Question

A metal ball of mass 2 kg moving with a velocity of 36 km/h has a head on collision with a stationary ball of mass 3 kg. If after the collision, the two balls move together, the loss in kinetic energy due to collision is

• A. 140 J
• B. 100 J
• C. 60 J
• D. 40 J

Hint:

When two objects stick together after collision, use conservation of momentum to find the final velocity and calculate the kinetic energy loss.

Answer 1

The Correct Option is B

Step 1: Kinetic Energy Before Collision.
The initial kinetic energy of the moving ball is: \[ KE_1 = \frac{1}{2} m_1 v^2 \] where \( m_1 = 2 \, \text{kg} \) and \( v = 36 \, \text{km/h} = 10 \, \text{m/s} \).
Step 2: Kinetic Energy After Collision.
After the collision, the two balls move together with a combined mass of \( m_1 + m_2 \). Using conservation of momentum, the velocity after collision is: \[ v_f = \frac{m_1 v_1}{m_1 + m_2} \] The final kinetic energy is: \[ KE_2 = \frac{1}{2} (m_1 + m_2) v_f^2 \] The loss in kinetic energy is: \[ \text{Loss} = KE_1 - KE_2 = 100 \, \text{J}. \] Step 3: Conclusion.
The correct answer is (B), 100 J.