| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Particle on slope with pulley |
| Difficulty | Standard +0.3 This is a standard two-body pulley system requiring resolution of forces on inclined planes and friction at limiting equilibrium. The question guides students through part (i) by telling them the answer to show, and part (ii) is a straightforward application of F=μR. While it involves multiple components (two slopes, friction, pulley), the method is routine and well-practiced in A-level mechanics courses. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Component of weight down the plane: \(4.7g\sin 60°\) | B1 | AG — Award if seen |
| Equilibrium equation: \(T = 4.7g\sin 60° = 39.889...\), so \(T = 39.9\) to 3 sf | E1 [2] | Must be clear that 39.9 N is the tension and not just component of weight |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Resolve perpendicular to slope: \(N = 4g\cos 25°\) | B1 | Need not be evaluated here \([\approx 35.5]\) |
| Resolve up the slope: \(T - F - 4g\sin 25° = 0\) | M1, A1 | Allow only sign errors; \(F\) need not be evaluated here \([\approx 23.3]\) |
| On point of sliding: \(F = \mu N = \mu \times 4g\cos 25°\) | M1 | Do not allow \(F \leq \mu N\) unless used subsequently |
| \(\mu = \frac{4.7g\sin 60° - 4g\sin 25°}{4g\cos 25°} = 0.656\) to 3 sf | A1 [5] | FT their values (answer is 0.657 if 39.9 used for \(T\)) |
# Question 11:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Component of weight down the plane: $4.7g\sin 60°$ | B1 | AG — Award if seen |
| Equilibrium equation: $T = 4.7g\sin 60° = 39.889...$, so $T = 39.9$ to 3 sf | E1 [2] | Must be clear that 39.9 N is the tension and not just component of weight |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Resolve perpendicular to slope: $N = 4g\cos 25°$ | B1 | Need not be evaluated here $[\approx 35.5]$ |
| Resolve up the slope: $T - F - 4g\sin 25° = 0$ | M1, A1 | Allow only sign errors; $F$ need not be evaluated here $[\approx 23.3]$ |
| On point of sliding: $F = \mu N = \mu \times 4g\cos 25°$ | M1 | Do not allow $F \leq \mu N$ unless used subsequently |
| $\mu = \frac{4.7g\sin 60° - 4g\sin 25°}{4g\cos 25°} = 0.656$ to 3 sf | A1 [5] | FT their values (answer is 0.657 if 39.9 used for $T$) |
---11 Fig. 11 shows two blocks at rest, connected by a light inextensible string which passes over a smooth pulley. Block A of mass 4.7 kg rests on a smooth plane inclined at $60 ^ { \circ }$ to the horizontal. Block B of mass 4 kg rests on a rough plane inclined at $25 ^ { \circ }$ to the horizontal. On either side of the pulley, the string is parallel to a line of greatest slope of the plane. Block B is on the point of sliding up the plane.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{904025c9-6d68-4344-bd41-8c0fccfcf92f-07_332_931_443_575}
\captionsetup{labelformat=empty}
\caption{Fig. 11}
\end{center}
\end{figure}
(i) Show that the tension in the string is 39.9 N correct to 3 significant figures.\\
(ii) Find the coefficient of friction between the rough plane and Block B.
\hfill \mbox{\textit{OCR MEI Paper 1 2018 Q11 [7]}}