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Question

Physics Question on work, energy and power

A bullet looses (1n)th\left(\frac{1}{n}\right)^{th} of its velocity passing through one plank. The number of such planks that are required to stop the bullet can be :

A

n22n1\frac{n^{2}}{2n - 1}

B

2n2n1\frac{2n^{2}}{n - 1}

C

Infinite

D

n

Answer

n22n1\frac{n^{2}}{2n - 1}

Explanation

Solution


(11n)2V2=V22as\left(1-\frac{1}{ n }\right)^{2} V ^{2}= V ^{2}-2 as
2as=V2(1(n1n)2)=V2(2n1n2)2 as = V ^{2}\left(1-\left(\frac{ n -1}{ n }\right)^{2}\right)= V ^{2}\left(\frac{2 n -1}{ n ^{2}}\right)
O=V22ansO = V ^{2}-2 ans
n=V22as=V2V2(2n1n2)=n22n1n =\frac{ V ^{2}}{2 as }=\frac{ V ^{2}}{ V ^{2}\left(\frac{2 n -1}{ n ^{2}}\right)}=\frac{ n ^{2}}{2 n -1}