A bullet loses left( frac{1}{n} right)^{th} o | vedclass

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(A) Let $u$ be the initial velocity of the bullet of mass $m$. After passing through a plank of thickness $x$,its velocity decreases to $v$.Given that

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$\frac{n^2}{2n - 1}$

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(A) Let $u$ be the initial velocity of the bullet of mass $m$. After passing through a plank of thickness $x$,its velocity decreases to $v$.Given that the bullet loses $\frac{1}{n}$ of its velocity,the final velocity $v$ is:$v = u - \frac{u}{n} = u \left( \frac{n - 1}{n} \right)$Using the work-energy theorem for one plank,where $F$ is the retarding force:$Fx = \frac{1}{2} m u^2 - \frac{1}{2} m v^2$$Fx = \frac{1}{2} m u^2 - \frac{1}{2} m \left( u \frac{n - 1}{n} \right)^2$$Fx = \frac{1}{2} m u^2 \left[ 1 - \frac{(n - 1)^2}{n^2} \right] = \frac{1}{2} m u^2 \left[ \frac{n^2 - (n^2 - 2n + 1)}{n^2} \right] = \frac{1}{2} m u^2 \left( \frac{2n - 1}{n^2} \right)$Let $P$ be the number of planks required to stop the bullet. The total distance traveled is $Px$,and the final velocity is $0$:$F(Px) = \frac{1}{2} m u^2 - 0$$P(Fx) = \frac{1}{2} m u^2$Substituting the value of $Fx$:$P \left[ \frac{1}{2} m u^2 \left( \frac{2n - 1}{n^2} \right) \right] = \frac{1}{2} m u^2$$P = \frac{n^2}{2n - 1}$

$A$ bullet loses $\left( \frac{1}{n} \right)^{th}$ of its velocity while passing through one plank. The number of such planks required to stop the bullet is:

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