Question

A bullet losses $1/n$ of its velocity in passing through a plank. What is the least number of planks required to stop the bullet? (Assuming constant retardation)

Answer 1

Assume that the energy loss per plank is constant. This is not the same as the bullet losing $\frac{1}{n^{t}h}$of its velocity per plank.

According to this assumption, the energy loss becomes

ΔE= $\frac{1}{2}\times m{v}^2-[\frac{1}{2}\times m(v-\frac{v}{n}{)}^2]$

so the number of planks N is,

N= $\frac{1}{2}\times \frac{m{v}^2}{△E}$ = $\frac{n^2}{2n-1}$

Number of such planks required to stop the bullet are $\frac{n^2}{2n-1}$