| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Particle on rough horizontal surface, particle hanging |
| Difficulty | Standard +0.3 This is a standard two-particle pulley system with friction, requiring Newton's second law applied to both particles, calculation of friction force, and then basic kinematics. It's slightly above average difficulty due to the friction component and multi-step nature, but follows a well-established method taught in M1 with no novel insight required. |
| Spec | 3.03o Advanced connected particles: and pulleys3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F = 0.5(0.6g)\) | B1 | |
| M1 | For applying Newton's second law to A or to B | |
| \(0.4g - T = 0.4a\) | A1 | |
| \(T - F = 0.6a\) | A1 | Alternative: \(0.4g - F = (0.4 + 0.6)a\) — B1 |
| M1 | For substituting for F and solving for \(a\) or for T | |
| Acceleration is \(1 \text{ ms}^{-2}\) and tension is 3.6 N | A1 [6] | SR (max mark 1/3): \(0.4g - T = 0.4ga\) and \(T - F = 0.6ga\) — B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using \(s = (0) + \frac{1}{2}at^2\) | |
| Time taken is 2.45 s | A1ft [2] | ft \(t = (6/a)^{\frac{1}{2}}\) |
## Question 5:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = 0.5(0.6g)$ | B1 | |
| | M1 | For applying Newton's second law to A or to B |
| $0.4g - T = 0.4a$ | A1 | |
| $T - F = 0.6a$ | A1 | Alternative: $0.4g - F = (0.4 + 0.6)a$ — B1 |
| | M1 | For substituting for F and solving for $a$ or for T |
| Acceleration is $1 \text{ ms}^{-2}$ and tension is 3.6 N | A1 [6] | SR (max mark 1/3): $0.4g - T = 0.4ga$ and $T - F = 0.6ga$ — B1 |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $s = (0) + \frac{1}{2}at^2$ |
| Time taken is 2.45 s | A1ft [2] | ft $t = (6/a)^{\frac{1}{2}}$ |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{ee138c3f-51e1-4a69-9750-9eb49ac87e22-3_314_867_1457_639}
A block $B$ of mass 0.6 kg and a particle $A$ of mass 0.4 kg are attached to opposite ends of a light inextensible string. The block is held at rest on a rough horizontal table, and the coefficient of friction between the block and the table is 0.5 . The string passes over a small smooth pulley $C$ at the edge of the table and $A$ hangs in equilibrium vertically below $C$. The part of the string between $B$ and $C$ is horizontal and the distance $B C$ is 3 m (see diagram). $B$ is released and the system starts to move.\\
(i) Find the acceleration of $B$ and the tension in the string.\\
(ii) Find the time taken for $B$ to reach the pulley.
\hfill \mbox{\textit{CAIE M1 2008 Q5 [8]}}