vector |
| Total: 5 | ||
| Question Total: (10) |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\lambda^2 + 2\lambda - 3 = 0\ (=(\lambda+3)(\lambda-1))$ | M1 | Set $\mathbf{j}$ component $= 0$ and solve for $\lambda$ |
| $\Rightarrow \lambda = 1$ | A1 | Only. Seen or implied. Accept $t=1$ |
| Use $\mathbf{a} = \dfrac{d\mathbf{v}}{dt}$ | M1 | Attempt derivative of both components with respect to $t$. Powers going down. Condone errors in dealing with signs/indices for the square root. Answer must be a vector |
| $= \dfrac{-1}{2\sqrt{5-t}}\mathbf{i} + (2t+2)\mathbf{j}$ | A1 | Any equivalent form |
| $= -\dfrac{1}{4}\mathbf{i} + 4\mathbf{j}$ | A1 | Only. Any equivalent form. ISW if they go on to find the magnitude |
| **Total: 5** | | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $\mathbf{s} = \int \mathbf{v}\, dt$ | M1 | Attempt integral of both components. (M0 if they have assumed that one component is zero.) Powers going up. Condone errors in dealing with signs/indices for the square root |
| $\mathbf{s} = \left(-\dfrac{2}{3}(5-t)^{\frac{3}{2}}(+A)\right)\mathbf{i} + \left(\dfrac{1}{3}t^3 + t^2 - 3t(+B)\right)\mathbf{j}$ | A1 | Unsimplified expression with error in at most one term |
| | A1 | Correct unsimplified expression. Allow with no constant(s) of integration |
| Use $t=1$, $\mathbf{s} = -2\mathbf{i}+\mathbf{j}$ | DM1 | Use of initial condition to find constant(s) of integration. Dependent on the previous M1 |
| $\mathbf{s} = \left(-\dfrac{2}{3}(5-T)^{\frac{3}{2}}+\dfrac{10}{3}\right)\mathbf{i}+\left(\dfrac{1}{3}T^3+T^2-3T+\dfrac{8}{3}\right)\mathbf{j}$ | A1 | Any equivalent form for the position vector |
| **Total: 5** | | |
| **Question Total: (10)** | | |
---4. At time $t$ seconds $( 0 \leqslant t < 5 )$, a particle $P$ has velocity $\mathbf { v m s } ^ { - 1 }$, where
$$\mathbf { v } = ( \sqrt { 5 - t } ) \mathbf { i } + \left( t ^ { 2 } + 2 t - 3 \right) \mathbf { j }$$
When $t = \lambda$, particle $P$ is moving in a direction parallel to the vector $\mathbf { i }$.
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of $P$ when $t = \lambda$
The position vector of $P$ is measured relative to the fixed point $O$ When $t = 1$, the position vector of $P$ is $( - 2 \mathbf { i } + \mathbf { j } ) \mathrm { m }$.
Given that $1 \leqslant T < 5$
\item find, in terms of $T$, the position vector of $P$ when $t = T$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2022 Q4 [10]}}