A bullet moving with a velocity of 100, m/s c | vedclass

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Question

Given initial velocity of the bullet in first case $\left(\mathrm{u}_{1}\right)=100 \mathrm{m} / \mathrm{s} .$ Initial number of planks $\left(\mathrm

Vedclass · Accepted Answer

$8$

Vedclass · Answer

Given initial velocity of the bullet in first case $\left(\mathrm{u}_{1}ight)=100 \mathrm{m} / \mathrm{s} .$ Initial number of planks $\left(\mathrm{n}_{1}ight)=2$
Initial stopping distance $\left(s_{1}ight)=n_{1} x=2 x$ (where $\mathrm{x}$ is the thickness of one plank).
Initial velocity of the bullet in second case $\left(\mathrm{u}_{2}ight)=200 \mathrm{m} / \mathrm{s} .$ We know that relation for the stopping distance $(s)$ is $:$
$\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}$
since, the bullet is just able to penetrate the planks, therefore its final velocity $\mathrm{v}=0$
Thus,
$2 \mathrm{as}=-\mathrm{u}^{2} 	ext { or } \mathrm{s} \propto \mathrm{u}^{2}$
Therefore, $\quad \frac{s_{1}}{s_{2}}=\left(\frac{u_{1}}{u_{2}}ight)^{2}=\frac{1}{4}$
$	ext { or } \quad s_{2}=4 s_{1}=4 	imes 2 \mathrm{x}=8 \mathrm{x}$
Thus, final number of plank $\left(\mathrm{n}_{2}ight)=\frac{\mathrm{s}_{2}}{\mathrm{x}}=\frac{8 \mathrm{x}}{\mathrm{x}}=8$

A bullet moving with a velocity of $100\, m/s$ can just penetrate two planks of equal thickness. The number of such planks penetrated by the same bullet, when the velocity is doubled, will be

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