CAIE M1 2007 June — Question 4 7 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2007
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNewton's laws and connected particles
TypeAtwood machine, vertical strings
DifficultyModerate -0.8 This is a standard two-particle pulley system requiring straightforward application of Newton's second law to find acceleration and tension, followed by basic kinematics. The setup is routine with clear given values, requiring only F=ma for each particle, solving simultaneous equations, and using s=ut+½at². No novel insight or complex problem-solving needed—slightly easier than average A-level mechanics.
Spec3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension3.03k Connected particles: pulleys and equilibrium

4 \includegraphics[max width=\textwidth, alt={}, center]{f7a22c07-44e3-4891-be60-cbab772f45df-3_702_709_269_719} Particles \(P\) and \(Q\), of masses 0.6 kg and 0.2 kg respectively, are attached to the ends of a light inextensible string which passes over a smooth fixed peg. The particles are held at rest with the string taut. Both particles are at a height of 0.9 m above the ground (see diagram). The system is released and each of the particles moves vertically. Find
  1. the acceleration of \(P\) and the tension in the string before \(P\) reaches the ground,
  2. the time taken for \(P\) to reach the ground.

Question 4:
Part (i):
AnswerMarks Guidance
Working/AnswerMark Guidance
M1For applying Newton's second law to P or to Q (3 terms)
\(0.6g - T = 0.6a\)A1
\(T - 0.2g = 0.2a\)A1 Allow B1 for \(0.6g - 0.2g = (0.6 + 0.2)a\) as an alternative for either of the above A marks
Acceleration is \(5\ \text{ms}^{-2}\)B1
Tension is 3 NA1
Part (ii):
AnswerMarks Guidance
Working/AnswerMark Guidance
\([0.9 = \frac{1}{2} \cdot 5t^2]\)M1 For using \(s = ut + \frac{1}{2}at^2\)
Time taken is 0.6 sA1ft ft \(\sqrt{1.8/a}\)
## Question 4:

### Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| | M1 | For applying Newton's second law to P or to Q (3 terms) |
| $0.6g - T = 0.6a$ | A1 | |
| $T - 0.2g = 0.2a$ | A1 | Allow B1 for $0.6g - 0.2g = (0.6 + 0.2)a$ as an alternative for either of the above A marks |
| Acceleration is $5\ \text{ms}^{-2}$ | B1 | |
| Tension is 3 N | A1 | |

### Part (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $[0.9 = \frac{1}{2} \cdot 5t^2]$ | M1 | For using $s = ut + \frac{1}{2}at^2$ |
| Time taken is 0.6 s | A1ft | ft $\sqrt{1.8/a}$ |

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4\\
\includegraphics[max width=\textwidth, alt={}, center]{f7a22c07-44e3-4891-be60-cbab772f45df-3_702_709_269_719}

Particles $P$ and $Q$, of masses 0.6 kg and 0.2 kg respectively, are attached to the ends of a light inextensible string which passes over a smooth fixed peg. The particles are held at rest with the string taut. Both particles are at a height of 0.9 m above the ground (see diagram). The system is released and each of the particles moves vertically. Find\\
(i) the acceleration of $P$ and the tension in the string before $P$ reaches the ground,\\
(ii) the time taken for $P$ to reach the ground.

\hfill \mbox{\textit{CAIE M1 2007 Q4 [7]}}