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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 ? 11 years ago

Dear mohammod

Let the thickness of one plank = d
and the acceleration provided by the plank = a

If n planks are required to stop the bullet, then
0^2 = vo^2 + 2a*nd
2and = -vo^2

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

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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11 years ago

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

v=u/20 after penetrating 0ne plank

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

n=1

3 years ago
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....
one year ago
Dear Student,

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 * 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

Thanks and Regards