Question

The bob A of a simple pendulum is released when the string makes an angle of $45^{\circ}$ with the vertical. It hits another bob B of the same material and same mass kept at rest on the table. If the collision is elastic}

• A. both A and B rise to the same height
• B. both A and B come to rest at B
• C. both A and B move with the same velocity of A
• D. A comes to rest and B moves with the velocity of A

Hint:

For equal masses in elastic collision, velocities are exchanged.

Answer 1

The Correct Option is D

The problem describes an elastic collision between two identical bobs A and B. Here is the step-by-step explanation:

Concepts Involved:

• Elastic Collision: In an elastic collision, both momentum and kinetic energy are conserved.
• Conservation of Energy: Potential energy at the max height is converted into kinetic energy at the lowest point.
• Conservation of Momentum: For two bodies of equal mass, during an elastic collision, the velocities are exchanged.

Step 1: Calculate the Initial Velocity of Bob A

When bob A is released from an angle of \(45^{\circ}\), it swings down converting its potential energy into kinetic energy.

At the top, the potential energy is \(mgh\) where \(h\) is the height. At bottom, all energy is kinetic:

\(\frac{1}{2}mv^2 = mg \cdot (L - L \cos(45^\circ))\), where \(L\) is the length of the pendulum.

Simplifying, \(v = \sqrt{2gL(1 - \cos(45^\circ))}\)

\(v = \sqrt{2gL \cdot \left(1 - \frac{\sqrt{2}}{2}\right)}\)

Step 2: During the Collision

In an elastic collision between two objects of equal mass, the velocities are exchanged:

Before collision: velocity of A = \(v\), velocity of B = 0

After collision: velocity of A = 0, velocity of B = \(v\)

Conclusion:

Bob A comes to rest, and Bob B moves with the velocity that A had before the collision.

Thus, the correct answer is: A comes to rest and B moves with the velocity of A.