A bullet when fired at a target with velocity of $100\ ms^{- 1}$ penetrates 1 m into it. If the bullet is fired at a similar target with a thickness 0.5m, then it will emerge from it with a velocity of

Question

Dr. Abubacker · Accepted Answer

(a) Let $v$ be the velocity with which the bullet will emerge
Now, change in kinetic energy = work done
For first case, $\frac{1}{2}m(100)^{2} - \frac{1}{2}m 	imes 0 = F$
For second case, $\frac{1}{2}m(100)^{2} - \frac{1}{2}mv^{2} = F 	imes 0.5$
Dividing eq. (ii) by Eq. (i), we get
$\frac{(100)^{2} - (v)^{2}}{(100)^{2}} = \frac{0.5}{1} = \frac{1}{2}\ \ or\ \ v = \frac{100}{\sqrt{2}} = 50\sqrt{2}ms^{- 1}$

Q. A bullet when fired at a target with velocity of $100\ ms^{- 1}$ penetrates 1 m into it. If the bullet is fired at a similar target with a thickness 0.5m, then it will emerge from it with a velocity of

Explanation

Reviewed by Dr. Abubacker

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