A bullet is fired normally on an immovable plank of wood. It loses 25% of its kinetic energy in penetrating a thicknees x of the plank. What is the total thickness penetrated by the bullet?

Let ‘u’ be the initial velocity of the bullet of mass ‘m’.

Initial kinetic energy = ½ mu^{2}

If ‘v’ is the final velocity then,

Final kinetic energy = ½ mv^{2} = ½ mu^{2} – ½ mu^{2} × 25/100

=> v^{2} = 0.75u^{2}

Suppose, the acceleration of the bullet is ‘a’.

Then, v^{2} = u^{2} + 2ax

=> a = (v^{2} – u^{2})/(2x)

=> a = (0.75u^{2} – u^{2})/2x

=> a = -0.25u^{2}/(2x)

With this acceleration the bullet penetrates until it comes to rest. If, ‘s’ is the total penetration then,

Using,

v^{2} = u^{2} + 2as

=> 0 = u^{2} + 2{-0.25u^{2}/(2x)}s

=> s = x/0.25 = 4x

So, the total penetration of the bullet is 4x.

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