| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Multi-stage motion: particle reaches ground/pulley causing string to go slack |
| Difficulty | Standard +0.3 This is a standard two-stage pulley problem requiring Newton's second law with friction, followed by kinematics with energy considerations. While it involves multiple steps and careful bookkeeping of distances, the techniques are routine for M1: setting up force equations, finding acceleration, using SUVAT, then analyzing motion after B hits the floor. The problem is slightly above average due to the two-stage nature and need to track string constraint carefully, but requires no novel insight. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03k Connected particles: pulleys and equilibrium3.03t Coefficient of friction: F <= mu*R model6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | For applying Newton's second law to A or to B or using \((M + m)a = Mg − F\) | |
| \(0.45a = 0.45g − T\) and \(0.2a = T − F\) or \((0.45 + 0.2)a = 0.45g − F\) | A1 | |
| F = 0.3 × 0.2g | B1 | |
| M1 | For substituting for F and solving for a | |
| Acceleration is 6 ms⁻² | A1 | |
| \([v^2 = 2 \times 6 \times [2 − (2.8 − 2.1)]]\) | M1 | For using \(v^2 = (0^2) + 2as\) (s must be less than 2) |
| Speed is 3.95 ms⁻¹ | A1 | AG |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.2a_2 = −0.06g\) | B1ℹ | ℹ Incorrect F |
| M1 | For using \(v^2 = 3.95^2 + 2a_2[2.1 − \text{distance moved by B}]\) | |
| \(v^2 = 15.6 + 2(−3)(0.8)\) | A1 | |
| Speed is 3.29 ms⁻¹ | A1 | |
| [4] |
| Answer | Marks |
|---|---|
| WD against friction = 0.06g × [2.1 − (2 − 0.7)] | B1 |
| M1 | For using KE loss = WD against friction |
| \(\frac{1}{2}(0.2 \times 3.95^2 − \frac{1}{2}(0.2)v^2 = 0.48\) | A1 |
| Speed is 3.29 ms⁻¹ | A1 |
## (i)
| M1 | For applying Newton's second law to A or to B or using $(M + m)a = Mg − F$
$0.45a = 0.45g − T$ and $0.2a = T − F$ or $(0.45 + 0.2)a = 0.45g − F$ | A1 |
F = 0.3 × 0.2g | B1 |
| M1 | For substituting for F and solving for a
Acceleration is 6 ms⁻² | A1 |
$[v^2 = 2 \times 6 \times [2 − (2.8 − 2.1)]]$ | M1 | For using $v^2 = (0^2) + 2as$ (s must be less than 2)
Speed is 3.95 ms⁻¹ | A1 | AG
| | [7]
## (ii)
$0.2a_2 = −0.06g$ | B1ℹ | ℹ Incorrect F
| M1 | For using $v^2 = 3.95^2 + 2a_2[2.1 − \text{distance moved by B}]$
$v^2 = 15.6 + 2(−3)(0.8)$ | A1 |
Speed is 3.29 ms⁻¹ | A1 |
| | [4]
# Question 6(ii) Alternative
WD against friction = 0.06g × [2.1 − (2 − 0.7)] | B1 |
| M1 | For using KE loss = WD against friction
$\frac{1}{2}(0.2 \times 3.95^2 − \frac{1}{2}(0.2)v^2 = 0.48$ | A1 |
Speed is 3.29 ms⁻¹ | A1 |\includegraphics{figure_6}
Particles $A$ and $B$, of masses $0.2 \text{ kg}$ and $0.45 \text{ kg}$ respectively, are connected by a light inextensible string of length $2.8 \text{ m}$. The string passes over a small smooth pulley at the edge of a rough horizontal surface, which is $2 \text{ m}$ above the floor. Particle $A$ is held in contact with the surface at a distance of $2.1 \text{ m}$ from the pulley and particle $B$ hangs freely (see diagram). The coefficient of friction between $A$ and the surface is $0.3$. Particle $A$ is released and the system begins to move.
\begin{enumerate}[label=(\roman*)]
\item Find the acceleration of the particles and show that the speed of $B$ immediately before it hits the floor is $3.95 \text{ m s}^{-1}$, correct to 3 significant figures. [7]
\item Given that $B$ remains on the floor, find the speed with which $A$ reaches the pulley. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2010 Q6 [11]}}