a bullet is fired normally on immovable wooden plank. it losses 25%of its momentum in penetrating a thickness 3.5 cm. find the total thickness penetrated by the bullet
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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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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.
If it'll helpful to you then mark me as a brainliest..
~~~ Thanks~~~~
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