A free body of mass 8 kg is moving at 2 m c | vedclass

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(B) Initial mass $M = 8 \ kg$,initial velocity $u = 2 \ m/s$. Initial momentum $P_i = M \cdot u = 8 	imes 2 = 16 \ kg \cdot m/s$. Initial kinetic ene

Vedclass · Accepted Answer

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

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(B) Initial mass $M = 8 \ kg$,initial velocity $u = 2 \ m/s$. Initial momentum $P_i = M \cdot u = 8 	imes 2 = 16 \ kg \cdot m/s$. Initial kinetic energy $K_i = \frac{1}{2} M u^2 = \frac{1}{2} 	imes 8 	imes 2^2 = 16 \ J$. After the explosion,the body splits into two equal parts of mass $m = 4 \ kg$ each. Let their velocities be $v_1$ and $v_2$. By conservation of momentum: $m v_1 + m v_2 = P_i \implies 4(v_1 + v_2) = 16 \implies v_1 + v_2 = 4$. The final kinetic energy $K_f = K_i + \Delta E = 16 + 16 = 32 \ J$. Also,$K_f = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 = \frac{1}{2} 	imes 4 	imes (v_1^2 + v_2^2) = 2(v_1^2 + v_2^2) = 32 \implies v_1^2 + v_2^2 = 16$. Solving $v_1 + v_2 = 4$ and $v_1^2 + v_2^2 = 16$: $(v_1 + v_2)^2 = v_1^2 + v_2^2 + 2 v_1 v_2 \implies 16 = 16 + 2 v_1 v_2 \implies v_1 v_2 = 0$. This means either $v_1 = 0$ or $v_2 = 0$. If $v_1 = 0$,then $v_2 = 4 \ m/s$. Thus,one part comes to rest and the other moves in the same direction as the original body.

$A$ free body of mass $8 \ kg$ is moving at $2 \ m \cdot s^{-1}$ along a straight line. It splits into two equal parts due to an internal explosion releasing $16 \ J$ of energy,and neither part deviates from the original line of motion. What happens to the two parts?

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