Question

As shown below, bob A of a pendulum having a massless string of length \( R \) is released from 60° to the vertical. It hits another bob B of half the mass that is at rest on a frictionless table in the center. Assuming elastic collision, the magnitude of the velocity of bob A after the collision will be (take \( g \) as acceleration due to gravity):

• A. \( \frac{2}{3} \sqrt{Rg} \)
• B. \( \sqrt{Rg} \)
• C. \( \frac{1}{3} \sqrt{Rg} \)
• D. \( \frac{4}{3} \sqrt{Rg} \)

Hint:

In problems involving elastic collisions, use both conservation of energy and conservation of momentum to solve for unknown velocities.

Answer 1

The Correct Option is C

Velocity of $ A $ just before hitting:

The velocity of $ A $ just before it hits $ B $ is given by:

$ u = \sqrt{2g \frac{R}{2}} = \sqrt{gR} $

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

By Conservation of Momentum (COM):
The total momentum before and after the collision must be conserved. Thus:

$ mu = mv_1 + \frac{m}{2}v_2 $

Simplifying:

$ 2v_1 + v_2 = 2u \quad \dots (i) $

Using the Coefficient of Restitution ($ e $):
The coefficient of restitution is given as $ e = 1 $. By definition:

$ e = \frac{v_2 - v_1}{u} $

Substituting $ e = 1 $:

$ v_2 - v_1 = u \quad \dots (ii) $

Solving Equations (i) and (ii):
From equation (ii):

$ v_2 = v_1 + u $

Substitute $ v_2 = v_1 + u $ into equation (i):

$ 2v_1 + (v_1 + u) = 2u $

Simplify:

$ 3v_1 + u = 2u $

$ 3v_1 = u $

$ v_1 = \frac{u}{3} $

Substitute $ u = \sqrt{gR} $:

$ v_1 = \frac{\sqrt{gR}}{3} $

Final Answer:
The velocity of $ A $ just after the collision is:

$ \boxed{\frac{1}{3} \sqrt{gR}} $