JEE Physics 2025 Question with Solution

Given: Bob A of mass $m$ is released from $60^\circ$ to the vertical on a pendulum of length $R$. Bob B has mass $\frac{m}{2}$ and is initially at rest. The collision is elastic.

Find: The magnitude of velocity of bob A after collision.

First find the speed of A just before collision using conservation of mechanical energy. The vertical drop from $60^\circ$ is $R - R\cos 60^\circ = \frac{R}{2}$, so

Hence, the speed just before impact is

Let the velocities of A and B just after collision be $v_1$ and $v_2$ respectively.

Using conservation of momentum along the line of impact,

which gives

Since the collision is elastic, coefficient of restitution $e = 1$. Therefore,

so

From equation $\text{(ii)}$,

Substitute into equation $\text{(i)}$:

Now substitute $\nu = \sqrt{gR}$:

Therefore, the magnitude of velocity of bob A after collision is $\frac{1}{3}\sqrt{gR}$. The correct option is A.

The solution's lists option C in the answer key, but the extracted solution working and final boxed result give $\frac{1}{3}\sqrt{gR}$, so the solution has been followed.

Given: The pendulum bob starts from rest at $60^\circ$ to the vertical and strikes a lighter bob at the lowest point.

Find: Speed of bob A after an elastic collision.

The height fallen by bob A is obtained from geometry:

So just before collision,

Now apply the two collision relations:

1. Momentum conservation:

2. Elastic collision condition:

These two equations are sufficient to determine $v_1$ and $v_2$.

Multiply the momentum equation by $\frac{2}{m}$:

Using $v_2 = v_1 + \nu$,

Hence, bob A moves after collision with speed $\frac{1}{3}\sqrt{gR}$.