If linear momentum is increased by 50% then how much will the kinetic energy increase by, and why?

The kinetic energy increases by 125 %.

Let initial momentum be ##p## and kinetic energy be ##K##.
##p = mv##
If ##p## increases by 50% (0.50##p##)
the new momentum ##p’ = p + 0.50p = 1.50p##
The initial ##K = 1/2 mv^2 = (mv)^2/(2m) = p^2/(2m)##
The new kinetic energy ##K’ = (p’)^2/(2m)##
So ##(K’)/K = (p’^2)/ p^2##
##(K’)/K = (1.50p)^2/p^2 = 1.50^2/1 = 2.25##
##K’ = 2.25 K##
% change = ##(K’ – K)/K × 100 % = (2.25 K – K)/K × 100 % = 125 %##
% increase in kinetic energy = 125%

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