Answer:
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\[({{x}_{cm}},0)\].
Now, \[{{y}_{cm}}=\frac{{{m}_{1}}{{y}_{1}}+{{m}_{2}}{{y}_{2}}}{{{m}_{1}}+{{m}_{2}}}=0\]
\[\therefore \]\[{{m}_{1}}{{y}_{1}}+{{m}_{2}}{{y}_{2}}=0\]
or \[{{m}_{2}}=\frac{-{{m}_{1}}{{y}_{1}}}{{{y}_{2}}}\]\[{{y}_{2}}=\frac{-{{m}_{1}}{{y}_{1}}}{{{m}_{2}}}=\frac{-m/4}{3m/4}\times
15=-5cm\]
\[\therefore
\] Larger fragment will be at y = -5 cm; along x-axis.
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