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if we increase kinetic energy by 20%, then find percentage increase in linear momentum
Asked by kaziryan.05 | 19 May, 2021, 07:43: AM
If v is speed of particle and m is mass of particle , then Kinetic energy K is given as

K = (1/2) m v2

Momentum p = ( m v )

If we eliminate v and write kinetic energy in terms of momentum , we get

K = p2 / ( 2 m )

If we take logarithm on both side and differentiate , we get

ln(K) = ln(1/2m) +  2 ln(p)

dK/K  = 2 ( dp/p )

Hence if change in kinetic energy is 20%  ,  dK/K = 0.2 .

Hence change in momentum ( dp/p ) = 0.1  or change in momentum 10%
Answered by Thiyagarajan K | 19 May, 2021, 09:11: AM
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