| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| 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 students to analyze motion in two phases: connected motion with both particles, then independent motion after the heavier particle hits the ground. It demands Newton's second law application, kinematic equations, and energy/momentum considerations across discontinuous motion, going beyond routine single-phase pulley exercises. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03k Connected particles: pulleys and equilibrium6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| M1 | For applying Newton's Second Law to one particle or for using \(m_1g - m_2g = (m_1+m_2)a\) | |
| \(1.3g - T = 1.3a\) and \(T - 0.7g = 0.7a\) or \(1.3g - 0.7g = (1.3+0.7)a\) and either \(1.3g - T=1.3a\) or \(T - 0.7g=0.7a\) | A1 | |
| Tension is 9.1 N | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Acceleration is 3 ms\(^{-2}\) | B1 | |
| \([2 = \frac{1}{2} \times 3 \times t^2]\) | M1 | For using \(s = \frac{1}{2}at^2\) |
| Time taken is 1.15 seconds | A1 | [6 marks] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([v^2 = 2 \times 3 \times 2]\) | M1 | For using \(v^2 = u^2 + 2as\) to find the speed on reaching plane |
| \(v = \sqrt{12(3.464)}\) | A1\(\sqrt{}\) | ft \(\sqrt{(4a)}\) or \(at\) from (i) |
| \([0 = 12 - 2gs \rightarrow s = ...]\) | M1 | For using \(v^2 = u^2 + 2as\) to find the distance 0.7 kg particle continues upwards |
| Greatest height is 4.6 m | A1 | [4 marks] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([1.3g \times 2 = \frac{1}{2}(1.3)v^2 + 9.1 \times 2]\) or \([9.1 \times 2 = \frac{1}{2}(0.7)v^2 + 0.7g \times 2]\) | M1 | For using PE loss \(=\) KE gain \(+\) WD\(_T\) for 1.3 kg or for using WD\(_T\) \(=\) KE gain \(+\) PE gain for 0.7 kg |
| \(v = \sqrt{12(3.464)}\) | A1\(\sqrt{}\) | ft \(\sqrt{(4a)}\) or \(at\) from (i) |
| \([\frac{1}{2} \times 0.7v^2 = 0.7gs \rightarrow s = ...]\) | M1 | For using KE loss \(=\) PE gain |
| Greatest height is 4.6 m | A1 | [4 marks] |
## Question 6:
### Part (i)(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| | M1 | For applying Newton's Second Law to one particle or for using $m_1g - m_2g = (m_1+m_2)a$ |
| $1.3g - T = 1.3a$ and $T - 0.7g = 0.7a$ or $1.3g - 0.7g = (1.3+0.7)a$ and either $1.3g - T=1.3a$ or $T - 0.7g=0.7a$ | A1 | |
| Tension is 9.1 N | B1 | |
### Part (i)(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Acceleration is 3 ms$^{-2}$ | B1 | |
| $[2 = \frac{1}{2} \times 3 \times t^2]$ | M1 | For using $s = \frac{1}{2}at^2$ |
| Time taken is 1.15 seconds | A1 | **[6 marks]** |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[v^2 = 2 \times 3 \times 2]$ | M1 | For using $v^2 = u^2 + 2as$ to find the speed on reaching plane |
| $v = \sqrt{12(3.464)}$ | A1$\sqrt{}$ | ft $\sqrt{(4a)}$ or $at$ from (i) |
| $[0 = 12 - 2gs \rightarrow s = ...]$ | M1 | For using $v^2 = u^2 + 2as$ to find the distance 0.7 kg particle continues upwards |
| Greatest height is 4.6 m | A1 | **[4 marks]** |
#### Alternative for Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[1.3g \times 2 = \frac{1}{2}(1.3)v^2 + 9.1 \times 2]$ or $[9.1 \times 2 = \frac{1}{2}(0.7)v^2 + 0.7g \times 2]$ | M1 | For using PE loss $=$ KE gain $+$ WD$_T$ for 1.3 kg or for using WD$_T$ $=$ KE gain $+$ PE gain for 0.7 kg |
| $v = \sqrt{12(3.464)}$ | A1$\sqrt{}$ | ft $\sqrt{(4a)}$ or $at$ from (i) |
| $[\frac{1}{2} \times 0.7v^2 = 0.7gs \rightarrow s = ...]$ | M1 | For using KE loss $=$ PE gain |
| Greatest height is 4.6 m | A1 | **[4 marks]** |
---6 Two particles of masses 1.3 kg and 0.7 kg are connected by a light inextensible string that passes over a fixed smooth pulley. The particles are held at the same vertical height with the string taut. The distance of each particle above a horizontal plane is 2 m , and the distance of each particle below the pulley is 4 m . The particles are released from rest.\\
(i) Find
\begin{enumerate}[label=(\alph*)]
\item the tension in the string before the particle of mass 1.3 kg reaches the plane,
\item the time taken for the particle of mass 1.3 kg to reach the plane.\\
(ii) Find the greatest height of the particle of mass 0.7 kg above the plane.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2016 Q6 [10]}}