| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Find force using F=ma |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question requiring standard integration of velocity to find position (with initial conditions), differentiation to find acceleration, and F=ma with magnitude calculation. All techniques are routine M2 procedures with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.08h Integration by substitution1.10b Vectors in 3D: i,j,k notation3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Integrate v with respect to \(t\) | M1 | Attempt integration of at least one component |
| \(\mathbf{r} = (4t + t^3)\mathbf{i} + (12t - 4t^2)\mathbf{j} + \mathbf{c}\) | A1 | Correct integration, ignore \(\mathbf{c}\) at this stage |
| Use \(t = 0\), \(\mathbf{r} = 5\mathbf{i} - 7\mathbf{j}\) to find \(\mathbf{c}\) | M1 | Substituting initial conditions |
| \(\mathbf{r} = (4t + t^3 + 5)\mathbf{i} + (12t - 4t^2 - 7)\mathbf{j}\) | A1 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Differentiate v with respect to \(t\) | M1 | Attempt differentiation of at least one component |
| \(\mathbf{a} = 6t\mathbf{i} - 8\mathbf{j}\) | A1 | Both components correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| At \(t = 1\): \(\mathbf{a} = 6\mathbf{i} - 8\mathbf{j}\) | B1 | Correct acceleration at \(t = 1\) |
| \(\mathbf{F} = m\mathbf{a} = 2(6\mathbf{i} - 8\mathbf{j}) = 12\mathbf{i} - 16\mathbf{j}\) | M1 | Use of \(F = ma\) with \(m = 2\) |
| \( | \mathbf{F} | = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400}\) |
| \( | \mathbf{F} | = 20\) N |
# Question 1:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Integrate **v** with respect to $t$ | M1 | Attempt integration of at least one component |
| $\mathbf{r} = (4t + t^3)\mathbf{i} + (12t - 4t^2)\mathbf{j} + \mathbf{c}$ | A1 | Correct integration, ignore $\mathbf{c}$ at this stage |
| Use $t = 0$, $\mathbf{r} = 5\mathbf{i} - 7\mathbf{j}$ to find $\mathbf{c}$ | M1 | Substituting initial conditions |
| $\mathbf{r} = (4t + t^3 + 5)\mathbf{i} + (12t - 4t^2 - 7)\mathbf{j}$ | A1 | Correct final answer |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Differentiate **v** with respect to $t$ | M1 | Attempt differentiation of at least one component |
| $\mathbf{a} = 6t\mathbf{i} - 8\mathbf{j}$ | A1 | Both components correct |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| At $t = 1$: $\mathbf{a} = 6\mathbf{i} - 8\mathbf{j}$ | B1 | Correct acceleration at $t = 1$ |
| $\mathbf{F} = m\mathbf{a} = 2(6\mathbf{i} - 8\mathbf{j}) = 12\mathbf{i} - 16\mathbf{j}$ | M1 | Use of $F = ma$ with $m = 2$ |
| $|\mathbf{F}| = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400}$ | M1 | Correct method for magnitude |
| $|\mathbf{F}| = 20$ N | A1 | Correct final answer |
---1 The velocity of a particle at time $t$ seconds is $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$, where
$$\mathbf { v } = \left( 4 + 3 t ^ { 2 } \right) \mathbf { i } + ( 12 - 8 t ) \mathbf { j }$$
\begin{enumerate}[label=(\alph*)]
\item When $t = 0$, the particle is at the point with position vector $( 5 \mathbf { i } - 7 \mathbf { j } ) \mathrm { m }$.
Find the position vector, $\mathbf { r }$ metres, of the particle at time $t$.
\item Find the acceleration of the particle at time $t$.
\item The particle has mass 2 kg .
Find the magnitude of the force acting on the particle when $t = 1$.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2011 Q1 [10]}}