Question

A block of mass \( m \) moving on a frictionless surface at speed \( V \) collides elastically with a block of same mass, initially at rest. Now the first block moves at an angle \( \theta \) with its initial direction and has speed \( V_1 \). The speed of the second block after collision is:

• A. \( \sqrt{V^2 + V_1^2} \)
• B. \( \sqrt{V - V_1} \)
• C. \( \sqrt{V^2 - V_1^2} \)
• D. \( \sqrt{V_1^2 - V^2} \)

Hint:

In elastic collisions, both momentum and kinetic energy are conserved. The final velocities can be determined using these conservation laws.

Answer 1

The Correct Option is C

Step 1: Conservation of Momentum.
In elastic collisions, the total momentum before and after the collision is conserved. Since both blocks have the same mass, the velocities are related by the following equation: \[ V_2 = \sqrt{V^2 - V_1^2} \] where \( V_2 \) is the velocity of the second block after the collision.
Step 2: Final Answer.
Thus, the speed of the second block is given by \( \sqrt{V^2 - V_1^2} \).