question_answer

An isolated particle of mass m is moving in a horizontal plane \[(x,-y)\] along the x-axis at a certain height above the ground. It explodes suddenly into two fragments of masses m/4 and 3 m/4. An instant later, the smaller fragment is at y = + 15 cm. What is the position of larger fragment at this instant?

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.