| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| 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 multi-stage pulley problem requiring: (i) standard connected particles acceleration, (ii) kinematics with constant acceleration, (iii-iv) analysis after string goes slack with particle continuing under gravity alone, requiring careful tracking of motion phases and understanding that particle rises to 1.3m (twice the original height) before returning. The conceptual jump of analyzing post-impact motion with slack string and synthesizing a complete velocity-time graph elevates this above routine M1 questions, though the individual calculations are standard. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([T - 0.12g = 0.12a\ \&\ 0.38g - T = 0.38a;\ a = \frac{0.38 - 0.12}{0.38 + 0.12}g]\) | M1 | For using Newton's second law for A and B or for using \(a = \frac{M-m}{M+m}g\) |
| Acceleration is \(5.2\ \text{ms}^{-2}\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([v^2 = 2 \times 5.2 \times 0.65;\ 0.65 = \frac{1}{2} \times 5.2T_B^2]\) | M1 | For using \(v^2 = 2ah\) or \(s = \frac{1}{2}at^2\) |
| Speed of B is \(2.6\ \text{ms}^{-1}\) or \(T_B = 0.5\) | A1ft | ft incorrect \(a\) |
| \(T_B = 0.5\) or Speed of B is \(2.6\ \text{ms}^{-1}\) | B1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([-2.6 = 2.6 - 10(T - 0.5)]\) | M1 | For using \(-V = V - g(T - T_B)\) or equivalent |
| \(T = 1.02\) | A1ft | ft incorrect \(V\) and/or \(T_B\) |
| Correct graph for \(0 < t < 1.02\); ft incorrect values of \(V\), \(T\) and \(T_B\) | B1ft [3] | Graph: rises to peak at \(t=0.5\), falls back to zero at \(t=1.02\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([0.65 + 0.5(1.02 - 0.5) \times 2.6]\) | M1 | For using total distance \(= \frac{1}{2}(VT_B) + 2 \times \frac{1}{2}\frac{T_A - T_B}{2}V\) |
| Total distance is \(1.326\) m (accept \(1.33\)) | A1 [2] |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[T - 0.12g = 0.12a\ \&\ 0.38g - T = 0.38a;\ a = \frac{0.38 - 0.12}{0.38 + 0.12}g]$ | M1 | For using Newton's second law for A and B or for using $a = \frac{M-m}{M+m}g$ |
| Acceleration is $5.2\ \text{ms}^{-2}$ | A1 [2] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[v^2 = 2 \times 5.2 \times 0.65;\ 0.65 = \frac{1}{2} \times 5.2T_B^2]$ | M1 | For using $v^2 = 2ah$ or $s = \frac{1}{2}at^2$ |
| Speed of B is $2.6\ \text{ms}^{-1}$ or $T_B = 0.5$ | A1ft | ft incorrect $a$ |
| $T_B = 0.5$ or Speed of B is $2.6\ \text{ms}^{-1}$ | B1 [3] | |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[-2.6 = 2.6 - 10(T - 0.5)]$ | M1 | For using $-V = V - g(T - T_B)$ or equivalent |
| $T = 1.02$ | A1ft | ft incorrect $V$ and/or $T_B$ |
| Correct graph for $0 < t < 1.02$; ft incorrect values of $V$, $T$ and $T_B$ | B1ft [3] | Graph: rises to peak at $t=0.5$, falls back to zero at $t=1.02$ |
### Part (iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[0.65 + 0.5(1.02 - 0.5) \times 2.6]$ | M1 | For using total distance $= \frac{1}{2}(VT_B) + 2 \times \frac{1}{2}\frac{T_A - T_B}{2}V$ |
| Total distance is $1.326$ m (accept $1.33$) | A1 [2] | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{918b65cc-617d-4942-8d96-b02eef21e417-4_506_471_255_836}
Two particles $A$ and $B$ have masses 0.12 kg and 0.38 kg respectively. The particles are attached to the ends of a light inextensible string which passes over a fixed smooth pulley. $A$ is held at rest with the string taut and both straight parts of the string vertical. $A$ and $B$ are each at a height of 0.65 m above horizontal ground (see diagram). $A$ is released and $B$ moves downwards. Find\\
(i) the acceleration of $B$ while it is moving downwards,\\
(ii) the speed with which $B$ reaches the ground and the time taken for it to reach the ground.\\
$B$ remains on the ground while $A$ continues to move with the string slack, without reaching the pulley. The string remains slack until $A$ is at a height of 1.3 m above the ground for a second time. At this instant $A$ has been in motion for a total time of $T \mathrm {~s}$.\\
(iii) Find the value of $T$ and sketch the velocity-time graph for $A$ for the first $T \mathrm {~s}$ of its motion.\\
(iv) Find the total distance travelled by $A$ in the first $T$ s of its motion.
\hfill \mbox{\textit{CAIE M1 2012 Q7 [10]}}