| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Multi-part pulley system, subsequent motion |
| Difficulty | Standard +0.3 This is a standard connected particles problem requiring systematic application of Newton's second law to two masses, followed by kinematics calculations. While it has multiple parts and requires careful bookkeeping of the motion phases (before and after Q hits ground), the techniques are routine M1 content with no novel problem-solving insight needed. The geometry is straightforward (sin θ = 40/160 = 1/4), and each part follows predictably from the previous one. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For applying Newton's \(2^{nd}\) law to P or to Q | |
| For \(T - (40 \div 160) \times 0.76g = 0.76a\) or \(0.49g - T = 0.49a\) | A1 | |
| For \(0.49g - T = 0.49a\) or \(T - (40 \div 160) \times 0.76g = 0.76a\) or \(0.49g - (400 \div 160) \times 0.76g = (0.49 + 0.76)a\) | B1 | |
| Acceleration is \(2.4\text{ ms}^{-2}\) and tension is \(3.72\text{ N}\) \((3.724\text{ exact})\) | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([v^2 = 2 \times 2.4 \times 0.3]\) | M1 | For using \(v^2 = 0 + 2as\) |
| Speed is \(1.20\text{ ms}^{-1}\) | A1ft | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using \(v^2 = u^2 + 2as\) with \(v = 0\) and \(a = -(40 \div 160)g\) | |
| Distance while Q is on the ground \(= (2 \times 2.4 \times 0.3) \div 2(40g \div 160)\) \((= 0.288\text{ m})\) | A1ft | ft a from (i) and/or \(s = 30\) |
| Distance travelled is \(0.588\text{ m}\) | A1 | [3] |
## Question 7(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For applying Newton's $2^{nd}$ law to P or to Q |
| For $T - (40 \div 160) \times 0.76g = 0.76a$ or $0.49g - T = 0.49a$ | A1 | |
| For $0.49g - T = 0.49a$ or $T - (40 \div 160) \times 0.76g = 0.76a$ or $0.49g - (400 \div 160) \times 0.76g = (0.49 + 0.76)a$ | B1 | |
| Acceleration is $2.4\text{ ms}^{-2}$ and tension is $3.72\text{ N}$ $(3.724\text{ exact})$ | A1 | [4] |
---
## Question 7(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[v^2 = 2 \times 2.4 \times 0.3]$ | M1 | For using $v^2 = 0 + 2as$ |
| Speed is $1.20\text{ ms}^{-1}$ | A1ft | [2] | ft a from (i) $(a \neq \pm g)$ |
---
## Question 7(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $v^2 = u^2 + 2as$ with $v = 0$ and $a = -(40 \div 160)g$ |
| Distance while Q is on the ground $= (2 \times 2.4 \times 0.3) \div 2(40g \div 160)$ $(= 0.288\text{ m})$ | A1ft | ft a from (i) and/or $s = 30$ |
| Distance travelled is $0.588\text{ m}$ | A1 | [3] |7\\
\includegraphics[max width=\textwidth, alt={}, center]{139371b7-e142-4ed6-bff3-3ca4c32b9c6b-4_342_1257_255_445}
A smooth inclined plane of length 160 cm is fixed with one end at a height of 40 cm above the other end, which is on horizontal ground. Particles $P$ and $Q$, of masses 0.76 kg and 0.49 kg respectively, are attached to the ends of a light inextensible string which passes over a small smooth pulley fixed at the top of the plane. Particle $P$ is held at rest on the same line of greatest slope as the pulley and $Q$ hangs vertically below the pulley at a height of 30 cm above the ground (see diagram). $P$ is released from rest. It starts to move up the plane and does not reach the pulley. Find\\
(i) the acceleration of the particles and the tension in the string before $Q$ reaches the ground,\\
(ii) the speed with which $Q$ reaches the ground,\\
(iii) the total distance travelled by $P$ before it comes to instantaneous rest.
\hfill \mbox{\textit{CAIE M1 2014 Q7 [9]}}