Q. The average force necessary to stop a bullet of mass $20\ g$ moving with a speed of $250\ m/s$, as it penetrates into the wood for a distance of $12\ cm$ is
A) $2.2 \times 10^{3}\ N$
B) $3.2 \times 10^{3}\ N$
C) $4.2 \times 10^{3}\ N$
D) $5.2 \times 10^{3}\ N$
Explanation
(d) $u = 250\ m/s,\ v = 0,\ s = 0.12\ metre$
$F = ma = m\left( \frac{u^{2} - v^{2}}{2s} \right) = \frac{20 \times 10^{- 3} \times (250)^{2}}{2 \times 0.12}$
$\therefore F = 5.2 \times 10^{3}N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.