The answer is 11 planks
Solution:
Let us assume that the thickness of the plank be s and acceleration provided by each plank is a. Using equation of motion we have,
v2−u2=2as
As the bullet stops finally, therefore above equation can be rewritten as,
0−u2=2as
Now, let the number of planks required be n, therefore the equation can be written as,
−u2=2asn
or,
n=−u2/2as ... (1)
Now, question says that the bullet loses its speed by 1/20 on passing through plank, so the final speed of the bullet when it leaves one plank is,
v=u−u/20
v=19u/20
Again using equation of motion we have,
v2−u2=2as
Substituting for v in above equation we get,
(19u/20)2−u2=2as
361u2/400−u2=2as
−39u2/400=2as
or,
2as=−39u2/400 ... (2)
substituting equation (2) in equation (1) we get,
So, the number of planks required is 11 because 0.26 can be considered as one plank.
