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

6 years ago


Answers : (2)


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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6 years ago

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


 solving 1and 2 


6 years ago

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