Help with Perfectly Elastic Collision problem.

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SUMMARY

The discussion centers on solving the perfectly elastic collision problem involving a 20g ball and a 100g ball. The initial speed of the 20g ball, denoted as vi, is calculated using the conservation of momentum and energy principles. The participant initially derived an incorrect speed of 5.9 m/s, while the correct answer is 7.9 m/s, as confirmed by the textbook. The key to solving the problem lies in correctly applying the conservation of momentum and energy equations, particularly recognizing that the kinetic energy of the second ball converts to potential energy at its maximum height.

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ChodeNode
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Homework Statement


A 20g ball is fired horizontally with initial speed vi toward a 100g ball that is hanging motionless from a 1.0m long string. The balls undergo a head-on, perfectly elastic collision after which the 100g ball swings out to a maximum angle of 50degrees. What was vi?


Homework Equations


Conservation of momentum:
m1vi1 + m2vi2 = m1vf1 + m2vf2

Conservation of energy:
1/2m1(vi1^2) = m2gy
or
1/2m1(vi1^2) = m2gy + 1/2m1(vf1^2)


The Attempt at a Solution


My prof says all the kinetic energy converts to potential gravitational energy, which I initially agreed with. My prof is consistently giving answers different from the book though. When I do it his way, I end up with an answer (5.9m/s) different from the book (7.9m/s). His way completely ignores the need for the conservation of momentum equation and, when I thought about it, the balls aren't equal mass so the shot ball should bounce back a little bit in addition to the larger ball swinging on the string.

So I figured my conservation of energy equation should be the second one. When I do it this way though, I end up with all three velocities unknown from the conservation of momentum equation. With only two equations, I can't solve for the velocities.

Can somebody point out to me where I've gone wrong?

Thanks for any help.
 
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Hi CodeNode, welcome to PF.
Calculating m2gY, you can find vf2.
Substitute this value in the conservation of momentum equation. Using this equation and the conservation of energy equation solve for v1i and v1f.
 
rl.bhat said:
Hi CodeNode, welcome to PF.
Calculating m2gY, you can find vf2.
Substitute this value in the conservation of momentum equation. Using this equation and the conservation of energy equation solve for v1i and v1f.
Okay, so you're saying that because I know that all the energy ends up as m2gy, then all the kinetic energy of ball two immediately after impact should be equal to that and therefore:

m2gy = 1/2m2(vf2^2)

Is that right?

EDIT: Thanks for the response, btw.

EDIT 2: Okay, that got me what I needed, thank you. Apparently my algebra was going wrong somewhere with my substitution. The book was kind enough to provide a simplification of the velocity variables based on the substitution and I got the answer the book got.
 
Last edited:

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