What is the method for calculating velocities in a perfectly elastic collision?

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The discussion focuses on calculating the velocities of two balls involved in a perfectly elastic collision. A 1kg ball, after falling from a height of 1.2m, strikes a stationary 1.5kg ball. The initial velocity of the 1kg ball is calculated to be 4.85 m/s using the equation mgh = 1/2mv^2. Applying the conservation of momentum and the coefficient of restitution, the final velocities are determined to be V1 = -0.97 m/s for the 1kg ball and V2 = 3.88 m/s for the 1.5kg ball.

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1. a 1kg ball traverse a frictionless tube. the ball after falling through a height of 1.2m, strikes a1.5kg ball which initially at rest. find the velocities of the two balls, if the collision is perfectly elastic. V1= -0.97m/s V2= 3.88m/s



2. i found out the initial velocity of 1kg ball
mgh = 1/2mv^2
1(9.81)1.2= 1/2(1)v^2
v= 4.85m/s

after that, i use mu+mu = mv+mv
1(4.85)+ 1.5(0)= 1v +1.5v2

after that, what should i do to get the answer?
help me...please...
 
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In a perfectly elastic collision KE is also conserved. This gives you a second equation in the two unknowns (velocities of the balls after collision).
 
what u mean?
but i use Newton's law of restitution to get the answer.
e=(v1-v2)/(u2-u1)

-4.85=v1-v2---------second equation
and i finally get the answer.
 

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