A projectile of mass \(M\) is fired so that the horizontal range is \(4\, km\). At the highest point the projectile explodes in two parts of masses \(M/4\) and \(3M/4\) respectively and the heavier part starts falling down vertically with zero initial speed. The horizontal range (distance from point of firing) of the lighter part is .................. \(\mathrm{km}\)

\(QR=x\;km\)

The centre of mass will hit the ground at point \(Q\). As the haviour mass comes to rest after bracking

\({x_{cm}} = \frac{{{m_1}{x_1} + {m_2}{x_2}}}{{{m_1} + {m_2}}}\)

\(\;Here\;{m_{1\;}} = \frac{{3M}}{4},{m_{2\;}} = \frac{M}{4},\;\)\(\;{x_{1\;}} = OP = 2,\;{x_{2\;}} = OR\; = \;4 + x\)

So,

\(4 = \frac{{\frac{{3M}}{4} \times 2 + \frac{M}{4} \times \left( {4 + x} \right)}}{M}\)

\(4M = \frac{{3M}}{2} + M\left( {1 + \frac{x}{4}} \right)1 + \frac{x}{4} \)\(= \frac{5}{2}\frac{x}{4} = \frac{3}{2}x = 6\;km\)

Hence horizontal range of lighter particle is

\(OR = OQ + QR = 4+6 = 10\; km\)