Question

a bullet loses 1/20 of its velocity on passing through a plank . the least no.of required to stop bullet is

Answer 1

Let the thickness of one plank = d
and the acceleration provided by the plank = a

v^2 = vo^2 + 2ad
If n planks are required to stop the bullet, then
0^2 = vo^2 + 2a*nd
2and = -vo^2
n = vo^2/(-2ad) -----------------(1)

v = vo - vo/20 = 19 vo/20 in passing through one plank
(19 vo/20)^2 = vo^2 + 2ad
361/400 * vo^2 = vo^2 + 2ad
-2ad = vo^2(1 - 361/400)
-2ad = vo^2 * 39/400

Substituting this value of -2ad into equation (1):
n = vo^2/(vo^2 * 39/400) = 400/39
The minimum number of planks needed = smallest integer greater than 400/39 = 11
Ans: 11

Answer 2

Hey !

Let v be the speed of the bullet incident on the first plank.

Its speed after it passes the plank = 19/20 v

If x is the thickness of the plank, the deceleration a due to resistance of plank is given by,

           (19/20)² v² - v² = 2ax

=> 2ax = -39/400 v²

Let the bullet is stopped after passing through n such planks. Then the distance covered by bullet is x

      => 0 - (19/20)² v² = 2anx

     => -(19/20)v² = n × -39/400 v²

     => n = 361/39 = 9.25 ≈ 9

GOOD LUCK !