Question

A metal ball of mass 2 kg moving with a speed of 10 m/s had a head-on collision with a stationary ball of mass 3 kg. If after collision, both the balls move together, then the loss in kinetic energy due to collision is

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

Hint:

The loss in kinetic energy during a collision can be found by calculating the initial and final kinetic energies and subtracting them.

Answer 1

The Correct Option is A

Step 1: Using conservation of momentum.
Since there is no external force, the total momentum before and after the collision is conserved. Let the initial velocity of the first ball be \( v_1 = 10 \, \text{m/s} \) and the mass of the first ball be \( m_1 = 2 \, \text{kg} \), and the mass of the second ball be \( m_2 = 3 \, \text{kg} \) which is initially at rest. After the collision, both balls move together with velocity \( v_f \). The conservation of momentum gives: \[ m_1 v_1 = (m_1 + m_2) v_f \] Substituting the given values: \[ 2 \times 10 = (2 + 3) v_f \] \[ v_f = \frac{20}{5} = 4 \, \text{m/s} \]
Step 2: Finding the initial and final kinetic energies.
The initial kinetic energy is: \[ KE_{\text{initial}} = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} \times 2 \times 10^2 = 100 \, \text{J} \] The final kinetic energy is: \[ KE_{\text{final}} = \frac{1}{2} (m_1 + m_2) v_f^2 = \frac{1}{2} \times 5 \times 4^2 = 40 \, \text{J} \]
Step 3: Calculating the loss in kinetic energy.
The loss in kinetic energy is: \[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}} = 100 - 40 = 60 \, \text{J} \]
Step 4: Conclusion.
Thus, the loss in kinetic energy due to the collision is 60 J, corresponding to option (A).