Problem:

A bullet fired from a rifle loses 20% of its speed while passing through a wooden plank. Then, minimum number of wooden planks required to completely stop the bullet is?

Solution:

To solve this problem, we need to use the concept of the speed of the bullet after passing through each wooden plank. Let's assume that the initial speed of the bullet is 'v' and the final speed of the bullet after passing through a wooden plank is 'u'.

Step 1: Calculate the speed of the bullet after passing through one wooden plank.

As given in the problem, the bullet loses 20% of its speed while passing through a wooden plank. Therefore, the speed of the bullet after passing through one wooden plank can be calculated as:

u = (80/100) * v
u = 0.8v

Step 2: Calculate the number of wooden planks required to completely stop the bullet.

Let's assume that the bullet is completely stopped after passing through 'n' wooden planks. Then, we can write:

(0.8v)^n = 0
0.8^n * v^n = 0

Since the velocity of the bullet cannot be negative, we can ignore the second equation. Therefore, we have:

0.8^n = 0
n*log(0.8) = log(0)
n = log(0)/log(0.8)
n = infinity

This means that the speed of the bullet will never become zero and it will keep on decreasing after passing through each wooden plank. Therefore, the bullet cannot be completely stopped by any finite number of wooden planks.

Conclusion:

The minimum number of wooden planks required to completely stop the bullet is infinity.

Assume thickness of each wooden plank is x
initial speed u then speed of bullet after passing through one wooden plank is, v=u-20/100u=4/5u
according to 2nd equation of motion, v² = u²-2as
putting the value in above equation then
(4/5u)²-u²=2ax
2ax=-9/25u²
let minimum no. of wooden plank is n, and here v=0
(0)²=u²-2a(nx)
(0)²-u²=2a(nx); n=-u²/2ax
-u²/[ -9/25 u²]=n
n=2.78
n=3( by round off)