| 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 +0.8 This is a multi-stage pulley problem requiring Newton's second law with friction, kinematics across two phases of motion, and careful geometric reasoning about string lengths. While the individual techniques are standard M1 content, the two-phase analysis and the need to determine distances and apply energy/kinematics correctly across stages elevates this above routine exercises. |
| 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 |
|---|---|---|
| (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 | |
| \(F = 0.3 \times 0.2g\) | B1 | |
| M1 | For substituting for F and solving for \(a\) | |
| Acceleration is \(6 \text{ ms}^{-2}\) | 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 \(3.95 \text{ ms}^{-1}\) | A1 | AG |
| [7] | ||
| (ii) \(0.2a_2 = -0.06g\) | B1ft | ft incorrect F |
| M1 | For using \(v^2 = 3.95^2 + 2a_s[2.1 - \text{ distance moved by B}]\) | |
| \(v^2 = 15.6 + 2(-3)(0.8)\) | A1 | |
| Speed is \(3.29 \text{ ms}^{-1}\) | A1 | |
| [4] |
| Answer | Marks |
|---|---|
| WD against friction \(= 0.06g \times [2.1 - (2 - 0.7)]\) | B1 |
| M1 | For using KE loss = WD against friction |
| \(\frac{1}{2}0.2 \times 3.95^2 - \frac{1}{2}0.2v^2 = 0.48\) | A1 |
| Speed is \(3.29 \text{ ms}^{-1}\) | 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 |
$F = 0.3 \times 0.2g$ | B1 |
| | M1 | For substituting for F and solving for $a$
Acceleration is $6 \text{ ms}^{-2}$ | 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 $3.95 \text{ ms}^{-1}$ | A1 | AG
| | | **[7]**
**(ii)** $0.2a_2 = -0.06g$ | B1ft | ft incorrect F
| | M1 | For using $v^2 = 3.95^2 + 2a_s[2.1 - \text{ distance moved by B}]$
$v^2 = 15.6 + 2(-3)(0.8)$ | A1 |
Speed is $3.29 \text{ ms}^{-1}$ | A1 |
| | | **[4]**
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## Question 6 Alternative for (ii)
WD against friction $= 0.06g \times [2.1 - (2 - 0.7)]$ | B1 |
| | M1 | For using KE loss = WD against friction
$\frac{1}{2}0.2 \times 3.95^2 - \frac{1}{2}0.2v^2 = 0.48$ | A1 |
Speed is $3.29 \text{ ms}^{-1}$ | A1 |
---\includegraphics{figure_6}
Particles $A$ and $B$, of masses 0.2 kg and 0.45 kg respectively, are connected by a light inextensible string of length 2.8 m. The string passes over a small smooth pulley at the edge of a rough horizontal surface, which is 2 m above the floor. Particle $A$ is held in contact with the surface at a distance of 2.1 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 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]}}