$QR=x\;km$
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
$OR = OQ + QR = 4+6 = 10\; km$
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| $(a)$ Gravitational constant $(G)$ | $(i)$ $\left[ L ^{2} T ^{-2}\right]$ |
| $(b)$ Gravitational potential energy | $(ii)$ $\left[ M ^{-1} L ^{3} T ^{-2}\right]$ |
| $(c)$ Gravitational potential | $(iii)$ $\left[ LT ^{-2}\right]$ |
| $(d)$ Gravitational intensity | $(iv)$ $\left[ ML ^{2} T ^{-2}\right]$ |
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