| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Lighter particle on surface released, heavier hangs |
| Difficulty | Standard +0.3 This is a standard two-particle pulley problem requiring Newton's second law to find acceleration, then kinematics for maximum height and time. The setup is straightforward with clear given values, requiring routine application of F=ma and SUVAT equations across three parts totaling about 8-10 marks. Slightly easier than average due to the simple vertical motion and explicit problem structure. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Part (i) | ||
| \(T - 0.3g = 0.3a\) and \(0.7g - T = 0.7a\) | M1 | For applying Newton's second law to A or to B or for using \((M + m)a = (M - m)g\) |
| or \((0.7 + 0.3)a = (0.7 - 0.3)g\) | A1 | |
| Acceleration is 4 m s\(^{-2}\) | A1 | 3 |
| Part (ii) | ||
| \(s_1 = 1.6/(2 \times 4)\) | B1ft | ft acceleration |
| M1 | For using \(0^2 = 1.6^2 - 2gs_2\) | |
| Height is 0.448 m | A1 | 3 |
| Part (iii) | ||
| \(t_1 = 1.6/4\) | B1ft | ft acceleration (can be scored in (ii)) |
| M1 | For using \(0 = 1.6 - gt_2\) | |
| Time taken is 0.56 s | A1 | 3 |
| (Alternative for part (iii)) | ||
| \(\Rightarrow t_1 + t_2 = (s_1 + s_2)/0.8\) | M1 | For observing that the average speed is the same for each of the two phases and equal to \((0 + 1.6)/2\) m s\(^{-1}\) |
| Time taken is 0.56 s | A1 | 3 |
| (Alternatively for parts ii and iii using v–t graph) | ||
| M1 | Use of gradient to find \(t_1\) or \(t_2\) | |
| \(t_1 = 1.6/4\) and \(t_2 = 1.6/10\) | A1 | |
| Time taken is 0.56s | A1 | |
| M1 | For use of area to find \(s_1\) or \(s_2\) or \(s_1 + s_2\) | |
| \(s_1 = 0.4 \times 1.6/2\) or \(s_2 = 0.16 \times 1.6/2\) or \(s_1 + s_2 = (0.4 + 0.16) \times 1.6/2\) | A1 | |
| Height is 0.448m | A1 | 6 |
| **Part (i)** | | |
|---|---|---|
| $T - 0.3g = 0.3a$ and $0.7g - T = 0.7a$ | M1 | For applying Newton's second law to A or to B or for using $(M + m)a = (M - m)g$ |
| or $(0.7 + 0.3)a = (0.7 - 0.3)g$ | A1 | |
| Acceleration is 4 m s$^{-2}$ | A1 | 3 |
| **Part (ii)** | | |
|---|---|---|
| $s_1 = 1.6/(2 \times 4)$ | B1ft | ft acceleration |
| | M1 | For using $0^2 = 1.6^2 - 2gs_2$ |
| Height is 0.448 m | A1 | 3 | From $s_1 + s_2 = 0.32 + 0.128$ |
| **Part (iii)** | | |
|---|---|---|
| $t_1 = 1.6/4$ | B1ft | ft acceleration (can be scored in (ii)) |
| | M1 | For using $0 = 1.6 - gt_2$ |
| Time taken is 0.56 s | A1 | 3 | From $t_1 + t_2 = 0.4 + 0.16$ |
| **(Alternative for part (iii))** | | |
|---|---|---|
| $\Rightarrow t_1 + t_2 = (s_1 + s_2)/0.8$ | M1 | For observing that the average speed is the same for each of the two phases and equal to $(0 + 1.6)/2$ m s$^{-1}$ |
| Time taken is 0.56 s | A1 | 3 |
| **(Alternatively for parts ii and iii using v–t graph)** | | |
|---|---|---|
| | M1 | Use of gradient to find $t_1$ or $t_2$ |
| $t_1 = 1.6/4$ and $t_2 = 1.6/10$ | A1 | |
| Time taken is 0.56s | A1 | |
| | M1 | For use of area to find $s_1$ or $s_2$ or $s_1 + s_2$ |
| $s_1 = 0.4 \times 1.6/2$ or $s_2 = 0.16 \times 1.6/2$ or $s_1 + s_2 = (0.4 + 0.16) \times 1.6/2$ | A1 | |
| Height is 0.448m | A1 | 6 |6\\
\includegraphics[max width=\textwidth, alt={}, center]{efa7175f-832b-4cd3-82ab-52e402115081-4_686_511_269_817}
Particles $A$ and $B$, of masses 0.3 kg and 0.7 kg respectively, are attached to the ends of a light inextensible string which passes over a smooth fixed pulley. Particle $A$ is held on the horizontal floor and particle $B$ hangs in equilibrium. Particle $A$ is released and both particles start to move vertically.\\
(i) Find the acceleration of the particles.
The speed of the particles immediately before $B$ hits the floor is $1.6 \mathrm {~ms} ^ { - 1 }$. Given that $B$ does not rebound upwards, find\\
(ii) the maximum height above the floor reached by $A$,\\
(iii) the time taken by $A$, from leaving the floor, to reach this maximum height.
\hfill \mbox{\textit{CAIE M1 2009 Q6 [9]}}