| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Particle on rough horizontal surface, particle hanging |
| Difficulty | Moderate -0.3 This is a standard mechanics pulley problem with friction, requiring routine application of Newton's second law and kinematics. The question guides students through each step (force diagram, equations of motion, finding acceleration, then using SUVAT), making it slightly easier than average. The calculation is straightforward once equations are set up correctly, with no novel problem-solving required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03a Force: vector nature and diagrams3.03b Newton's first law: equilibrium3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Diagram with R upward, \(2gN\) and \(0.8gN\) downward, T and F horizontal | B1 (1.1a) | Both weights correct. Allow weight of box and sphere but not if both marked weight |
| B1 (1.1a) | Common tension in the right directions. Allow \(T_1\) and \(T_2\) provided they are clearly shown equal | |
| B1 (1.1a) [3] | Friction and normal reaction and no extra forces. For \(F\), allow \(0.35R\), \(0.35\times 2g\), \(0.7g\) or \(6.86\) N |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T - F = 2a\) | B1 (1.1a) | Allow any expression for F. For \(F\), allow \(0.35R\), \(0.35\times 2g\), \(0.7g\) or \(6.86\) N |
| \(0.8g - T = 0.8a\) | B1 (1.1a) [2] | Allow distinct tensions if consistent with diagram |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Vertically for the block: \(R = 2g\) | M1 (3.1b) | Attempt to use \(\mu\) to evaluate friction. Some of this work may already have been seen in previous part |
| Friction \(F = 0.35R = 0.7g\) | A1 (2.1) | Correct value for F |
| Add equations: \(0.8g - F = 2.8a\) | M1 (2.1) | Eliminate \(T\) from their equations |
| \(0.8g - 0.7g = 2.8a\), \(a = 0.35\) ms\(^{-2}\) | A1 (2.1) [4] | AG must follow from correct work |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use \(s = 0.5\), \(u = 0\), \(a = 0.35\): \(0.5 = \frac{1}{2}\times 0.35 \times t^2\) | M1 (1.1a) | Using *suvat* equation(s) leading to a value for \(t\) |
| \(t = 1.69\) s | A1 (1.1) [2] | Do not allow \(\pm 1.69\) |
# Question 11:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Diagram with R upward, $2gN$ and $0.8gN$ downward, T and F horizontal | B1 (1.1a) | Both weights correct. Allow weight of box and sphere but not if both marked weight |
| | B1 (1.1a) | Common tension in the right directions. Allow $T_1$ and $T_2$ provided they are clearly shown equal |
| | B1 (1.1a) [3] | Friction and normal reaction and no extra forces. For $F$, allow $0.35R$, $0.35\times 2g$, $0.7g$ or $6.86$ N |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T - F = 2a$ | B1 (1.1a) | Allow any expression for F. For $F$, allow $0.35R$, $0.35\times 2g$, $0.7g$ or $6.86$ N |
| $0.8g - T = 0.8a$ | B1 (1.1a) [2] | Allow distinct tensions if consistent with diagram |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Vertically for the block: $R = 2g$ | M1 (3.1b) | Attempt to use $\mu$ to evaluate friction. Some of this work may already have been seen in previous part |
| Friction $F = 0.35R = 0.7g$ | A1 (2.1) | Correct value for F |
| Add equations: $0.8g - F = 2.8a$ | M1 (2.1) | Eliminate $T$ from their equations |
| $0.8g - 0.7g = 2.8a$, $a = 0.35$ ms$^{-2}$ | A1 (2.1) [4] | AG must follow from correct work |
## Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use $s = 0.5$, $u = 0$, $a = 0.35$: $0.5 = \frac{1}{2}\times 0.35 \times t^2$ | M1 (1.1a) | Using *suvat* equation(s) leading to a value for $t$ |
| $t = 1.69$ s | A1 (1.1) [2] | Do not allow $\pm 1.69$ |
---11 A block of mass 2 kg is placed on a rough horizontal table. A light inextensible string attached to the block passes over a smooth pulley attached to the edge of the table. The other end of the string is attached to a sphere of mass 0.8 kg which hangs freely.
The part of the string between the block and the pulley is horizontal. The coefficient of friction between the table and the block is 0.35 . The system is released from rest.
\begin{enumerate}[label=(\alph*)]
\item Draw a force diagram showing all the forces on the block and the sphere.
\item Write down the equations of motion for the block and the sphere.
\item Show that the acceleration of the system is $0.35 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\item Calculate the time for the block to slide the first 0.5 m . Assume the block does not reach the pulley.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2020 Q11 [11]}}