Question

When a bullet is fired with a velocity of $150~\text{m s}^{-1}$ at a target of thickness 50 cm, it emerges with a velocity of $100~\text{m s}^{-1}$. If another bullet of same mass is fired with same velocity at a second target of thickness 80 cm, then the velocity with which the bullet emerges from the second target is (Retarding forces are equal in both the cases)

• A. $60~\text{m s}^{-1}$
• B. $75~\text{m s}^{-1}$
• C. $50~\text{m s}^{-1}$
• D. $40~\text{m s}^{-1}$

Hint:

When bullets pass through targets with same retarding force, energy lost is proportional to thickness. Use $v_f = \sqrt{v_i^2 - 2a s}$ or proportion of KE lost.

Answer 1

The Correct Option is A

1. Energy lost due to target: $\Delta E \propto \text{thickness}$. For first target: $v_1 = 150, v_2 = 100 \implies \Delta KE_1 = \frac{1}{2} m (150^2 - 100^2)$.
2. For second target of 80 cm, proportionally: $\Delta KE_2 = \Delta KE_1 \times \frac{80}{50}$.
3. Solve for $v_f$: $\frac{1}{2} m (150^2 - v_f^2) = \Delta KE_2 \implies v_f = 60~\text{m s}^{-1}$.