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A rifle bullet loses 1/20th of its velocity in passing through a plank. The least no. of such planks required to stop bullet:

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The answer is 11 planks


Let us assume that the thickness of the plank be s and acceleration provided by each plank is a. Using equation of motion we have,
As the bullet stops finally, therefore above equation can be rewritten as,
Now, let the number of planks required be n, therefore the equation can be written as,
 n=−u2/2as  ... (1)

Now, question says that the bullet loses its speed by 1/20 on passing through plank, so the final speed of the bullet when it leaves one plank is,

Again using equation of motion we have,
Substituting for v in above equation we get,


 2as=−39u2/400  ... (2)
substituting equation (2) in equation (1) we get,

So, the number of planks required is 11 because 0.26 can be considered as one plank.

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