Question

As shown in fig, a bob of mass $m$ is attached by one end of string of length $1m$ and makes an angle ${60}^{\circ }$ with vertical. When it is released it strikes perfectly elastically with block of mass $3m$ (which is rest on smooth horizontal table). The height of table is also $1m$. The range obtained by $3m$ block from vertical side $CD$ of table is

Answer 1

$u_1$ is the velocity of $m$ just before collision and
$v_2$ is the velocity of $3m$ just after collision.
From Law of conservation of energy
$\frac{1}{2}m{u}_1^2=mgL(1-cos{60}^{\circ })u_1^2=2gL(1-\frac{1}{2})u_1^2=2gL\times \frac{1}{2}=gL{u}_1=\sqrt{gL}\text{As collision is elastic, by conservation of Linear momentum and energy we get}{v}_2=\frac{2m_1{u}_1}{m_1+m_2}=\frac{2m\times \sqrt{gL}}{4m}{v}_2=\frac{1}{2}\sqrt{gL}\text{Equations of motion in Vertical direction for projectile motion of}3mH=1=\frac{1}{2}g{t}^2\Rightarrow t=\sqrt{\frac{2}{g}}\text{Equations of motion in Horizontal direction for projectile motion of}3mR=v_2\times t=\frac{1}{2}\sqrt{g\times 1}\times \sqrt{\frac{2}{g}}R=\frac{1}{\sqrt{2}}metre$.