Question

A metal ball of mass 2 kg moving with speed of 36 km/h has a head on collision with a stationary ball of mass 3 kg. If after collision, both the balls move as a single mass, then the loss in K.E. due to collision is

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

Answer 1

The Correct Option is D

To solve this problem, we need to analyze the collision step-by-step. The problem provides the initial conditions and asks us to find the loss in kinetic energy (K.E.) after a collision.

1. First, convert the speed of the moving metal ball to meters per second. Since the speed is given as 36 km/h, we convert it using the formula: v = \frac{36 \times 1000}{60 \times 60} \, \text{m/s} = 10 \, \text{m/s}.
2. Using this speed, calculate the initial kinetic energy of the moving ball using the formula for kinetic energy: K.E_1 = \frac{1}{2} m v^2, where m = 2 \, \text{kg} and v = 10 \, \text{m/s}. K.E_1 = \frac{1}{2} \times 2 \times 10^2 = 100 \, \text{J}.
3. The stationary ball has initial K.E. of 0 \, \text{J} since it is not moving.
4. After the collision, the two balls move together as a single mass of 2 + 3 = 5 \, \text{kg}. Using the law of conservation of momentum, find the velocity v_f of the combined mass: m_1v_1 + m_2v_2 = (m_1 + m_2)v_f, where v_2 = 0 (the second ball is initially stationary). 2 \times 10 + 0 = 5v_f, which gives v_f = 4 \, \text{m/s}.
5. Calculate the combined kinetic energy of both balls after collision: K.E_2 = \frac{1}{2} \times 5 \times 4^2 = 40 \, \text{J}.
6. The loss in kinetic energy is the difference between the initial and final kinetic energies: \Delta K.E = K.E_1 - K.E_2 = 100 - 40 = 60 \, \text{J}.

Therefore, the loss in kinetic energy due to the collision is 60 J, which corresponds to the correct answer according to the options provided.