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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 ?
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
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,Askiitians ExpertsPrudhvi Teja Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian
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,Askiitians ExpertsPrudhvi Teja
Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian
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
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