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An isolated particle of mass M is moving in a horizontal plane (xy), along the x-axis at a certain height above the ground. It suddenly explodes into two fragments of mass m/4 & 3m/4. And instant later the smaller fragment is at Y=15cm.The larger fragment at this instance is at :

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Solution

as the particle of mass m is moving along x axis its potential as well as kinetic energy about y axis is O. Hence, by conservation of energy mgh= (m/4)g15 + (3m/4)gy =0 { because h along y axis is 0} or , (-mg ) x 15=3 x mgy hence y= -5cm Or
As the particle is exploded only due to its internal energy. So net external force during this process is 0. So the centre mass will not change.
Let the particle while explosion was above the origin of the co-ordinate system. i e just before explosion X=0 and Y=0

Centre of mass will be at X=0 and Y=0

after explosion Centre of mass will be at X=0 and Y=0

Ycm= [(m/4 * 15) + (3m/4 * y)]/m =0

solving y=-5cm

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