| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Lighter particle on surface released, heavier hangs |
| Difficulty | Standard +0.3 This is a standard two-particle pulley system requiring Newton's second law for connected particles, followed by kinematics to find maximum height. The setup is straightforward with clear vertical motion, standard masses, and typical multi-part structure. Slightly easier than average due to simple numbers and well-practiced technique, but requires proper understanding of the system becoming slack after A hits the floor. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For applying Newton's second law to \(A\) or to \(B\) or for using \(m_Ag - m_Bg = (m_A + m_B)a\) | |
| \(0.35g - T = 0.35a\) \(T - 0.15g = 0.15a\) \((0.35 - 0.15)g = (0.35 + 0.15)a\) | A1 | Two of the three equations |
| Acceleration is \(4\ \text{ms}^{-2}\) | B1 | |
| Tension is 2.1 N | B1 | Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v_1^2 = 0 + 8 \times 1.6\ (= 12.8)\) | M1 | For using \(v_1^2 = 0 + 2a \times 1.6\) |
| \(H = 1.6 + (-12.8) \div (-20)\) | M1 | For using \(H = 1.6 + (0 - v_1^2)/(-2g)\) or for using \(h = (0 - v_1^2)/(-2g)\) |
| Greatest height is 2.24 m | A1 | Total: 3 marks |
## Question 4(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For applying Newton's second law to $A$ or to $B$ or for using $m_Ag - m_Bg = (m_A + m_B)a$ |
| $0.35g - T = 0.35a$ $T - 0.15g = 0.15a$ $(0.35 - 0.15)g = (0.35 + 0.15)a$ | A1 | Two of the three equations |
| Acceleration is $4\ \text{ms}^{-2}$ | B1 | |
| Tension is 2.1 N | B1 | Total: 4 marks |
## Question 4(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v_1^2 = 0 + 8 \times 1.6\ (= 12.8)$ | M1 | For using $v_1^2 = 0 + 2a \times 1.6$ |
| $H = 1.6 + (-12.8) \div (-20)$ | M1 | For using $H = 1.6 + (0 - v_1^2)/(-2g)$ or for using $h = (0 - v_1^2)/(-2g)$ |
| Greatest height is 2.24 m | A1 | Total: 3 marks |
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\includegraphics[max width=\textwidth, alt={}, center]{f23ea8e7-9b81-4192-8c20-8c46aabfecca-3_442_495_255_826}
Particles $A$ and $B$, of masses 0.35 kg and 0.15 kg respectively, are attached to the ends of a light inextensible string which passes over a fixed smooth pulley. The system is at rest with $B$ held on the horizontal floor, the string taut and its straight parts vertical. $A$ is at a height of 1.6 m above the floor (see diagram). $B$ is released and the system begins to move; $B$ does not reach the pulley. Find\\
(i) the acceleration of the particles and the tension in the string before $A$ reaches the floor,\\
(ii) the greatest height above the floor reached by $B$.
\hfill \mbox{\textit{CAIE M1 2015 Q4 [7]}}