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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 ?

7 years ago

Prudhvi teja
83 Points

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

Please feel free to post as many doubts on our discussion forum as you can.we will get you the answer and detailed  solution very  quickly.

All the best.

Regards,
Prudhvi Teja

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7 years ago
510 Points

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

7 years ago
Susharitha
17 Points
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....

2 months ago
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