Question

A small bob A of mass m is attached to a massless rigid rod of length 1 m pivoted at point P and kept at an angle of 60° with vertical. At 1 m below P, bob B is kept on a smooth surface. If bob B just manages to complete the circular path of radius R after being hit elastically by A, then radius R is_______ m :

• A. 3/5
• B. (2 - √3)/5
• C. 1/5
• D. (2 + √3)/5

Hint:

In elastic collisions between two equal masses, if one is at rest, they simply swap velocities. This is a huge time-saver in JEE!

Answer 1

The Correct Option is C

To solve this problem, we need to determine the radius \( R \) of the circular path bob B manages to complete after being struck by bob A. The setup consists of two bobs of equal mass \( m \), with bob A initially at a height, hitting bob B elastically.

Step 1: Calculate Potential Energy

Initially, bob A is at a height of 1 m from the pivot point P. Given the angle of 60° from the vertical, the vertical height \( h \) can be calculated as:

\(h = 1 \times \cos(60^\circ) = 1 \times \frac{1}{2} = \frac{1}{2} \, \text{m}\)

The potential energy of bob A is:

\(PE = m \cdot g \cdot h = m \cdot g \cdot \frac{1}{2}\)

Step 2: Convert Potential Energy to Kinetic Energy

Just before collision, the potential energy of bob A converts into kinetic energy:

\(KE = \frac{1}{2} m v^2 = m \cdot g \cdot \frac{1}{2}\)

From this, solve for velocity \( v \):

\(\frac{1}{2} m v^2 = m \cdot g \cdot \frac{1}{2} \implies v^2 = g \implies v = \sqrt{g}\)

Step 3: Elastic Collision

During the elastic collision between the two bobs of equal mass, the velocity of bob A is transferred to bob B. Therefore, bob B also achieves velocity \( v = \sqrt{g} \).

Step 4: Bob B Completes the Circular Path

For bob B to just complete a circular path of radius \( R \), it must possess sufficient kinetic energy to reach the top of the circle against gravitational pull. At the top of the path, the centripetal force required is provided by gravity:

\(m \cdot \frac{v^2}{R} = m \cdot g\)

From the kinetic energy, \(\frac{1}{2} m v^2\), applying the condition at the highest point where velocity is zero:

\(v^2 = 5gR \implies g = 5gR \implies R = \frac{1}{5}\, \text{m}\)

Conclusion

Thus, the radius \( R \) of the circular path is \( \frac{1}{5} \) meters.