Question

A bullet is fired into a fixed target looses one third of its velocity after travelling 4 cm. It penetrates further $D \times 10^{-3}$ m before coming to rest. The value of $D$ is:

• A. 3
• B. 2
• C. 4
• D. 5

Answer 1

Answer: (D)

5

Answer 2

Let the initial velocity be $v_0$ and the deceleration be $a$. Using $v^2 = u^2 + 2as$: Phase 1: $v_1 = \frac{1}{3}v_0$, $s_1 = 0.04$ m. $(\frac{1}{3}v_0)^2 = v_0^2 + 2as_1 \implies -\frac{8}{9}v_0^2 = 2as_1$. Phase 2: $v_2 = 0$, $u = v_1 = \frac{1}{3}v_0$, $s_2 = D \times 10^{-3}$ m. $0^2 = (\frac{1}{3}v_0)^2 + 2as_2 \implies -\frac{1}{9}v_0^2 = 2as_2$. Dividing Phase 2 by Phase 1: $\frac{-\frac{1}{9}v_0^2}{-\frac{8}{9}v_0^2} = \frac{2as_2}{2as_1} \implies \frac{1}{8} = \frac{s_2}{s_1}$. $s_2 = \frac{1}{8}s_1 = \frac{1}{8}(0.04 \text{ m}) = 0.005 \text{ m}$. $D \times 10^{-3} = 0.005 \implies D = 5$.