Question

Two unequal masses, $1kg$ on the left and $2kg$ on the right are connected on two sides of a light string passing over a light and smooth pulley as shown in figure. The system is released from rest. The larger mass is stopped after $1.0$ second after the system is set into motion, and immediately released again. The time elapsed before the string is tight again is $(g=10{\text{m/s}}^2)$

• A. $\frac{1}{4}\text{second}$
• B. $\frac{1}{2}\text{second}$
• C. $\frac{2}{3}\text{second}$
• D. $\frac{1}{3}\text{second}$

Answer 1

The correct option is D $\frac{1}{3}\text{second}$

Net pulling force $=2g-1g=10\text{N}$
Mass being pulled $=2+1=3\text{kg}$
$\therefore$ Acceleration of the system is $a=\frac{10}{3}{\text{m/s}}^2$
Velocity of both the blocks at $t=1$ second will be
$v_0=at=\frac{10}{3}\times 1=\frac{10}{3}\text{m/s}$
Now, at this moment, velocity of $2\text{kg}$ block becomes zero, while that of $1\text{kg}$ block is $\frac{10}{3}\text{m/s}$ upwards.
Hence, string becomes tight again when, displacement of $1\text{kg}$ block = displacement of $2\text{kg}$ block.
or $v_{o}t-\frac{1}{2}g{t}^2=\frac{1}{2}g{t}^2$
$\Rightarrow t=\frac{v_0}{g}=\frac{\left(10/3\right)}{10}=\frac{1}{3}\text{second}$
Hence, the correct answer is option (d).