| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Lighter particle on surface released, heavier hangs |
| Difficulty | Standard +0.3 This is a standard two-stage pulley problem requiring application of Newton's second law to find acceleration, then kinematics for speeds and distances. While it involves multiple phases (connected motion, then A alone after B lands), each phase uses routine M1 techniques with no novel insight required. Slightly above average due to the two-part nature and careful tracking of the system state. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| *EITHER:* \([T - 0.35g = 0.35a\) or \(0.45g - T = 0.45a\) or \(0.45g - 0.35g = 0.8a]\) | (M1 | Applies Newton's Second Law to one of the particles or forms system equation in \(a\ (m_Bg - m_Ag = (m_A + m_B)a)\) |
| \([0.45g - T = 0.45a\) or \(T - 0.35g = 0.35a] \rightarrow a = \ldots\) | M1 | Applies Newton's Second Law to form second equation in \(T\) and \(a\) and solves for \(a\), or solves system equation for \(a\) |
| \(a = 1.25\ \text{m s}^{-2}\) | A1 | |
| \([v^2 = 2 \times 1.25 \times 0.64]\ (= 1.6)\) | M1 | Using \(v^2 = u^2 + 2as\) |
| Velocity \(= 1.26\ \text{ms}^{-1}\) | A1) | |
| *OR:* \([\text{PE loss} = 0.45g \times 0.64 - 0.35g \times 0.64]\) | (M1 | Attempts PE loss |
| \([\text{KE gain} = \frac{1}{2}(0.35 + 0.45)v^2]\) | M1 | Attempts KE gain |
| PE loss \(= 0.45g \times 0.64 - 0.35g \times 0.64\) and KE gain \(= \frac{1}{2}(0.35 + 0.45)v^2\) | A1 | |
| \([\frac{1}{2}(0.8)v^2 = 0.1g \times 0.64]\ (v^2 = 1.6)\) | M1 | Using PE loss \(=\) KE gain |
| Velocity \(= 1.26\ \text{ms}^{-1}\) | A1) | |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| *EITHER:* \([0 = 1.6 - 2gs]\ (s = 0.08)\) | (M1 | Using \(v^2 = u^2 + 2as\) |
| Distance \(= 0.16\) m | A1) | |
| *OR:* \([0.35gh = \frac{1}{2}(0.35) \times 1.6]\ (h = 0.08)\) | (M1 | Using PE gain \(=\) KE loss for particle A |
| Distance \(= 0.16\) m | A1) | |
| 2 |
# Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| *EITHER:* $[T - 0.35g = 0.35a$ or $0.45g - T = 0.45a$ or $0.45g - 0.35g = 0.8a]$ | **(M1** | Applies Newton's Second Law to one of the particles or forms system equation in $a\ (m_Bg - m_Ag = (m_A + m_B)a)$ |
| $[0.45g - T = 0.45a$ or $T - 0.35g = 0.35a] \rightarrow a = \ldots$ | **M1** | Applies Newton's Second Law to form second equation in $T$ and $a$ and solves for $a$, or solves system equation for $a$ |
| $a = 1.25\ \text{m s}^{-2}$ | **A1** | |
| $[v^2 = 2 \times 1.25 \times 0.64]\ (= 1.6)$ | **M1** | Using $v^2 = u^2 + 2as$ |
| Velocity $= 1.26\ \text{ms}^{-1}$ | **A1)** | |
| *OR:* $[\text{PE loss} = 0.45g \times 0.64 - 0.35g \times 0.64]$ | **(M1** | Attempts PE loss |
| $[\text{KE gain} = \frac{1}{2}(0.35 + 0.45)v^2]$ | **M1** | Attempts KE gain |
| PE loss $= 0.45g \times 0.64 - 0.35g \times 0.64$ and KE gain $= \frac{1}{2}(0.35 + 0.45)v^2$ | **A1** | |
| $[\frac{1}{2}(0.8)v^2 = 0.1g \times 0.64]\ (v^2 = 1.6)$ | **M1** | Using PE loss $=$ KE gain |
| Velocity $= 1.26\ \text{ms}^{-1}$ | **A1)** | |
| | **5** | |
# Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| *EITHER:* $[0 = 1.6 - 2gs]\ (s = 0.08)$ | **(M1** | Using $v^2 = u^2 + 2as$ |
| Distance $= 0.16$ m | **A1)** | |
| *OR:* $[0.35gh = \frac{1}{2}(0.35) \times 1.6]\ (h = 0.08)$ | **(M1** | Using PE gain $=$ KE loss for particle A |
| Distance $= 0.16$ m | **A1)** | |
| | **2** | |
---4 Two particles $A$ and $B$ have masses 0.35 kg and 0.45 kg respectively. The particles are attached to the ends of a light inextensible string which passes over a small fixed smooth pulley which is 1 m above horizontal ground. Initially particle $A$ is held at rest on the ground vertically below the pulley, with the string taut. Particle $B$ hangs vertically below the pulley at a height of 0.64 m above the ground. Particle $A$ is released.\\
(i) Find the speed of $A$ at the instant that $B$ reaches the ground.\\
(ii) Assuming that $B$ does not bounce after it reaches the ground, find the total distance travelled by $A$ between the instant that $B$ reaches the ground and the instant when the string becomes taut again.\\
\hfill \mbox{\textit{CAIE M1 2017 Q4 [7]}}