Question

A bullet looses (1/n)th of its velocity passing through one plank. The number of such planks that are required to stop the bullet can be?

Answer 1

let initial velocity of bullet =v

so final velocity =v-v/n=v (n-1)/n

Let plank thickness=l

now use formula

v^2=u^2+2as

{v(n-1)/n}^2=v^2+2al

v^2 {(n-1)^2/n^2-1}=2al

a=v^2/2l {(1-2n)/n} ----------(1)

at a time bullet loose their speed

i.e final vel=0

initial vel=v

use formula again

v^2=u^2+2as
0=v^2-2aS (because bullet loose
speed apply retardation by plank)
v^2/2S=a----------(2)

now (1) divided by (2)
S/l=n^2/(2n-1)
S/l is here maximum number of plank penetrated by bullet
so , number of plank=n^2/(2n-1)