| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Particle on slope with pulley |
| Difficulty | Standard +0.3 This is a standard A-level mechanics problem involving connected particles and friction on an inclined plane. It requires resolving forces, applying Newton's second law, and understanding limiting friction, but follows well-established procedures with no novel insights needed. The multi-part structure and algebraic manipulation place it slightly above average difficulty, but it remains a typical exam question testing core mechanics concepts. |
| Spec | 3.03o Advanced connected particles: and pulleys3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| \(R = mg\cos 30\) | B1 | Resolving perpendicular to the plane |
| \(T = \frac{1}{4}mg\) | B1 | Resolving vertically for \(B\) |
| \(T + F - mg\sin 30 = 0\) | M1, A1 | Resolving parallel to the plane – three terms |
| \(F = \mu(mg\cos 30)\) | M1 | Use of \(F = \mu R\) |
| \(\frac{1}{4}mg + \mu\left(\frac{mg\sqrt{3}}{2}\right) - \frac{1}{2}mg = 0 \Rightarrow \mu = \ldots\) | M1 | Deriving equation in \(\mu\) and attempt to solve |
| \(\mu = \frac{\sqrt{3}}{6}\) | A1 [7] |
| Answer | Marks | Guidance |
|---|---|---|
| \(F = mg\sin 30 - \lambda mg\ (> 0)\) | M1 | Resolving parallel to the plane with \(\lambda mg\) |
| \(F > 0 \Rightarrow \lambda < \frac{1}{2}\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(T - F - mg\sin 30 = m\left(\frac{1}{4}g\right)\) | M1 | N, II parallel to the plane – four terms |
| \(2mg - T = 2m\left(\frac{1}{4}g\right)\) | B1 | N, II for \(B\) |
| \(2mg - F - mg\sin 30 = \frac{3}{4}mg\) | A1 | Correct method for eliminating \(T\) |
| \(2mg - \mu(mg\cos 30) - mg\sin 30 = \frac{3}{4}mg\) | A1 | Correct use of \(F = \mu R\) and \(R = mg\cos 30\) |
| \(\mu = \frac{\sqrt{3}}{2}\) | A1 [5] |
## Question 12:
### Part (i)(a):
$R = mg\cos 30$ | B1 | Resolving perpendicular to the plane |
$T = \frac{1}{4}mg$ | B1 | Resolving vertically for $B$ |
$T + F - mg\sin 30 = 0$ | M1, A1 | Resolving parallel to the plane – three terms |
$F = \mu(mg\cos 30)$ | M1 | Use of $F = \mu R$ |
$\frac{1}{4}mg + \mu\left(\frac{mg\sqrt{3}}{2}\right) - \frac{1}{2}mg = 0 \Rightarrow \mu = \ldots$ | M1 | Deriving equation in $\mu$ and attempt to solve |
$\mu = \frac{\sqrt{3}}{6}$ | A1 [7] | |
### Part (i)(b):
$F = mg\sin 30 - \lambda mg\ (> 0)$ | M1 | Resolving parallel to the plane with $\lambda mg$ |
$F > 0 \Rightarrow \lambda < \frac{1}{2}$ | A1 [2] | |
### Part (ii):
$T - F - mg\sin 30 = m\left(\frac{1}{4}g\right)$ | M1 | N, II parallel to the plane – four terms |
$2mg - T = 2m\left(\frac{1}{4}g\right)$ | B1 | N, II for $B$ |
$2mg - F - mg\sin 30 = \frac{3}{4}mg$ | A1 | Correct method for eliminating $T$ |
$2mg - \mu(mg\cos 30) - mg\sin 30 = \frac{3}{4}mg$ | A1 | Correct use of $F = \mu R$ and $R = mg\cos 30$ |
$\mu = \frac{\sqrt{3}}{2}$ | A1 [5] | |12 One end of a light inextensible string is attached to a particle $A$ of mass $m \mathrm {~kg}$. The other end of the string is attached to a second particle $B$ of mass $\lambda m \mathrm {~kg}$, where $\lambda$ is a constant. Particle $A$ is in contact with a rough plane inclined at $30 ^ { \circ }$ to the horizontal. The string is taut and passes over a small smooth pulley $P$ at the top of the plane. The part of the string from $A$ to $P$ is parallel to a line of greatest slope of the plane. The particle $B$ hangs freely below $P$ (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{d5ab20c8-afd5-473e-8238-96762bd3786d-8_405_670_493_685}
The coefficient of friction between $A$ and the plane is $\mu$.\\
(i) It is given that $A$ is on the point of moving down the plane.
\begin{enumerate}[label=(\alph*)]
\item Find the exact value of $\mu$ when $\lambda = \frac { 1 } { 4 }$.
\item Show that the value of $\lambda$ must be less than $\frac { 1 } { 2 }$.\\
(ii) Given instead that $\lambda = 2$ and that the acceleration of $A$ is $\frac { 1 } { 4 } g \mathrm {~ms} ^ { - 2 }$, find the exact value of $\mu$.
\section*{END OF QUESTION PAPER}
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 2018 Q12 [14]}}