Find change in ke of body when its momentum is increased by 50%
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Kinetic energy K and linear momentum P are related as
K=P22m………(1)K=P22m………(1)
The new momentum is P+50% 0f P =(3/2)P.
If the new kinetic energy is K’ then
K′=(3/2)P22m=94p22m………(2)K′=(3/2)P22m=94p22m………(2)
Divide equation (2) by Equation (1) to get
K′=(9/4)KK′=(9/4)K
Now the % change in K is
K′−KK100=(9/4)K−KK100=K′−KK100=(9/4)K−KK100= 125%
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