If the momentum of a body is increased by 100% the percentage, we get
p_2=2p_1
The kinetic energy
K=(mv^2)/2=p^2/2m
So
K_2=(p_2^2)/2m=(4p_1^2)/2m=4K_1
Thus, the kinetic energy of a body is increased by four times or by 300%.
Answer: 300%
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