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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

Solution:

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,
 v2−u2=2as
As the bullet stops finally, therefore above equation can be rewritten as,
 0−u2=2as
Now, let the number of planks required be n, therefore the equation can be written as,
 −u2=2asn
or,
 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,
 v=u−u/20

v=19u/20
Again using equation of motion we have,
 v2−u2=2as
Substituting for v in above equation we get,
 (19u/20)2−u2=2as

361u2/400−u2=2as

−39u2/400=2as
or,
 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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