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 mu2

If ‘v’ is the final velocity then,

Final kinetic energy = ½ mv2 = ½ mu2 – ½ mu2 × 25/100

=> v2 = 0.75u2

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

Then, v2 = u2 + 2ax

=> a = (v2 – u2)/(2x)

=> a = (0.75u2 – u2)/2x

=> a = -0.25u2/(2x)

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

Using,

v2 = u2 + 2as

=> 0 = u2 + 2{-0.25u2/(2x)}s

=> s = x/0.25 = 4x

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

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