| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Acceleration from velocity differentiation |
| Difficulty | Moderate -0.8 This is a straightforward calculus mechanics question requiring routine differentiation of an exponential function for acceleration, basic analysis of the resulting expression, and integration for displacement. All techniques are standard M2 procedures with no problem-solving insight needed, making it easier than average but not trivial due to the exponential terms. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Height of \(P\) above \(Q\): \(h = 4 - 4\cos60° = 4 - 2 = 2\) m | M1 | Correct height difference |
| \(KE = mgh = 32 \times 9.8 \times 2\) | M1 | Use of energy conservation |
| \(KE = 627.2\) J | A1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}mv^2 = 627.2\) | M1 | |
| \(v = \sqrt{\frac{2 \times 627.2}{32}} = 6.26\) ms\(^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F = \mu R = \mu \times 32g\) | M1 | Normal reaction on horizontal surface |
| Work done by friction \(= \mu \times 32g \times 5\) | M1 | |
| \(627.2 - \mu \times 32 \times 9.8 \times 5 = 0\) | A1 | Energy equation |
| \(\mu = \frac{627.2}{1568} = 0.4\) | A1 |
# Question 3:
## Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Height of $P$ above $Q$: $h = 4 - 4\cos60° = 4 - 2 = 2$ m | M1 | Correct height difference |
| $KE = mgh = 32 \times 9.8 \times 2$ | M1 | Use of energy conservation |
| $KE = 627.2$ J | A1 | Correct value |
## Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 = 627.2$ | M1 | |
| $v = \sqrt{\frac{2 \times 627.2}{32}} = 6.26$ ms$^{-1}$ | A1 | |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F = \mu R = \mu \times 32g$ | M1 | Normal reaction on horizontal surface |
| Work done by friction $= \mu \times 32g \times 5$ | M1 | |
| $627.2 - \mu \times 32 \times 9.8 \times 5 = 0$ | A1 | Energy equation |
| $\mu = \frac{627.2}{1568} = 0.4$ | A1 | |
---3 A particle moves in a straight line and at time $t$ has velocity $v$, where
$$v = 2 t - 12 \mathrm { e } ^ { - t } , \quad t \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find an expression for the acceleration of the particle at time $t$.
\item State the range of values of the acceleration of the particle.
\end{enumerate}\item When $t = 0$, the particle is at the origin.
Find an expression for the displacement of the particle from the origin at time $t$.\\
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA M2 Q3}}