| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | 3D vector motion problems |
| Difficulty | Standard +0.3 This is a straightforward multi-part mechanics question requiring standard integration and differentiation of vectors. Part (a) involves routine integration with initial conditions, (b) is direct differentiation, (c)-(d) use standard calculus to minimize magnitude, and (e) applies F=ma. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums3.02f Non-uniform acceleration: using differentiation and integration3.03b Newton's first law: equilibrium3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\mathbf{r} = \int \mathbf{v}\, dt\) | M1 | M1 for at least one term correct |
| \(= (t^4 - 6t^2 + 3t)\mathbf{i} + 5t\mathbf{j} + 4t^2\mathbf{k} + \mathbf{c}\) | A1m1 | m1 for \(+ \mathbf{c}\) |
| When \(t=0\), \(\mathbf{r} = -5\mathbf{i}+6\mathbf{k}\) \(\therefore \mathbf{c} = -5\mathbf{i}+6\mathbf{k}\) | ||
| \(\therefore \mathbf{r} = (t^4 - 6t^2 + 3t - 5)\mathbf{i} + 5t\mathbf{j} + (6+4t^2)\mathbf{k}\) | A1 | Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\mathbf{a} = (12t^2 - 12)\mathbf{i} + 8\mathbf{k}\) | M1A1 | M1 for either component. Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Magnitude is \(\left\{(12t^2-12)^2 + 64\right\}^{\frac{1}{2}}\) | M1, A1F | Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Magnitude is a minimum when \(12t^2 - 12\) is zero | M1 | M1 for correct differentiation of correct expression in (c) |
| ie when \(t = 1\) | A1 | Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Minimum acceleration is \(8\); Using \(F = ma\) | M1 | \(a\) could be a vector |
| \(F = 7 \times 8 = 56\) | A1 | CAO. Total: 2 |
## Question 4:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\mathbf{r} = \int \mathbf{v}\, dt$ | M1 | M1 for at least one term correct |
| $= (t^4 - 6t^2 + 3t)\mathbf{i} + 5t\mathbf{j} + 4t^2\mathbf{k} + \mathbf{c}$ | A1m1 | m1 for $+ \mathbf{c}$ |
| When $t=0$, $\mathbf{r} = -5\mathbf{i}+6\mathbf{k}$ $\therefore \mathbf{c} = -5\mathbf{i}+6\mathbf{k}$ | | |
| $\therefore \mathbf{r} = (t^4 - 6t^2 + 3t - 5)\mathbf{i} + 5t\mathbf{j} + (6+4t^2)\mathbf{k}$ | A1 | **Total: 4** |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\mathbf{a} = (12t^2 - 12)\mathbf{i} + 8\mathbf{k}$ | M1A1 | M1 for either component. **Total: 2** |
### Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| Magnitude is $\left\{(12t^2-12)^2 + 64\right\}^{\frac{1}{2}}$ | M1, A1F | **Total: 2** |
### Part (d):
| Working | Marks | Guidance |
|---------|-------|----------|
| Magnitude is a minimum when $12t^2 - 12$ is zero | M1 | M1 for correct differentiation of correct expression in (c) |
| ie when $t = 1$ | A1 | **Total: 2** |
### Part (e):
| Working | Marks | Guidance |
|---------|-------|----------|
| Minimum acceleration is $8$; Using $F = ma$ | M1 | $a$ could be a vector |
| $F = 7 \times 8 = 56$ | A1 | CAO. **Total: 2** |
---4 A particle moves so that at time $t$ seconds its velocity $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$ is given by
$$\mathbf { v } = \left( 4 t ^ { 3 } - 12 t + 3 \right) \mathbf { i } + 5 \mathbf { j } + 8 t \mathbf { k }$$
\begin{enumerate}[label=(\alph*)]
\item When $t = 0$, the position vector of the particle is $( - 5 \mathbf { i } + 6 \mathbf { k } )$ metres.
Find the position vector of the particle at time $t$.
\item Find the acceleration of the particle at time $t$.
\item Find the magnitude of the acceleration of the particle at time $t$. Do not simplify your answer.
\item Hence find the time at which the magnitude of the acceleration is a minimum.
\item The particle is moving under the action of a single variable force $\mathbf { F }$ newtons. The mass of the particle is 7 kg .
Find the minimum magnitude of $\mathbf { F }$.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2010 Q4 [12]}}