AIIMS 2018 Physics Elastic and Inelastic Collisions MCQ Question

To find the velocity of the bullet after it exits the block, we can use the principle of conservation of energy. The block rises to a height of 0.2 m, which means it gains potential energy (PE = mgh). For the block of mass 4 kg, this potential energy is $PE = 4 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 0.2 \, \text{m} = 7.848 \, \text{J}$. Using conservation of momentum before and after the collision, we can set up the equation $m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'$, where $m_1 = 0.02 \, \text{kg}$ (bullet), $v_1 = 600 \, \text{m/s}$, $m_2 = 4 \, \text{kg}$, $v_2 = 0$, and $v_1'$ is the bullet's velocity after the collision. Solving these equations leads to $v_1' = 200 \, \text{m/s}$, confirming option A as correct.

Options B (150 m/s), C (400 m/s), and D (300 m/s) do not satisfy the