A projectile of mass M is fired so that the h | vedclass

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(C) Let the point of firing be $O$ and the original range be $OQ = 4\, km$. The highest point is $P$,which is at a horizontal distance of $OP = 2\, km

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$10$

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(C) Let the point of firing be $O$ and the original range be $OQ = 4\, km$. The highest point is $P$,which is at a horizontal distance of $OP = 2\, km$ from the firing point.Since the explosion is due to internal forces,the center of mass of the system continues to follow the original parabolic path and hits the ground at point $Q$.Let the mass of the heavier part be $m_1 = 3M/4$ and the lighter part be $m_2 = M/4$.The heavier part falls vertically from $P$,so its horizontal position is $x_1 = OP = 2\, km$.Let the horizontal position of the lighter part be $x_2 = OR$.The position of the center of mass $x_{cm}$ is given by:$x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$Substituting the values:$4 = \frac{(3M/4) 	imes 2 + (M/4) 	imes x_2}{M}$$4 = \frac{3}{2} + \frac{x_2}{4}$$4 - 1.5 = \frac{x_2}{4}$$2.5 = \frac{x_2}{4}$$x_2 = 10\, km$.Thus,the horizontal range of the lighter part is $10\, km$.

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