Question

A block of mass $m_1$ lies on a smooth horizontal table and is connected to another freely hanging block of mass $m_2$ by a light inextensible string passing over a smooth fixed pulley situated at the edge of the table as shown in the figure. Initially the system is at rest with $m_1$ at a distance $d$ from the pulley. The time taken for $m_1$ to reach the pulley is

• A. $\frac{m_{2}g}{m_1+m_2}$
• B. $\sqrt{\frac{2d\left(m_1+m_2\right)}{m_{2}g}}$
• C. $\sqrt{\frac{2m_{2}d}{\left(m_1+m_2\right)g}}$
• D. $\sqrt{\frac{2m_{1}d}{\left(m_1+m_2\right)g}}$

Answer 1

Answer: (B)

Let $a$ be the common acceleration of the system. The free body diagrams of two blocks are as shown in the figure. Their equations of motion are
$T=m_{1}a\dots$ (i)
$m_{2}g-T=m_{2}a....(i)$
From (i) and (ii), we get
$a=\frac{m_{2}g}{m_1+m_2}....(ii)$
Using, $s=ut+\frac{1}{2}a{t}^2$
$\therefore d=0\times t+\frac{1}{2}\frac{m_{2}g}{m_1+m_2}{t}^2(Using(iii))$
or $t=\sqrt{\frac{2d\left(m_1+m_2\right)}{m_{2}g}}$