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, $1s$ after the system is set into motion. The time elapsed before the string is tight again is :

• A. $1/4s$
• B. $1/2s$
• C. $2/3s$
• D. $1/3s$

Answer 1

The correct option is D $1/3s$

Net pulling force $=2g-1g=10N$

Mass being pulled $=2+1=3kg$

$\therefore$ Acceleration of the system is $a=\frac{10}{3}m/s^2$

$\therefore$ Velocity of both the blocks at $t=1s$ will be

$v_0=at=\left(\frac{10}{3}\right)\left(1\right)=\frac{10}{3}m/s$

Now at this moment, velocity of $2kg$ becomes zero while that of $1kg$ block is $10m/s$ upwards. Hence, string becomes tight again when displacement of $1kg$ block $=$ displacement of $2kg$ block.

$\Rightarrow v_{0}t-\frac{1}{2}g{t}^2=\frac{1}{2}g{t}^2$

$\Rightarrow t=\frac{v_0}{g}=\frac{\frac{10}{3}}{10}=\frac{1}{3}s$