How Does the Velocity of Two Colliding Balls Change After Impact?

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The discussion focuses on calculating the velocities of two colliding balls during a double-ball bounce scenario. Initially, both balls are dropped from a height of 2 meters, and the velocity just before impact is determined to be approximately 2.8 m/s. After the lower ball hits the ground, its velocity is calculated to be around 6.325 m/s, assuming a perfectly elastic collision. The interaction between the two balls is considered, with the upper ball potentially gaining significant velocity due to energy transfer during the collision. The complexities of real-world physics are acknowledged, particularly regarding the timing of the bounces.
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Homework Statement



4. The double-ball bounce
Consider 2 balls on top of another that are being dropped from a height h = 2 m,
see the Figure below. They will hit the ground and bounce back and we want to
compute the velocity of the upper ball after the bounce.
(a) (b)
We can model this as a sequence of two collisions. Let us assume that the upper ball
has a mass of 100 g, and that the lower ball has a mass of 10 kg. We rst consider
the collision between the lower ball and the ground.
(a) What is the velocity of the balls just before the lower ball hits the ground?

(b) Assume that the collision between the ball and the ground is perfectly elastic,
what is the velocity of the lower ball just after it hits the ground?
Now let us consider the collision of the upper and the lower ball.
(c) Assume that the collision is elastic with a coecient of restitution e = 0:9.
Compute the velocity of the upper ball after the collision.
(d) What is the maximum height the upper ball reaches after the collision?



The Attempt at a Solution



For this question for part a where two balls are falling and you work out the velocity, because its on top to i count it as one system? or do i calculate the velocities individually?
Just need to clarify this so i can do it correctly :)
 
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Just realized they will both have same velocity.. lol
and velocity will be about 2.8 m/s i believe
 
Ok so far i have velocity of balls before hitting the ground using d = 1/2at^2 + Vit
this works out as 2= 5t^2
therefor t = 0.28 seconds
And then using V = a.t i get V at 2.8 seconds (g = 10m/s^2) for this example

Then for velocity after collision i find the Ep (potential energy) = m.g.h = 200
And then use that for Ek = 1/2mv^2
which means 200 = 1/2 10. V^2
so V^2 = 200/5
V = Sqaure root of 40
= 6.325 m/s
Is that correct so far?
 
This problem is more complicated in the real world. One issue is if the upper ball bounces off the lower ball before, at the same time, or after the lower ball bounces off the ground. You're probably supposed to assume that the upper ball bounces off the lower ball at the same time the lower ball bounces off the ground. The upper ball is going to end up with much more velocity than the lower ball, and may end up bouncing higher than the initial height it was dropped from, since some of the energy is transferred from the lower ball to the upper ball during a bounce.
 
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