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`        A bullet loses 1/n of its velocity in passing through a plank . what is the least no of planks required to stop the bullet?`
2 years ago

```							Dear AditiLet   a = acceleration/deceleration of the bullet inside the plank.let  s = thickness of the planklet the initial velocity of the bullet before hitting the plank: = ufinal velocity = v = u - u/n = u (n-1)/nv²  = u² + 2 a su² (n-1)² / n² = u² + 2 a s2 a s = u² [ (n-1)² - n² ] / n²    =  - (2n - 1) u² / n²a = - (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) / n²N = 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 ] + 1where [ 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. RegardsArun (askIITians forum expert)
```
2 years ago
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