Question

A bullet loses 1/20th of its velocity passing through a plank. Find the least number of planks required to just stop the bullet. (Explain in simple terms to me)

Answer 1

Let velocity of bullet is V
and number of planks = n
and each plank's width = L


bullet loses 1/20th speed passing through a plank
e.g
Distance L covered then in plank by bullet then bullet loses 1/20th velocity this phenomena continue again n again .

so, we have require retardation ,
use, kinematics equation ,
v² = u² + 2as
here,
v = V- V/20 = 19V/20
u = V
s = L

(19v/20)² = v² +2aL

361v²/400 - v² = 2aL

a = -39v²/800L ------(1)

now, bullet will be stop , penetrating after n planks e.g final velocity = 0
initial velocity = v
acceleration = -39v²/800L
distance = nL

use, again, same equation ,
v² = u² + 2aS
0 = v² - 2×39v²/800L × nL
v²/ = 78v²n/800
n = 800/78 = 10.25

hence, effective number of planks = 11


Trick :- when bullet loses 1/nth speed after passing through a plank then number of plank = (n/2 + 1)

here , 1/20th
hence n = 20
so, effective number of planks = (20/2+1) = 11