A free body of mass 8 , kg is travelling at 2 | vedclass

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(B) As the body splits into two equal parts due to an internal explosion,the momentum of the system remains conserved.Initial momentum = Final momentu

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One part comes to rest and the other moves in the same direction as that of the original body.

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(B) As the body splits into two equal parts due to an internal explosion,the momentum of the system remains conserved.Initial momentum = Final momentum$8 	imes 2 = 4v_1 + 4v_2$$\Rightarrow v_1 + v_2 = 4$ ...$(i)$By the law of conservation of energy:Initial kinetic energy + Energy released = Final kinetic energy$\frac{1}{2} 	imes 8 	imes (2)^2 + 16 = \frac{1}{2} 	imes 4 	imes v_1^2 + \frac{1}{2} 	imes 4 	imes v_2^2$$16 + 16 = 2v_1^2 + 2v_2^2$$v_1^2 + v_2^2 = 16$ ...(ii)From $(i)$,$v_2 = 4 - v_1$. Substituting this into (ii):$v_1^2 + (4 - v_1)^2 = 16$$v_1^2 + 16 - 8v_1 + v_1^2 = 16$$2v_1^2 - 8v_1 = 0$$2v_1(v_1 - 4) = 0$So,$v_1 = 0$ or $v_1 = 4 \, m/s$.If $v_1 = 0$,then $v_2 = 4 \, m/s$. If $v_1 = 4$,then $v_2 = 0$.Thus,one part comes to rest and the other moves in the same direction as that of the original body.

$A$ free body of mass $8 \, kg$ is travelling at $2 \, m/s$ in a straight line. At a certain instant,the body splits into two equal parts due to an internal explosion which releases $16 \, J$ of energy. Neither part leaves the original line of motion. Finally:

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