If kinetic energy of a body is increased by 44%,its momentum will increase by?
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K.E is increased by 44%.
K.E = 1/2mv^2.
K.E’ =1+44% =1+0.44 =1.44
1.44= 1/2mv^2
K.E = p^2÷2m
Let the initial momentum be K.E and increased momentum be K.E’
K.E’ = 1.44K.E
p’^2/2m= 1.44p^2/2m
p’^2= 1.44p^2
p'= √1.44p^2
p'= 1.2p
Let the initial momentum be 1
Increase in momentum = (p'-p)÷p ×100%
(1.2–1)÷1 ×100% = 0.2×100%=20%
Therefore the kinetic energy has been increased by 20%.
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Amrendra Oraon, B. Tech from National Institute of Technology, Calicut (2012)
Answered Sep 14
If we consider this question, there is a logical way to arrive at solution.
We know that Kinetic Energy is propostional to square of velocity v2v2 and mometum is propostional to velocity vv. Also, only physical quantity that changes here is velocity as mass is constant.
Thus, we can only consider velocity for this problem.
Given that new kinetic energy is 44% more than its previous value. This also means that if old v2v2 is 100 then new v2v2 is 144.
Now, this settles down to fact that if old vv=10, then new vv=12.
Thus, there is increase in 20% in velocity. And hence, there will be increase of 20% in momentum as well, because momentum is directly propotional to velocity as stated earlier.
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Ankit Maharshi, former E-tutor at Unacademy
Answered Sep 12
We know that
Kinetic energy=p^2/2m
Here p is momentum of particle and m is the mass of particle.
From here we can say that kinetic energy of a particle is proportional to the square of momentum.
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