| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | String breaks during motion |
| Difficulty | Standard +0.8 This is a two-stage pulley problem requiring students to analyze motion before and after string breaks, then calculate the time difference for particles hitting the ground. It demands careful tracking of velocities at the break point, applying SUVAT equations in both stages, and coordinating two separate motion scenarios—significantly more complex than standard single-stage pulley questions. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.03k Connected particles: pulleys and equilibrium |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Particle A: \([1.3g - T = 1.3a]\) or Particle B: \([T - 0.7g = 0.7a]\) | M1 | Use of Newton's Second Law for A or B or use of \(a = \frac{(m_A - m_B)g}{(m_A + m_B)}\) |
| \(1.3g - T = 1.3a\) and \(T - 0.7g = 0.7a\) OR \(a = \frac{(1.3-0.7)g}{(1.3+0.7)}\) and \(1.3g - T = 1.3a\) or \(T - 0.7g = 0.7a\) | A1 | Two correct equations |
| \([6 = 2a, a=3]\) or \([\frac{1.3g-T}{1.3} = \frac{T-0.7g}{0.7}, T = 9.1]\) | M1 | Solves for \(a\) or for \(T\) |
| \(a = 3 \text{ ms}^{-2}\) and \(T = 9.1\text{ N}\) | A1 | \((a=3)\) |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Distance while connected \(= 0.375\text{ m}\) | B1 | |
| \([v^2 = 0^2 + 2 \times 3 \times 0.375 \rightarrow v = ...]\) | M1 | Use of suvat to find \(v\) at 'break' \((v^2 = 2as)\) |
| \(v = 1.5 \text{ ms}^{-1}\) | A1 | Correct value or expression for \(v\) |
| \([A: 1.375 = 1.5t + \frac{1}{2}gt^2 \rightarrow t = 0.395...]\) | M1 | Finds one time 'from break to floor' |
| \([B: 1.375 = -1.5t + \frac{1}{2}gt^2\) or \(-1.375 = 1.5t - \frac{1}{2}gt^2 \rightarrow t = 0.695...]\) | M1 | Finds second time 'from break to floor' |
| Difference in times \(= 0.3\text{ s}\) | A1 | |
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([u_B = 1.5, v_B = 0, a = -g, 0 = 1.5 - gt \rightarrow t = 0.15]\) | M1 | Finds \(t_B\) from 'break' to maximum height |
| Difference in times \(= 2 \times 0.15\) | M1 | |
| Difference in times \(= 0.3\text{ s}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([A: 0.375 = \frac{1}{2} \times 3t^2 \rightarrow t = 0.5; \quad 1.375 = 1.5t + \frac{1}{2}gt^2 \rightarrow t = 0.395...; \quad t_A \text{ total} = 0.5 + 0.395... = 0.895...\text{ s}]\) | M1 | Use of suvat to find total time for \(A\) |
| \([B: 0.375 = \frac{1}{2} \times 3t^2 \rightarrow t = 0.5; \quad 0 = 1.5 - gt \rightarrow t = 0.15, s = 1.5t - \frac{1}{2}gt^2 = 0.1125; \quad 1.4875 = \frac{1}{2} \times gt^2 \rightarrow t = 0.545...; \quad t_B \text{ total} = 1.195\text{ s}]\) | M1 | Use of suvat to find total time for \(B\) |
| Difference in times \(= 0.3\text{ s}\) | A1 |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Particle A: $[1.3g - T = 1.3a]$ or Particle B: $[T - 0.7g = 0.7a]$ | M1 | Use of Newton's Second Law for A or B or use of $a = \frac{(m_A - m_B)g}{(m_A + m_B)}$ |
| $1.3g - T = 1.3a$ and $T - 0.7g = 0.7a$ OR $a = \frac{(1.3-0.7)g}{(1.3+0.7)}$ and $1.3g - T = 1.3a$ or $T - 0.7g = 0.7a$ | A1 | Two correct equations |
| $[6 = 2a, a=3]$ or $[\frac{1.3g-T}{1.3} = \frac{T-0.7g}{0.7}, T = 9.1]$ | M1 | Solves for $a$ or for $T$ |
| $a = 3 \text{ ms}^{-2}$ and $T = 9.1\text{ N}$ | A1 | $(a=3)$ |
| | **4** | |
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Distance while connected $= 0.375\text{ m}$ | B1 | |
| $[v^2 = 0^2 + 2 \times 3 \times 0.375 \rightarrow v = ...]$ | M1 | Use of suvat to find $v$ at 'break' $(v^2 = 2as)$ |
| $v = 1.5 \text{ ms}^{-1}$ | A1 | Correct value or expression for $v$ |
| $[A: 1.375 = 1.5t + \frac{1}{2}gt^2 \rightarrow t = 0.395...]$ | M1 | Finds one time 'from break to floor' |
| $[B: 1.375 = -1.5t + \frac{1}{2}gt^2$ or $-1.375 = 1.5t - \frac{1}{2}gt^2 \rightarrow t = 0.695...]$ | M1 | Finds second time 'from break to floor' |
| Difference in times $= 0.3\text{ s}$ | A1 | |
| | **6** | |
**Alternative Method 1 for 4(ii) (last 3 marks):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[u_B = 1.5, v_B = 0, a = -g, 0 = 1.5 - gt \rightarrow t = 0.15]$ | M1 | Finds $t_B$ from 'break' to maximum height |
| Difference in times $= 2 \times 0.15$ | M1 | |
| Difference in times $= 0.3\text{ s}$ | A1 | |
**Alternative Method 2 for 4(ii) (last 3 marks):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[A: 0.375 = \frac{1}{2} \times 3t^2 \rightarrow t = 0.5; \quad 1.375 = 1.5t + \frac{1}{2}gt^2 \rightarrow t = 0.395...; \quad t_A \text{ total} = 0.5 + 0.395... = 0.895...\text{ s}]$ | M1 | Use of suvat to find total time for $A$ |
| $[B: 0.375 = \frac{1}{2} \times 3t^2 \rightarrow t = 0.5; \quad 0 = 1.5 - gt \rightarrow t = 0.15, s = 1.5t - \frac{1}{2}gt^2 = 0.1125; \quad 1.4875 = \frac{1}{2} \times gt^2 \rightarrow t = 0.545...; \quad t_B \text{ total} = 1.195\text{ s}]$ | M1 | Use of suvat to find total time for $B$ |
| Difference in times $= 0.3\text{ s}$ | A1 | |(i) Show that, before the string breaks, the magnitude of the acceleration of each particle is $3 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ and find the tension in the string.\\
(ii) Find the difference in the times that it takes the particles to hit the ground.\\
\hfill \mbox{\textit{CAIE M1 2019 Q4 [10]}}