| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Vector motion with components |
| Difficulty | Moderate -0.3 This is a straightforward vector mechanics question requiring standard differentiation of exponential and polynomial functions to find velocity and acceleration. All parts follow directly from the definitions (v = dr/dt, a = dv/dt, F = ma) with no problem-solving insight needed, though the exponential component and multiple routine calculations place it slightly below average difficulty for A-level. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02a Kinematics language: position, displacement, velocity, acceleration3.03b Newton's first law: equilibrium3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{v} = \frac{d\mathbf{r}}{dt} = \left(e^{\frac{1}{2}t} - 8\right)\mathbf{i} + (2t - 6)\mathbf{j}\) | M1 | Attempt differentiation |
| A1 | Correct i component | |
| A1 | Correct j component |
| Answer | Marks | Guidance |
|---|---|---|
| When \(t=3\): \(\mathbf{v} = (e^{1.5} - 8)\mathbf{i} + (0)\mathbf{j}\) | M1 | Substitute \(t=3\) |
| Speed \(= | e^{1.5} - 8 | = |
| Answer | Marks |
|---|---|
| Travelling west (negative i direction) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{a} = \frac{d\mathbf{v}}{dt} = \frac{1}{2}e^{\frac{1}{2}t}\mathbf{i} + 2\mathbf{j}\) | M1 | Attempt differentiation of velocity |
| When \(t=3\): \(\mathbf{a} = \frac{1}{2}e^{1.5}\mathbf{i} + 2\mathbf{j}\) | A1 | Correct i component |
| \(= 2.241\mathbf{i} + 2\mathbf{j}\) | A1 | Correct j component |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{F} = m\mathbf{a} = 7(2.241\mathbf{i} + 2\mathbf{j})\) | M1 | Use \(F = ma\) |
| \(= 15.687\mathbf{i} + 14\mathbf{j}\) | A1 | Correct force vector |
| \( | \mathbf{F} | = \sqrt{15.687^2 + 14^2} = \sqrt{246.08 + 196} = \sqrt{442.08}\) |
| \(= 21.0 \text{ N}\) | A1 | Correct answer |
# Question 3:
## Part (a):
| $\mathbf{v} = \frac{d\mathbf{r}}{dt} = \left(e^{\frac{1}{2}t} - 8\right)\mathbf{i} + (2t - 6)\mathbf{j}$ | M1 | Attempt differentiation |
|---|---|---|
| | A1 | Correct **i** component |
| | A1 | Correct **j** component |
## Part (b)(i):
| When $t=3$: $\mathbf{v} = (e^{1.5} - 8)\mathbf{i} + (0)\mathbf{j}$ | M1 | Substitute $t=3$ |
|---|---|---|
| Speed $= |e^{1.5} - 8| = |4.482 - 8| = 3.52 \text{ m s}^{-1}$ | A1 | Correct speed |
## Part (b)(ii):
| Travelling west (negative **i** direction) | B1 | |
|---|---|---|
## Part (c):
| $\mathbf{a} = \frac{d\mathbf{v}}{dt} = \frac{1}{2}e^{\frac{1}{2}t}\mathbf{i} + 2\mathbf{j}$ | M1 | Attempt differentiation of velocity |
|---|---|---|
| When $t=3$: $\mathbf{a} = \frac{1}{2}e^{1.5}\mathbf{i} + 2\mathbf{j}$ | A1 | Correct **i** component |
| $= 2.241\mathbf{i} + 2\mathbf{j}$ | A1 | Correct **j** component |
## Part (d):
| $\mathbf{F} = m\mathbf{a} = 7(2.241\mathbf{i} + 2\mathbf{j})$ | M1 | Use $F = ma$ |
|---|---|---|
| $= 15.687\mathbf{i} + 14\mathbf{j}$ | A1 | Correct force vector |
| $|\mathbf{F}| = \sqrt{15.687^2 + 14^2} = \sqrt{246.08 + 196} = \sqrt{442.08}$ | M1 | Correct magnitude method |
| $= 21.0 \text{ N}$ | A1 | Correct answer |
---3 A particle moves on a horizontal plane, in which the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are directed east and north respectively.
At time $t$ seconds, the position vector of the particle is $\mathbf { r }$ metres, where
$$\mathbf { r } = \left( 2 \mathrm { e } ^ { \frac { 1 } { 2 } t } - 8 t + 5 \right) \mathbf { i } + \left( t ^ { 2 } - 6 t \right) \mathbf { j }$$
\begin{enumerate}[label=(\alph*)]
\item Find an expression for the velocity of the particle at time $t$.
\item \begin{enumerate}[label=(\roman*)]
\item Find the speed of the particle when $t = 3$.
\item State the direction in which the particle is travelling when $t = 3$.
\end{enumerate}\item Find the acceleration of the particle when $t = 3$.
\item The mass of the particle is 7 kg .
Find the magnitude of the resultant force on the particle when $t = 3$.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2009 Q3 [12]}}