| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Energy methods for pulley systems |
| Difficulty | Standard +0.3 This is a standard two-particle pulley system problem requiring straightforward application of energy conservation. Part (i) involves routine calculation of potential energy changes using given masses and heights. Part (ii) applies conservation of energy with a simple kinetic energy formula. The 'show that' format and direct energy method make this slightly easier than average, though it requires careful bookkeeping of the two particles' energy changes. |
| Spec | 6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | PE loss = B's loss − A's gain | |
| Loss of PE \(= 2g \times 3.24 - 1.6g(3.24 \times 0.8)\) | A1 | |
| Loss is \(23.328\ \text{J}\) | A1 [3] | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}(1.6 + 2)v^2 = 23.328\) | B1 | |
| Speed is \(3.6\ \text{m s}^{-1}\) | B1 [2] | |
| SR (max 1/2) for using Newton's second law and \(v^2 = u^2 + 2as\): \(2g - T = 2a\) and \(T - 1.6g \times 0.8 = 1.6a\), \(a = 2\), \(v^2 = 2 \times 2 \times 3.24\), \(v = 3.6\) B1 |
## Question 2:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | PE loss = B's loss − A's gain |
| Loss of PE $= 2g \times 3.24 - 1.6g(3.24 \times 0.8)$ | A1 | |
| Loss is $23.328\ \text{J}$ | A1 [3] | AG |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}(1.6 + 2)v^2 = 23.328$ | B1 | |
| Speed is $3.6\ \text{m s}^{-1}$ | B1 [2] | |
| | | SR (max 1/2) for using Newton's second law and $v^2 = u^2 + 2as$: $2g - T = 2a$ and $T - 1.6g \times 0.8 = 1.6a$, $a = 2$, $v^2 = 2 \times 2 \times 3.24$, $v = 3.6$ B1 |
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\includegraphics[max width=\textwidth, alt={}, center]{ceb367ee-4e12-4cb2-9020-078ea5724d6e-2_529_691_529_726}
Particle $A$ of mass 1.6 kg and particle $B$ of mass 2 kg are attached to opposite ends of a light inextensible string. The string passes over a small smooth pulley fixed at the top of a smooth plane, which is inclined at angle $\theta$, where $\sin \theta = 0.8$. Particle $A$ is held at rest at the bottom of the plane and $B$ hangs at a height of 3.24 m above the level of the bottom of the plane (see diagram). $A$ is released from rest and the particles start to move.\\
(i) Show that the loss of potential energy of the system, when $B$ reaches the level of the bottom of the plane, is 23.328 J .\\
(ii) Hence find the speed of the particles when $B$ reaches the level of the bottom of the plane.
\hfill \mbox{\textit{CAIE M1 2013 Q2 [5]}}