1.

If kinetic energy of a body is increased by 300%, then percentage change in momentum will be

Answer»

SOLUTION :`K.E.=(p^(2))/(2m)` so `p=sqrt(2mk)`
INCREASE in K.E. `=300%` of `k=3k`
Final K.E. `k'=k+3k=4k`
Final MOMENTUM `p'=sqrt(2mk')=sqrt(2mxx4k)=2sqrt(2mk)`
`=2p`
% increase in momnetum `=(p'-p)/pxx100=100%`


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