Question

A steel ball of mass 0.5 kg is fastened to a cord 20 cm long and fixed at the far end and is released when the cord is horizontal. At the bottom of its path the ball strikes a 2.5 kg steel block initially at rest on a frictionless surface. The collision is elastic. The speed of the block just after the collision will be :

• A. $\frac{10}{3}m{s}^{-1}$
• B. $\frac{2}{3}m{s}^{-1}$
• C. $5m{s}^{-1}$
• D. $\frac{5}{3}m{s}^{-1}$

Answer 1

The correct option is B $\frac{2}{3}m{s}^{-1}$

Length of the cord $L=20$ cm $=0.2$ m

Let velocity of ball at the time of collision be $u$

Work-energy theorem for the ball : $W_g=\Delta K.E$

$m_{1}g\left(0.2\right)=\frac{1}{2}{m}_1{u}^2$

$\therefore$ $u=\sqrt{2g\left(0.2\right)}=\sqrt{2\times 10\times 0.2}=2$ $m{s}^{-1}$

Given : $m_1=0.5kg$ $m_2=2.5kg$ $u_2=0$ $e=1$

Let speed of the block just after the collision be $v_2$

Using $v_2=\frac{(m_2-e{m}_1)u_2+(1+e)m_{1}u}{m_1+m_2}$

$v_2=\frac{(2.5-1\times 0.5)\times 0+(1+1)\left(0.5\right)\left(2\right)}{0.5+2.5}=\frac{2}{3}$ $m{s}^{-1}$