| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Find force using F=ma |
| Difficulty | Standard +0.3 This is a straightforward M2 mechanics question requiring differentiation of velocity to find force (F=ma), then integration to find position. The trigonometric functions are standard (sin, cos) with simple coefficients, and all steps follow directly from applying learned formulas. Slightly easier than average due to the routine nature of the calculus and lack of problem-solving insight required. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07k Differentiate trig: sin(kx), cos(kx), tan(kx)3.02f Non-uniform acceleration: using differentiation and integration3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{a} = \frac{d\mathbf{v}}{dt} = -8\sin 2t\, \mathbf{i} + 3\cos t\, \mathbf{j}\) | M1 A1 | Differentiate velocity vector |
| \(\mathbf{F} = m\mathbf{a} = 4(-8\sin 2t\, \mathbf{i} + 3\cos t\, \mathbf{j})\) | M1 | Use \(\mathbf{F} = m\mathbf{a}\) with \(m = 4\) |
| \(\mathbf{F} = (-32\sin 2t\, \mathbf{i} + 12\cos t\, \mathbf{j})\) N | A1 | Correct final expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| At \(t = \pi\): \(\mathbf{F} = (-32\sin 2\pi)\mathbf{i} + (12\cos\pi)\mathbf{j} = 0\mathbf{i} - 12\mathbf{j}\) | M1 | Substitute \(t = \pi\) |
| \( | \mathbf{F} | = 12\) N |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{r} = \int \mathbf{v}\, dt = \frac{4\sin 2t}{2}\mathbf{i} - 3\cos t\, \mathbf{j} + \mathbf{c}\) | M1 A1 | Integrate velocity vector, at least one term correct |
| \(= 2\sin 2t\, \mathbf{i} - 3\cos t\, \mathbf{j} + \mathbf{c}\) | ||
| At \(t=0\): \(\mathbf{r} = 2\mathbf{i} - 14\mathbf{j}\), so \(0\mathbf{i} - 3\mathbf{j} + \mathbf{c} = 2\mathbf{i} - 14\mathbf{j}\) | M1 | Apply initial condition |
| \(\mathbf{c} = 2\mathbf{i} - 11\mathbf{j}\) | A1 | |
| \(\mathbf{r} = (2\sin 2t + 2)\mathbf{i} + (-3\cos t - 11)\mathbf{j}\) | A1 | Correct final position vector |
# Question 1:
## Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{a} = \frac{d\mathbf{v}}{dt} = -8\sin 2t\, \mathbf{i} + 3\cos t\, \mathbf{j}$ | M1 A1 | Differentiate velocity vector |
| $\mathbf{F} = m\mathbf{a} = 4(-8\sin 2t\, \mathbf{i} + 3\cos t\, \mathbf{j})$ | M1 | Use $\mathbf{F} = m\mathbf{a}$ with $m = 4$ |
| $\mathbf{F} = (-32\sin 2t\, \mathbf{i} + 12\cos t\, \mathbf{j})$ N | A1 | Correct final expression |
## Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| At $t = \pi$: $\mathbf{F} = (-32\sin 2\pi)\mathbf{i} + (12\cos\pi)\mathbf{j} = 0\mathbf{i} - 12\mathbf{j}$ | M1 | Substitute $t = \pi$ |
| $|\mathbf{F}| = 12$ N | A1 | Correct magnitude |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{r} = \int \mathbf{v}\, dt = \frac{4\sin 2t}{2}\mathbf{i} - 3\cos t\, \mathbf{j} + \mathbf{c}$ | M1 A1 | Integrate velocity vector, at least one term correct |
| $= 2\sin 2t\, \mathbf{i} - 3\cos t\, \mathbf{j} + \mathbf{c}$ | | |
| At $t=0$: $\mathbf{r} = 2\mathbf{i} - 14\mathbf{j}$, so $0\mathbf{i} - 3\mathbf{j} + \mathbf{c} = 2\mathbf{i} - 14\mathbf{j}$ | M1 | Apply initial condition |
| $\mathbf{c} = 2\mathbf{i} - 11\mathbf{j}$ | A1 | |
| $\mathbf{r} = (2\sin 2t + 2)\mathbf{i} + (-3\cos t - 11)\mathbf{j}$ | A1 | Correct final position vector |
---1 A particle, of mass 4 kg , moves in a horizontal plane under the action of a single force, $\mathbf { F }$ newtons. The unit vectors $\mathbf { i }$ and $\mathbf { j }$ are in the horizontal plane, perpendicular to each other.
At time $t$ seconds, the velocity of the particle, $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$, is given by
$$\mathbf { v } = 4 \cos 2 t \mathbf { i } + 3 \sin t \mathbf { j }$$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find an expression for the force, $\mathbf { F }$, acting on the particle at time $t$ seconds.
\item Find the magnitude of $\mathbf { F }$ when $t = \pi$.
\end{enumerate}\item When $t = 0$, the particle is at the point with position vector $( 2 \mathbf { i } - 14 \mathbf { j } )$ metres. Find the position vector, $\mathbf { r }$ metres, of the particle at time $t$ seconds.\\[0pt]
[5 marks]
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{691c50b4-50b2-4e3a-a7e0-60f8ec35ee3c-02_1346_1717_1361_150}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2015 Q1 [10]}}