| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | When moving parallel to given vector |
| Difficulty | Standard +0.3 This is a straightforward mechanics question requiring differentiation of position vectors twice for part (a), then equating velocity direction to a given vector for part (b). Both parts use standard M2 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(\mathbf{v} = \frac{d\mathbf{r}}{dt}\) | M1 | Powers going down by 1. At least 2 powers going down |
| \(\mathbf{v} = (3t^2 - 8)\mathbf{i} + (t^2 - 2t + 2)\mathbf{j}\) | A1 | Any equivalent form |
| Use of \(\mathbf{a} = \frac{d\mathbf{v}}{dt}\) | M1 | Powers going down by 1. At least 2 powers going down |
| \(\mathbf{a} = 6t\mathbf{i} + (2t-2)\mathbf{j}\) | A1 | Any equivalent form |
| \(= 24\mathbf{i} + 6\mathbf{j}\ (\text{ms}^{-2})\) | A1 | Must see acceleration stated as a correct simplified vector. ISW |
| Total [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Direction \(2\mathbf{i} + \mathbf{j}\) | M1 | Form equation in \(t\) or \(T\) only using direction. Condone use of 2 on the wrong side. Using their \(\mathbf{v}\) |
| \(\Rightarrow (3T^2 - 8) = 2(T^2 - 2T + 2)\), \((T^2 + 4T - 12 = 0)\) | A1ft | Correct unsimplified equation in \(t\) or \(T\). Solving not required for M1. Follow their \(\mathbf{v}\): \(\mathbf{i}\) component \(= 2(\mathbf{j}\) component\()\) |
| \(T = 2\) | A1 | Only. Do not need to see method of solution |
| Total [3] |
## Question 1:
**Part 1a:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $\mathbf{v} = \frac{d\mathbf{r}}{dt}$ | M1 | Powers going down by 1. At least 2 powers going down |
| $\mathbf{v} = (3t^2 - 8)\mathbf{i} + (t^2 - 2t + 2)\mathbf{j}$ | A1 | Any equivalent form |
| Use of $\mathbf{a} = \frac{d\mathbf{v}}{dt}$ | M1 | Powers going down by 1. At least 2 powers going down |
| $\mathbf{a} = 6t\mathbf{i} + (2t-2)\mathbf{j}$ | A1 | Any equivalent form |
| $= 24\mathbf{i} + 6\mathbf{j}\ (\text{ms}^{-2})$ | A1 | Must see acceleration stated as a correct simplified vector. ISW |
| **Total [5]** | | |
**Part 1b:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Direction $2\mathbf{i} + \mathbf{j}$ | M1 | Form equation in $t$ or $T$ only using direction. Condone use of 2 on the wrong side. Using their $\mathbf{v}$ |
| $\Rightarrow (3T^2 - 8) = 2(T^2 - 2T + 2)$, $(T^2 + 4T - 12 = 0)$ | A1ft | Correct unsimplified equation in $t$ or $T$. Solving not required for M1. Follow their $\mathbf{v}$: $\mathbf{i}$ component $= 2(\mathbf{j}$ component$)$ |
| $T = 2$ | A1 | Only. Do not need to see method of solution |
| **Total [3]** | | |
---\begin{enumerate}
\item At time $t$ seconds, $t \geqslant 0$, a particle $P$ has position vector $\mathbf { r }$ metres with respect to a fixed origin $O$, where
\end{enumerate}
$$\mathbf { r } = \left( t ^ { 3 } - 8 t \right) \mathbf { i } + \left( \frac { 1 } { 3 } t ^ { 3 } - t ^ { 2 } + 2 t \right) \mathbf { j }$$
(a) Find the acceleration of $P$ when $t = 4$
At time $T$ seconds, $T \geqslant 0 , P$ is moving in the direction of ( $2 \mathbf { i } + \mathbf { j }$ )\\
(b) Find the value of $T$
\hfill \mbox{\textit{Edexcel M2 2022 Q1 [8]}}