A bullet loses 1 n of its velocity
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A bullet loses 1/nth of its velocity passing through a plank.. how many planks are required to stop the bullet?
Answered by admin @
Let a = acceleration/deceleration of the bullet inside the plank.
let s = thickness of the plank
let the initial velocity of the bullet before hitting the plank: = u
final velocity = v = u - u/n = u (n-1)/n
v² = u² + 2 a s
u² (n-1)² / n² = u² + 2 a s
2 a s = u² [ (n-1)² - n² ] / n² = - (2n - 1) u² / n²
a = - (2 n - 1) u² / (n² 2 s)
let S be the thickness of the N planks kept one after another to stop the bullet. The deceleration is the same. The initial velocity is the same.
v² = 0 = u² - 2 S * (2 n - 1) u² / (n² 2 s)
1 = S/s * (2 n -1) / n²
N= S/s = n² / (2 n - 1) = 1/2 * [ (2n² -n) + n ] / (2n -1)
= 1/2 * [ n + n / (2n -1) ]
= n/2 + n/(4n-2)
the second term above is between 0 and 1.
N = number of planks = [ n/2 ] + 1
where [ n/2 ] is the greatest integer function for n/2.
=====================================
let n = 1, S/s = 1 : so one plank.
n = 2, S/s = 4/3: so two planks needed.
n = 3, S/s = 9/5 , so two planks needed.
n = 4, S/s = 16/7 , so three planks are needed.
n = 10, S/s = 100/19 : six needed.
n = 11, S/s = 121/21 : six needed.
n = 21. S/s = 21² / (41) = 10.75. so eleven planks needed.
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