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