If the kinetic energy of bodu increased by 300% its momentum is increa

1.	If the kinetic energy of bodu increased by 300% its momentum is increased by
Answer» Answer: 41.42% Explanation: Originally- K.E = $\frac{1}{2} <klux>MV</klux>^{2}$ MOMENTUM (P)= mv Now K.E. is increased by 300% We know that mass of a body will ALWAYS remain constant. So only the velocity of the body will change. So, NEW K.E. (KE') = K.E. + 300% of K.E.= KE + $\frac{300}{100}$ * KE= 4KE Therefore KE' = 4KE so, $\frac{1}{2} mv'^{2}$ = 4* $\frac{1}{2} mv^{2}$ There fore, $v'^{2}$ = 2 $v^{2}$ v' = $\sqrt{2}$ v So New momentum (P')= mv' P' = $\sqrt{2}$ mv So increase in momentum = P'-P = $\sqrt{2}$ mv- mv= ( $\sqrt{2}$ -1) mv Increase % ={ ( $\sqrt{2}$ -1)mv/mv}* 100 Increase %= 41.42% Hope you find this useful. Please rate the answer!