| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Particle on incline, hanging counterpart |
| Difficulty | Standard +0.3 This is a standard connected particles problem requiring resolution of forces on an inclined plane, application of F=ma to both particles, and use of constant acceleration equations. While it involves multiple steps (finding friction, setting up equations for both particles, solving simultaneously, then kinematics), all techniques are routine M1 content with no novel insight required. The given sin α simplifies calculations, making this slightly easier than average for M1. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R = 2.6 \times (12 \div 13)\ (= 2.4)\) | B1 | |
| \([F = 0.2 \times 2.4]\) | M1 | For using \(F = \mu R\) |
| \([T - 2.6(5 \div 13) - F = 0.26a,\ 5.4 - T = 0.54a]\) | M1 | For applying Newton's \(2^{nd}\) law to A or to B |
| For any two of \(T - 1 - 0.48 = 0.26a,\ 5.4 - T = 0.54a\) or \((5.4 - 1 - 0.48) = (0.54 + 0.26)a\) | A1 | |
| Acceleration is \(4.9\ ms^{-2}\) | B1 | |
| Tension is \(2.75\) N \((2.754\) exact\()\) | A1 | [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([s = \frac{1}{2} \times 4.9 \times 0.4^2]\) | M1 | For using \(s = \frac{1}{2}at^2\) |
| Distance is \(0.392\) m | A1 | [2] |
## Question 5:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = 2.6 \times (12 \div 13)\ (= 2.4)$ | B1 | |
| $[F = 0.2 \times 2.4]$ | M1 | For using $F = \mu R$ |
| $[T - 2.6(5 \div 13) - F = 0.26a,\ 5.4 - T = 0.54a]$ | M1 | For applying Newton's $2^{nd}$ law to A or to B |
| For any two of $T - 1 - 0.48 = 0.26a,\ 5.4 - T = 0.54a$ or $(5.4 - 1 - 0.48) = (0.54 + 0.26)a$ | A1 | |
| Acceleration is $4.9\ ms^{-2}$ | B1 | |
| Tension is $2.75$ N $(2.754$ exact$)$ | A1 | [6] |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[s = \frac{1}{2} \times 4.9 \times 0.4^2]$ | M1 | For using $s = \frac{1}{2}at^2$ |
| Distance is $0.392$ m | A1 | [2] |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{2c628138-0729-4160-a95c-d6ab0f199cc5-3_275_663_258_742}
A light inextensible string has a particle $A$ of mass 0.26 kg attached to one end and a particle $B$ of mass 0.54 kg attached to the other end. The particle $A$ is held at rest on a rough plane inclined at angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 5 } { 13 }$. The string is taut and parallel to a line of greatest slope of the plane. The string passes over a small smooth pulley at the top of the plane. Particle $B$ hangs at rest vertically below the pulley (see diagram). The coefficient of friction between $A$ and the plane is 0.2 . Particle $A$ is released and the particles start to move.\\
(i) Find the magnitude of the acceleration of the particles and the tension in the string.
Particle $A$ reaches the pulley 0.4 s after starting to move.\\
(ii) Find the distance moved by each of the particles.
\hfill \mbox{\textit{CAIE M1 2013 Q5 [8]}}