| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Velocity from acceleration by integration |
| Difficulty | Standard +0.3 This is a straightforward M3 mechanics question involving Newton's second law with a time-varying force. Part (i) requires integrating F=ma to find velocity (standard technique), then finding max/min of a simple trigonometric function. Part (ii) involves integrating velocity to find displacement, then calculating average velocity. All steps are routine applications of calculus and mechanics principles with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(7.5\sin 2t = v (+c)\) oe | M1*, *M1 | Use of \(F = ma\); condone wrong / missing 0.2 and wrong sign. Attempt to integrate, one side correct; condone missing \(c\), condone missing \(c\). |
| Answer | Marks | Guidance |
|---|---|---|
| 11.5 (ms⁻¹) and -3.5 (ms⁻¹) | A1, A1 4 | CAO. Depends on both M marks and fully correct working |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = -\frac{15}{4}\cos 2t + 4t (+c)\) | M1*, *M1 | Ft if (i) has sin or cos term |
| Answer | Marks | Guidance |
|---|---|---|
| 8.77 (ms⁻¹) | *M1, A1 4 | {\(x(\frac{π}{2}) - x(\pi)\); found; CAO Accept \(\frac{15}{π} + 4\) |
## (i)
$3\cos 2t = 0.2 \frac{dv}{dt}$
$7.5\sin 2t = v (+c)$ oe | M1*, *M1 | Use of $F = ma$; condone wrong / missing 0.2 and wrong sign. Attempt to integrate, one side correct; condone missing $c$, condone missing $c$.
$v = 7.5 \sin 2t + 4$
11.5 (ms⁻¹) and -3.5 (ms⁻¹) | A1, A1 4 | CAO. Depends on both M marks and fully correct working
## (ii)
$x = -\frac{15}{4}\cos 2t + 4t (+c)$ | M1*, *M1 | Ft if (i) has sin or cos term
Ave speed is their distance $\frac{π}{4}$
$\left(\frac{15}{4} + 6\pi\right) - \left(-\frac{15}{4} + 4\pi\right)$
8.77 (ms⁻¹) | *M1, A1 4 | {$x(\frac{π}{2}) - x(\pi)$; found; CAO Accept $\frac{15}{π} + 4$ | No need to find $c$ (3.75). $\left(\frac{15}{2} + 6\pi\right) - (+4\pi)$
---A particle $Q$ of mass $0.2$ kg is projected horizontally with velocity $4$ m s$^{-1}$ from a fixed point $A$ on a smooth horizontal surface. At time $t$ s after projection $Q$ is $x$ m from $A$ and is moving away from $A$ with velocity $v$ m s$^{-1}$. There is a force of $3\cos 2t$ N acting on $Q$ in the positive $x$-direction.
\begin{enumerate}[label=(\roman*)]
\item Find an expression for the velocity of $Q$ at time $t$. State the maximum and minimum values of the velocity of $Q$ as $t$ varies. [4]
\item Find the average velocity of $Q$ between times $t = \pi$ and $t = \frac{3}{2}\pi$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR M3 2016 Q2 [8]}}