Question:
Two unequal masses are connected on two sides of a light string passing over a light and smooth pulley as shown in the figure. The system is released from the rest. The larger mass is stopped for a moment, 1 s after the system is set into motion. The time elapsed before the string is tight again is
Two unequal masses are connected on two sides of a light string passing over a light and smooth pulley as shown in the figure. The system is released from the rest. The larger mass is stopped for a moment, 1 s after the system is set into motion. The time elapsed before the string is tight again is
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After the larger mass is stopped, the smaller mass moves as a projectile under gravity. Use kinematics to find when the displacements match.
Updated On: Apr 20, 2026
- 1/4 s
- 1/2 s
- 2/3 s
- 1/3 s
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
When the larger mass (2 kg) is stopped, the smaller mass (1 kg) continues upward and then decelerates under gravity. The string becomes taut again when the 1 kg mass returns to the same position.
Step 2: Detailed Explanation:
System acceleration: \(a = (2-1)g/(2+1) = 10/3\) m/s\(^2\). Velocity after 1 s: \(v_0 = 10/3\) m/s. After stopping 2 kg, the 1 kg mass moves up with deceleration \(g = 10\) m/s\(^2\), and 2 kg is stationary. Time for string to become taut again: \[ v_0 t - \frac{1}{2}g t^2 = \frac{1}{2}g t^2 \Rightarrow v_0 t = gt^2 \Rightarrow t = v_0/g = \frac{10/3}{10} = \frac{1}{3} \text{ s} \]
Step 3: Final Answer:
Time elapsed before string is tight again \(= \dfrac{1}{3}\) s.
When the larger mass (2 kg) is stopped, the smaller mass (1 kg) continues upward and then decelerates under gravity. The string becomes taut again when the 1 kg mass returns to the same position.
Step 2: Detailed Explanation:
System acceleration: \(a = (2-1)g/(2+1) = 10/3\) m/s\(^2\). Velocity after 1 s: \(v_0 = 10/3\) m/s. After stopping 2 kg, the 1 kg mass moves up with deceleration \(g = 10\) m/s\(^2\), and 2 kg is stationary. Time for string to become taut again: \[ v_0 t - \frac{1}{2}g t^2 = \frac{1}{2}g t^2 \Rightarrow v_0 t = gt^2 \Rightarrow t = v_0/g = \frac{10/3}{10} = \frac{1}{3} \text{ s} \]
Step 3: Final Answer:
Time elapsed before string is tight again \(= \dfrac{1}{3}\) s.
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