1. Speed of bob A just before collision

Bob $A$ is released from $60^\circ$ to the vertical, with string length $R$.

The vertical drop of the bob till the lowest point is $h = R - R\cos 60^\circ = R - \frac{R}{2} = \frac{R}{2}.$

By conservation of mechanical energy, $mgh = \frac{1}{2} m u_A^2.$ So, $mg\left(\frac{R}{2}\right) = \frac{1}{2} m u_A^2$ $u_A^2 = gR$ $u_A = \sqrt{gR}.$

Thus, just before collision, bob $A$ has speed $u_A = \sqrt{Rg}.$ Bob $B$ is initially at rest, so $u_B = 0.$

2. Apply 1D elastic collision formula

Let mass of bob $A$ be $m$. Then mass of bob $B$ is half of this: $m_B = \frac{m}{2}.$

For a head-on elastic collision, final velocity of mass $m_1$ is $v_1 = \frac{m_1-m_2}{m_1+m_2}u_1 + \frac{2m_2}{m_1+m_2}u_2.$

Here, $m_1=m,\quad m_2=\frac{m}{2},\quad u_1=\sqrt{Rg},\quad u_2=0.$

So, $v_A = \frac{m-\frac{m}{2}}{m+\frac{m}{2}}\sqrt{Rg}$ $v_A = \frac{\frac{m}{2}}{\frac{3m}{2}}\sqrt{Rg}$ $v_A = \frac{1}{3}\sqrt{Rg}.$

Since the question asks for the magnitude of velocity of bob $A$ after collision, $|v_A| = \frac{1}{3}\sqrt{Rg}.$

3. Check options

• A: $\frac{4}{3}\sqrt{Rg}$ — incorrect
• B: $\frac{1}{3}\sqrt{Rg}$ — correct
• C: $\sqrt{Rg}$ — incorrect
• D: $\frac{2}{3}\sqrt{Rg}$ — incorrect

Therefore, the correct answer is Option B.