Question

As shown in fig, a bob of mass ‘m’ is attached by one end of string of length 1 metre 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 1 metre. The range obtained by ‘3m’ block from vertical side ‘CD’ of table will be…………….

• A. $\sqrt{2}$ metre
• B. $\frac{1}{\sqrt{2}}$ metre
• C. $\frac{1}{2\sqrt{2}}$ metre
• D. $2\sqrt{2}$ metre

Answer 1

The correct option is B

$\frac{1}{\sqrt{2}}$ metre

$u_1$ is the velocity of m just before collision with 2m and $v_2$ is the velocity of 2m just after collision.
$\frac{1}{2}m{u}_1^2=mgL(1-cos{60}^{\circ })u_1^2=2gL(1-\frac{1}{2})u_1^2=2gL\times \frac{1}{2}=gL{u}_1=\sqrt{gL}{v}_2=\frac{2m_1{u}_1}{m_1+m_2}=\frac{2m\times \sqrt{gL}}{4m}{v}_2=\frac{1}{2}\sqrt{gL}H=1=\frac{1}{2}g{t}^{2}t=\sqrt{\frac{2}{g}}R=v_2\times t=\frac{1}{2}\sqrt{g\times 1}\times \sqrt{\frac{2}{g}}R=\frac{1}{\sqrt{2}}metre$.