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  A bullet looses \left ( \frac{1}{n} \right )^{th}of its velocity passing through one plank. The number of such planks that are required to stop the bullet can be : a)  \frac{n^{2}}{2n-1} b)  \frac{2n^{2}}{n-1} c)  Infinite d)  n Hide
Asked by Ranjeetgupta26068 | 17 May, 2019, 15:29: PM
answered-by-expert Expert Answer
Formula to be used :  v2 = u2 - 2aS  ; where v is final speed , u is initial speed, a is retardation and S is the distance travelled.
 
Let u be initial speed and u[ 1 - (1/n)] be the final speed after going through one plank.
 
then we have,   u2[ 1 - (1/n) ]2 = u2 - 2 a S ................(1)
 
where a is retardation and S is plank thickness.
 
from (1) we get,  retardation a = u2 [ 1 - { 1 - (1/n) }2 ] / (2S) = u2 [  (2/n) - (1/n)2 ] / (2S) = u2 (2n-1)/( 2S n2 )..............(2)
 
if m planks are required to stop the bullet,   then we have   u2 - 2[ ( u2 (2n-1) /(2S n2 )] m S ) = 0  .................(3)
 
Hence from eqn.(3), we get   m =  n2 / ( 2n - 1 )
Answered by Thiyagarajan K | 17 May, 2019, 22:44: PM

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