| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Three or more connected particles |
| Difficulty | Standard +0.3 This is a standard two-particle pulley system requiring Newton's second law to find acceleration, then kinematics for speed and position. While it has multiple parts and requires checking if Q reaches rest, the techniques are routine M1 content with straightforward application of F=ma and SUVAT equations. Slightly above average due to the multi-stage nature and the need to compare distances, but no novel insight required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([7g - T = 7a\) and \(T - 3g = 3a]\) or \([7g - 3g = 10a]\) | M1 | For applying Newton's second law to P and to Q or for using \(m_P g - m_Q g = (m_P + m_Q)a\) |
| Acceleration is \(4\text{ ms}^{-2}\) | A1 | |
| \([v^2 = 0 + 2 \times 4 \times 0.4]\ (v^2 = 3.2)\) | M1 | For using \(v^2 = u^2 + 2as\) |
| Speed is \(1.79\text{ ms}^{-1}\) | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([0 = 3.2 + 2 \times (-g) \times s]\ (s = 0.16)\) | M1 | For using \(0 = u^2 + 2(-g)s\) |
| \(0.16 + 0.4 = 0.56\), so particle \(Q\) does not come to rest before it reaches the pulley | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([v^2 = 3.2 + 2 \times (-g) \times 0.1]\) | M1 | For using \(v^2 = u^2 + 2(-g)(0.1)\) |
| \(v = \sqrt{1.2}\ (= 1.10)\), so particle \(Q\) does not come to rest before it reaches the pulley | A1 | [2] |
# Question 3:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[7g - T = 7a$ **and** $T - 3g = 3a]$ or $[7g - 3g = 10a]$ | M1 | For applying Newton's second law to P **and** to Q or for using $m_P g - m_Q g = (m_P + m_Q)a$ |
| Acceleration is $4\text{ ms}^{-2}$ | A1 | |
| $[v^2 = 0 + 2 \times 4 \times 0.4]\ (v^2 = 3.2)$ | M1 | For using $v^2 = u^2 + 2as$ |
| Speed is $1.79\text{ ms}^{-1}$ | A1 | [4] |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[0 = 3.2 + 2 \times (-g) \times s]\ (s = 0.16)$ | M1 | For using $0 = u^2 + 2(-g)s$ |
| $0.16 + 0.4 = 0.56$, so particle $Q$ does not come to rest before it reaches the pulley | A1 | [2] |
**Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[v^2 = 3.2 + 2 \times (-g) \times 0.1]$ | M1 | For using $v^2 = u^2 + 2(-g)(0.1)$ |
| $v = \sqrt{1.2}\ (= 1.10)$, so particle $Q$ does not come to rest before it reaches the pulley | A1 | [2] |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{94c11160-a718-4de5-867a-27c755051fa6-2_312_1207_1320_468}
Particles $P$ and $Q$, of masses 7 kg and 3 kg respectively, are attached to the two ends of a light inextensible string. The string passes over two small smooth pulleys attached to the two ends of a horizontal table. The two particles hang vertically below the two pulleys. The two particles are both initially at rest, 0.5 m below the level of the table, and 0.4 m above the horizontal floor (see diagram).\\
(i) Find the acceleration of the particles and the speed of $P$ immediately before it reaches the floor.\\
(ii) Determine whether $Q$ comes to instantaneous rest before it reaches the pulley directly above it.
\hfill \mbox{\textit{CAIE M1 2016 Q3 [6]}}