CAIE M1 2013 November — Question 2 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2013
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPulley systems
TypeLighter particle on surface released, heavier hangs
DifficultyStandard +0.3 This is a standard two-stage pulley problem requiring application of Newton's second law to find acceleration, then kinematics to find speed, followed by energy conservation after B hits the floor. While it involves multiple steps and two phases of motion, the techniques are routine M1 content with no novel insight required—slightly easier than average due to straightforward setup and clear problem structure.
Spec3.02d Constant acceleration: SUVAT formulae3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys6.02i Conservation of energy: mechanical energy principle

2 \includegraphics[max width=\textwidth, alt={}, center]{fd534430-2619-4078-ad0a-2355e656e121-2_569_519_676_813} Particle \(A\) of mass 0.2 kg and particle \(B\) of mass 0.6 kg are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley. \(B\) is held at rest at a height of 1.6 m above the floor. \(A\) hangs freely at a height of \(h \mathrm {~m}\) above the floor. Both straight parts of the string are vertical (see diagram). \(B\) is released and both particles start to move. When \(B\) reaches the floor it remains at rest, but \(A\) continues to move vertically upwards until it reaches a height of 3 m above the floor. Find the speed of \(B\) immediately before it hits the floor, and hence find the value of \(h\).

Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
M1For using \(a = (M-m)g/(M+m)\) or for applying Newton's 2nd law to A and to B and solving for \(a\)
\(a = 5\)A1
When B reaches the floor: \(v^2 = 2 \times 5 \times 1.6\); speed is \(4\ \text{ms}^{-1}\)B1ft ft \(a \neq g\), \(v = \sqrt{3.2a}\)
M1For using \(0 = u^2 - 2gs\) or PE gain = KE loss
\(0 = 16 - 20s \quad (s = 0.8)\)A1ft ft speed
\(h + 1.6 + 0.8 = 3 \Rightarrow h = 0.6\)B1 6 marks
## Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| | M1 | For using $a = (M-m)g/(M+m)$ or for applying Newton's 2nd law to A and to B and solving for $a$ |
| $a = 5$ | A1 | |
| When B reaches the floor: $v^2 = 2 \times 5 \times 1.6$; speed is $4\ \text{ms}^{-1}$ | B1ft | ft $a \neq g$, $v = \sqrt{3.2a}$ |
| | M1 | For using $0 = u^2 - 2gs$ or PE gain = KE loss |
| $0 = 16 - 20s \quad (s = 0.8)$ | A1ft | ft speed |
| $h + 1.6 + 0.8 = 3 \Rightarrow h = 0.6$ | B1 | 6 marks |

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\includegraphics[max width=\textwidth, alt={}, center]{fd534430-2619-4078-ad0a-2355e656e121-2_569_519_676_813}

Particle $A$ of mass 0.2 kg and particle $B$ of mass 0.6 kg are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley. $B$ is held at rest at a height of 1.6 m above the floor. $A$ hangs freely at a height of $h \mathrm {~m}$ above the floor. Both straight parts of the string are vertical (see diagram). $B$ is released and both particles start to move. When $B$ reaches the floor it remains at rest, but $A$ continues to move vertically upwards until it reaches a height of 3 m above the floor. Find the speed of $B$ immediately before it hits the floor, and hence find the value of $h$.

\hfill \mbox{\textit{CAIE M1 2013 Q2 [6]}}