A rifle bullet loses left(frac{1}{25}right)^{ | vedclass

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(D) Let $u$ be the initial velocity of the bullet. After passing through one plank,the velocity becomes $v = u - \frac{u}{25} = \frac{24u}{25}$.Using

Vedclass · Accepted Answer

$13$

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(D) Let $u$ be the initial velocity of the bullet. After passing through one plank,the velocity becomes $v = u - \frac{u}{25} = \frac{24u}{25}$.Using the kinematic equation $v^2 - u^2 = 2as$,where $s$ is the thickness of one plank and $a$ is the constant retardation:$\left(\frac{24u}{25}\right)^2 - u^2 = 2as$$\frac{576u^2}{625} - u^2 = 2as$$2as = -\frac{49u^2}{625}$.If $n$ planks are required to stop the bullet,the final velocity becomes $0$ after a total distance $ns$:$0^2 - u^2 = 2a(ns)$$-u^2 = n(2as)$$-u^2 = n \left(-\frac{49u^2}{625}\right)$$n = \frac{625}{49} \approx 12.75$.Since the number of planks must be an integer,we need at least $13$ planks to stop the bullet.

$A$ rifle bullet loses $\left(\frac{1}{25}\right)^{th}$ of its velocity in passing through a plank. The least number of such planks required just to stop the bullet is

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