What are the final velocities of two balls after an elastic head-on collision?

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Homework Help Overview

The discussion revolves around an elastic head-on collision between two balls of different masses and initial velocities. The original poster presents a scenario involving a 2.0 kg ball and a 3.0 kg ball, detailing their initial speeds and asking for the final velocities post-collision.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants examine the calculations for final velocities based on conservation of momentum and energy principles. There is a focus on verifying the correctness of the equations used and the signs of the velocities, with some questioning the assumptions made regarding directionality.

Discussion Status

The discussion is active, with participants providing feedback on calculations and encouraging checks against both momentum and energy conservation. There is an acknowledgment of errors in the original calculations, and some participants are exploring the implications of these corrections.

Contextual Notes

Participants note the importance of correctly assigning signs to velocities based on direction, as well as the need to satisfy both conservation laws in elastic collisions. There is an indication that the original poster's results suggested an increase in energy, which raised concerns about the validity of the calculations.

Arooj
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Homework Statement


A 2.0 kg ball is traveling east at 8.0 m/s when it has a perfectly elastic head-on collision with a 3.0 kg ball traveling west at 10.0 m/s. What are the final velocities of the two balls?


Homework Equations



http://www.nvcc.edu/home/tstantcheva/231files/G09_hw.pdf
 
I used the derived equation from page 4.

The Attempt at a Solution


For the 2.0 kg ball I got -13.6 m/s.
For the 3.0 kg ball I got -15.6 m/s.
 
Last edited by a moderator:
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Please show your work. The new speeds for both balls are faster than either initial speed. Elastic collisions conserve energy. Your speeds indicate energy was somehow created.
 
vf1 = ((m1)(v1) + m2((2*v2) - v1)) / m1 + m2
vf1 = (16 + 3(-20 - 8))/5
vf1 = -13.6 m/s

vf2 = v1 + vf1 - v2
vf2 = 8 + -13.6 - 10
vf2= -15.6 m/s

I'm assuming my problem is from setting the values of the speeds in the opposite direction to negative, but I thought this must be done?
 
You have a sign error for the second velocity. What is -v2?
 
Ah I see what I did wrong, v2 = 4.4 , and v1 = -13.6, substituting them into the conservation of momentum equation yields 14 = 14.
 
You are right. Since you're checking momentum, you should also check energy as well. There are two conservation theorems to satisfy.
 

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