| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2023 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Find velocity from position |
| Difficulty | Standard +0.3 This is a straightforward M2 mechanics question requiring differentiation of position vectors and identifying when the i-component of velocity equals zero. Part (a) involves standard calculus and solving a simple equation; part (b) is direct substitution. Slightly easier than average due to routine techniques. |
| 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 |
| Differentiate r (both components) | M1 | In each component at least one power going down by 1 |
| \(\mathbf{v} = (4t^3 - 16t)\mathbf{i} + (12t - 3\sqrt{t})\mathbf{j}\) | A1 | Accept as two separate components |
| Equate i component of v to zero and solve for \(t\) | DM1 | Dependent on first M1. Must start with a component of the vector for v. Can have more than one value at this stage. |
| Obtain \((24 - 3\sqrt{2})\mathbf{j}\ (\text{ms}^{-1})\) | A1 | Accept \(20\mathbf{j}\ (\text{ms}^{-1})\) or better. (19.757359...) Correct answer only. Answer must be a vector |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Differentiate v (both components) | M1 | For differentiating their v, even if method for obtaining it was incorrect. Their v must be a vector. In each component at least one power going down by 1 |
| \(\mathbf{a} = (12t^2 - 16)\mathbf{i} + \left(12 - \frac{3}{2}t^{-\frac{1}{2}}\right)\mathbf{j}\) | A1 | Any equivalent form for acceleration |
| Obtain \(176\mathbf{i} + \frac{45}{4}\mathbf{j}\ (\text{ms}^{-2})\) | A1 | Accept \(180\mathbf{i} + 11\mathbf{j}\ (\text{ms}^{-2})\) or better. ISW |
| Total: [3] | Overall: (7) |
# Question 1:
## Part 1a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiate **r** (both components) | M1 | In each component at least one power going down by 1 |
| $\mathbf{v} = (4t^3 - 16t)\mathbf{i} + (12t - 3\sqrt{t})\mathbf{j}$ | A1 | Accept as two separate components |
| Equate **i** component of **v** to zero and solve for $t$ | DM1 | Dependent on first M1. Must start with a component of the vector for **v**. Can have more than one value at this stage. |
| Obtain $(24 - 3\sqrt{2})\mathbf{j}\ (\text{ms}^{-1})$ | A1 | Accept $20\mathbf{j}\ (\text{ms}^{-1})$ or better. (19.757359...) Correct answer only. Answer must be a vector |
**Total: [4]**
## Part 1b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiate **v** (both components) | M1 | For differentiating their **v**, even if method for obtaining it was incorrect. Their **v** must be a vector. In each component at least one power going down by 1 |
| $\mathbf{a} = (12t^2 - 16)\mathbf{i} + \left(12 - \frac{3}{2}t^{-\frac{1}{2}}\right)\mathbf{j}$ | A1 | Any equivalent form for acceleration |
| Obtain $176\mathbf{i} + \frac{45}{4}\mathbf{j}\ (\text{ms}^{-2})$ | A1 | Accept $180\mathbf{i} + 11\mathbf{j}\ (\text{ms}^{-2})$ or better. ISW |
**Total: [3] | Overall: (7)**
---\begin{enumerate}
\item At time $t$ seconds, $t > 0$, a particle $P$ is at the point with position vector $\mathbf { r } \mathrm { m }$, where
\end{enumerate}
$$\mathbf { r } = \left( t ^ { 4 } - 8 t ^ { 2 } \right) \mathbf { i } + \left( 6 t ^ { 2 } - 2 t ^ { \frac { 3 } { 2 } } \right) \mathbf { j }$$
(a) Find the velocity of $P$ when $P$ is moving in a direction parallel to the vector $\mathbf { j }$\\
(b) Find the acceleration of $P$ when $t = 4$
\hfill \mbox{\textit{Edexcel M2 2023 Q1 [7]}}