An isolated particle of mass m is moving in a horizontal plane (x - y), along the x-axis at a certain height about the ground. It explodes suddenly into two fragments of masses $\frac{\text{m}}{4}$ and $3\frac{\text{m}}{4}.$ An instant later, the smaller fragment is at y = + 15 cm. What is the position of larger fragment at this instant?
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As isolated particle is moving along x-axis at a certain height above the ground, there is no motion along Y-axis. Further, the explosion is under internal forces only. Therefore, centre of mass remains stationary along Y-axis after collision. Let the co-ordinates of centre of mass be $\left(\mathrm{x}_{\mathrm{cm}}{ }^{\prime} 0\right)$ Now, $\text{y}_{\text{cm}}=\frac{\text{m}_1\text{y}_1+\text{m}_2\text{y}_2}{\text{m}_1+\text{m}_2}=0$
$\therefore\text{m}_1\text{y}_1+\text{m}_2\text{y}_2=0$
$\text{or }\text{y}_2=-\frac{\text{m}_1\text{y}_1}{\text{m}_2}$
$=-\frac{\frac{\text{m}}{4}}{3\frac{\text{m}}{4}}\times15=-5\text{cm}$
$\therefore$ Larger fragment will be at y = -5cm; along x-axis.
$\therefore\text{m}_1\text{y}_1+\text{m}_2\text{y}_2=0$
$\text{or }\text{y}_2=-\frac{\text{m}_1\text{y}_1}{\text{m}_2}$
$=-\frac{\frac{\text{m}}{4}}{3\frac{\text{m}}{4}}\times15=-5\text{cm}$
$\therefore$ Larger fragment will be at y = -5cm; along x-axis.
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