| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Two particles over pulley, vertical strings |
| Difficulty | Standard +0.3 This is a standard M1 pulley problem requiring Newton's second law for connected particles, followed by routine kinematics (SUVAT) and energy calculations. The multi-part structure and need to track motion after disconnection adds slight complexity, but all techniques are textbook applications with no novel insight required. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.3g - T = 0.3a\), \(T - 0.2g = 0.2a\) | M1, A1 | For applying Newton's second law to either particle |
| \(0.3g - 0.2g = 0.3a + 0.2a\) | M1 | For eliminating \(T\) |
| \(a = 2\) | A1 | |
| Alternatively: \(a = \frac{m_1 - m_2}{m_1 + m_2}g\) | M2, A2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v = 2 \times 1.2\); Speed is \(2.4\ \text{ms}^{-1}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(s_1 = \frac{1}{2}(0 + 2.4) \times 1.2\) | B1 | |
| \(2.4^2 = 2gs_2\) or PE gain while string is slack \(= \frac{1}{2} \times 0.2 \times 2.4^2\) | M1 | For using \(u^2 = 2gs\) or 'gain in PE = loss in KE' |
| \((s_1 + s_2) = 1.728\) or PE gain while string is slack \(= 0.576\) J | A1 | May be implied by final answer |
| Total PE gain \(= 0.2g \times 1.728\) (or PE gain while string is taut \(= 0.2g \times 1.44\)) | M1 | For using PE gain \(= mg(s_1 + s_2)\) (or PE gain while string is taut \(= mgs_1\), in the case where PE gain while string is slack is calculated separately) |
| Total PE gain \(= 3.456\) J | A1 |
## Question 7:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.3g - T = 0.3a$, $T - 0.2g = 0.2a$ | M1, A1 | For applying Newton's second law to either particle |
| $0.3g - 0.2g = 0.3a + 0.2a$ | M1 | For eliminating $T$ |
| $a = 2$ | A1 | |
| Alternatively: $a = \frac{m_1 - m_2}{m_1 + m_2}g$ | M2, A2 | |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = 2 \times 1.2$; Speed is $2.4\ \text{ms}^{-1}$ | B1 | |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $s_1 = \frac{1}{2}(0 + 2.4) \times 1.2$ | B1 | |
| $2.4^2 = 2gs_2$ or PE gain while string is slack $= \frac{1}{2} \times 0.2 \times 2.4^2$ | M1 | For using $u^2 = 2gs$ or 'gain in PE = loss in KE' |
| $(s_1 + s_2) = 1.728$ or PE gain while string is slack $= 0.576$ J | A1 | May be implied by final answer |
| Total PE gain $= 0.2g \times 1.728$ (or PE gain while string is taut $= 0.2g \times 1.44$) | M1 | For using PE gain $= mg(s_1 + s_2)$ (or PE gain while string is taut $= mgs_1$, in the case where PE gain while string is slack is calculated separately) |
| Total PE gain $= 3.456$ J | A1 | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{2026cad4-8494-4139-ad21-d8a17ac2b955-4_601_515_699_815}
Two particles $A$ and $B$, of masses 0.3 kg and 0.2 kg respectively, are attached to the ends of a light inextensible string which passes over a smooth fixed pulley. Particle $B$ is held on the horizontal floor and particle $A$ hangs in equilibrium. Particle $B$ is released and each particle starts to move vertically with constant acceleration of magnitude $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(i) Find the value of $a$.
Particle $A$ hits the floor 1.2 s after it starts to move, and does not rebound upwards.\\
(ii) Show that $A$ hits the floor with a speed of $2.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(iii) Find the gain in gravitational potential energy by $B$, from leaving the floor until reaching its greatest height.
\hfill \mbox{\textit{CAIE M1 2005 Q7 [10]}}