A rifle bullet loses left(frac{1}{20}right)^{ | vedclass

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(A) Let the initial velocity of the bullet be $v$. After passing through one plank, the velocity becomes $v' = v - \frac{1}{20}v = \frac{19}{20}v$.Usi

Vedclass · Accepted Answer

$11$

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(A) Let the initial velocity of the bullet be $v$. After passing through one plank, the velocity becomes $v' = v - \frac{1}{20}v = \frac{19}{20}v$.Using the work-energy theorem for one plank of thickness $x$ and retarding force $F$:$-Fx = \frac{1}{2}m(v')^2 - \frac{1}{2}mv^2 = \frac{1}{2}m \left[ \left( \frac{19}{20}v ight)^2 - v^2 ight] = \frac{1}{2}mv^2 \left( \frac{361}{400} - 1 ight) = -\frac{1}{2}mv^2 \left( \frac{39}{400} ight)$.So, $Fx = \frac{39}{800}mv^2$.Let $n$ be the number of planks required to stop the bullet. The final velocity will be $0$ after a distance $nx$:$-F(nx) = \frac{1}{2}m(0)^2 - \frac{1}{2}mv^2 = -\frac{1}{2}mv^2$.Substituting $Fx$:$n \left( \frac{39}{800}mv^2 ight) = \frac{1}{2}mv^2$.$n = \frac{800}{2 	imes 39} = \frac{400}{39} \approx 10.25$.Since the number of planks must be an integer, we need $11$ planks to stop the bullet completely.

$A$ rifle bullet loses $\left(\frac{1}{20}\right)^{th}$ of its velocity in passing through a plank. Assuming that the plank exerts a constant retarding force, the least number of such planks required just to stop the bullet is .............

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