A bullet loses 1/20 of its velocity after penetrating into a plank. How many planks are required to stop the bullet ?

Dear mohammod

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

 

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applying v^2=u^2+2ad  ,here d is the width of plank

 v=u/20 after penetrating 0ne plank

  2ad=399u^2/400 .........eq1 

  for n number of planks total distance after which bullet stops is nd

 again applying v^2=u^2+2adn  and putting v=0

  n=u^2/2ad......eq2

 solving 1and 2 

   n=1

The formula is m=n^2÷2n-1Substituting in this we will be getting 400÷39So the answer should be approximately 10Therefore the required number of planks is 10....

Dear Student,
Please find below the solution to your problem.

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

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