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K.E. of a body is increased by 300%. Fin...

K.E. of a body is increased by 300%. Find the % increase in its momentum.

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`K.E.=p^(2)/(2m) "so"p=sqrt(2mk)`
Increase in K.E.=300% of k=3k
`"Final K.E"., K^(.)+3K=4K`
`"Final momentum", p^(.)=sqrt(2mk^(.))=sqrt(2mxx4k)=sqrt(2mk)=2p`
`%"increase in momentum" =(p^(.)=p)/pxx100=100%`
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FULL MARKS-WORK, ENERGY AND POWER-Additional Question Solved (II. Short Answer Questions)
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