Q.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 suddenly explodes into two fragments of masses m/4 and 3m/4. An instant later,the smaller fragment is at y=15 cm.The larger fragment at this instant is at ?
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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
Answered by
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According to the information provided in the question, the internal energy of the particle was responsible for the explosion.
Therefore no external force was provided.
As a result of this, the centre of mass will be at X=0 and y=0.
Therefore solving y= [(m/4 x 15) + (3m/4 x y)]/m =0, we get y= 5cm.
Therefore the correct answer is 5cm.
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