| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Heavier particle hits ground, lighter continues upward - vertical strings |
| Difficulty | Standard +0.8 This is a two-phase pulley problem requiring students to analyze motion before and after impact, connect the phases using kinematics, and work backwards from the post-impact motion to find initial conditions. It demands careful application of Newton's laws, understanding of what happens when one particle hits the ground (string goes slack), and multi-step reasoning across different motion phases—significantly more sophisticated than standard single-phase pulley problems. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium |
| Answer | Marks |
|---|---|
| M1 | For applying Newton's second law to A or to B |
| Answer | Marks | Guidance |
|---|---|---|
| \((0.9 - 0.6)g = (0.9 + 0.6)a\) | A1 | |
| B1 | ||
| Acceleration is \(2 \text{ m s}^{-2}\) and tension is \(7.2 \text{ N}\) | A1 | 4 |
| Answer | Marks |
|---|---|
| M1 | For using \(0 = u - gt\) |
| \(u = 3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \([\frac{1}{2}(0.9 + 0.6)3^2 = (0.9 - 0.6)gh]\) | M1 | For using \(v^2 = 0^2 + 2ah\) with \(v_{\text{limit}} = u_{\text{slack}}\) or for using KE gain = PE loss while the string is in tension |
| Height is \(2.25 \text{ m}\) | A1 | 4 |
**(i)**
| M1 | For applying Newton's second law to A or to B
$0.9g - T = 0.9a$ or $T - 0.6g = 0.6a$ or
$T - 0.6g = 0.6a$ or $0.9g - T = 0.9a$ or
$(0.9 - 0.6)g = (0.9 + 0.6)a$ | A1 |
| B1 |
Acceleration is $2 \text{ m s}^{-2}$ and tension is $7.2 \text{ N}$ | A1 | 4
**(ii)**
| M1 | For using $0 = u - gt$
$u = 3$ | A1 |
$[3^2 = 2 \times 2 \text{ h}]$
$[\frac{1}{2}(0.9 + 0.6)3^2 = (0.9 - 0.6)gh]$ | M1 | For using $v^2 = 0^2 + 2ah$ with $v_{\text{limit}} = u_{\text{slack}}$ or for using KE gain = PE loss while the string is in tension
Height is $2.25 \text{ m}$ | A1 | 4
---5 Particles $A$ and $B$, of masses 0.9 kg and 0.6 kg respectively, are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley. The system is released from rest with the string taut, with its straight parts vertical and with the particles at the same height above the horizontal floor. In the subsequent motion, $B$ does not reach the pulley.\\
(i) Find the acceleration of $A$ and the tension in the string during the motion before $A$ hits the floor.
After $A$ hits the floor, $B$ continues to move vertically upwards for a further 0.3 s .\\
(ii) Find the height of the particles above the floor at the instant that they started to move.
\hfill \mbox{\textit{CAIE M1 2011 Q5 [8]}}