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In the non-relativistic regime, if the momentum is increased by 100%, the percentage increase in kinetic energy is?
Dear Palakshi Let the momentum of the particle be p ,p = m*v.where m is the mass of the particle and v is the velocity of the particle.Now , Kinetic Energy = 1/2 * ( m * v^2 ).K.E. = 1/2 * ( m^2 *v^2 )/m.K.E. = 1/2 * ( p^2 )/m.Now the mass of an object is constant. So the change in momentum is caused by change in velocity .The momentum is increased by 100 % .Therefore , the new momentum would be 2*p .And the new K.E. will beK.E. = 1/2 * ( 4 * p^2 ) / m. i.e., 4 times the old kinetic energy.So, the percentage increase in kinetic energy will be 300 % . RegardsArun (askIITians forum expert)
Let the momentum of the particle be p ,
p = m*v.
where m is the mass of the particle and v is the velocity of the particle.
Now , Kinetic Energy = 1/2 * ( m * v^2 ).
K.E. = 1/2 * ( m^2 *v^2 )/m.
K.E. = 1/2 * ( p^2 )/m.
Now the mass of an object is constant. So the change in momentum is caused by change in velocity .
The momentum is increased by 100 % .
Therefore , the new momentum would be 2*p .
And the new K.E. will be
K.E. = 1/2 * ( 4 * p^2 ) / m. i.e., 4 times the old kinetic energy.
So, the percentage increase in kinetic energy will be 300 % .
Regards
Arun (askIITians forum expert)
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