IIT JEE , AIEEE Forum - Physics , Mechanics

krishna.gopal

Hi friends. First of all i am sorry for such a delayed reply. I and the reason for this delay was the fact that i am yet not very sure of my solution as i am unable to generalize the problem and hence get an absolute maxima of velocity. Still i am trying to answer it using some general understanding of physics.

First thing that as the system already has a momemtum along orignal direction of motion the particle which has maximum velocity afer the burst must continue to move in that direction only (Largest increase in velocity may com if the impulse on this particle is along initial direction of motion)

In general after the bursts the other two pieces may have momentums along the orignal line of motion as well as perpendiculr to it. But to maximize velocity of third partice the momentum perpendicular to motion will be zero.(Total KE is constant. If there is some momentum in perpendicular to motion, it will take some KE and hence velocity of third particle is reduced.

Third point that i want to make that the remaining two particles should move with equall velocity (that will give them minimnum KE for same momentum)

On the basis of these three logics for maximum velocity one particle should have velocity V after collision and other two should have v

Let Orignal mass of particle be 3m
Momentum balance
3m*500 = m(V+2v)
V+2V=1500
1.5*(1/2)*3m*500^2=(1/2)*m*V^2+m*v^2

Solving these two we get V=1000m/s
and v= 250 m/s

So according to me maximum velocity can be 1000m.s

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