Q157 / 170 · Manipal (MET) 2015 · Physics

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 stoped for a moment, is after the system is set into motion. The time elapsed before the string is tight again is Figure

  • A
    $1 / 4 \mathrm{~s}$
  • B
    $1 / 2 \mathrm{~s}$
  • C
    $2 / 3 \mathrm{~s}$
  • Correct answer
    D
    $1 / 3 \mathrm{~s}$

Solution

Net pulling force $=2 g-1 g=10 \mathrm{~N}$ Mass being pulled $=2+1=3 \mathrm{~kg}$ $\therefore$ Acceleration of the system is $a=\frac{10}{3} \mathrm{~m} / \mathrm{s}^2$ $\therefore$ Velocity of both the blocks at $t=1 \mathrm{~s}$ will be $v_0=a t=\left(\frac{10}{3}\right)(1)=\frac{10}{3} \mathrm{~m} / \mathrm{s}$ Now at this moment, velocity of 2 kg becomes zero, while that of 1 kg block is $10 \mathrm{~m} / \mathrm{s}$ upwards. Hence, string becomes tight again when displacement of 1 kg block = displacement of 2 kg block. $\begin{array}{ll}\Rightarrow & v_0 t-\frac{1}{2} g t^2=\frac{1}{2} g t^2 \\ \Rightarrow & t=\frac{v_0}{g}=\frac{\frac{10}{3}}{10}=\frac{1}{3} \mathrm{~s}\end{array}$

Hence, the correct answer is option: 'D'.

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