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(A) Let the velocities of the fragments be $v_1$ and $v_2$ for the $3 \ kg$ and $6 \ kg$ masses respectively. Since the initial momentum is zero,by th

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$240$

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(A) Let the velocities of the fragments be $v_1$ and $v_2$ for the $3 \ kg$ and $6 \ kg$ masses respectively. Since the initial momentum is zero,by the law of conservation of linear momentum: $3v_1 = 6v_2$,which implies $v_1 = 2v_2$.The kinetic energy of the $6 \ kg$ fragment is given by $KE_2 = \frac{1}{2} 	imes 6 	imes v_2^2 = 120 \ J$.This simplifies to $3v_2^2 = 120$,so $v_2^2 = 40 \ (m/s)^2$.The kinetic energy of the $3 \ kg$ fragment is $KE_1 = \frac{1}{2} 	imes 3 	imes v_1^2$.Substituting $v_1 = 2v_2$,we get $KE_1 = \frac{1}{2} 	imes 3 	imes (2v_2)^2 = \frac{1}{2} 	imes 3 	imes 4v_2^2 = 6v_2^2$.Since $v_2^2 = 40$,$KE_1 = 6 	imes 40 = 240 \ J$.

$A$ bomb of mass $9 \ kg$ explodes into two fragments of masses $3 \ kg$ and $6 \ kg$. If the kinetic energy of the $6 \ kg$ fragment is $120 \ J$,find the kinetic energy of the $3 \ kg$ fragment in $J$.

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