1-D Perfectly Elastic Collison PLEASE CHECK THANKS

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Homework Help Overview

The discussion revolves around a one-dimensional perfectly elastic collision problem involving two masses, where one mass is initially in motion and the other is stationary. The original poster presents their calculations regarding the velocities and momentum before and after the collision.

Discussion Character

  • Exploratory, Assumption checking, Mixed

Approaches and Questions Raised

  • The original poster attempts to verify their calculations related to momentum and velocity after the collision. Some participants question the correctness of the original poster's results and suggest checking both momentum and kinetic energy conservation. Others express uncertainty and seek further confirmation from additional participants.

Discussion Status

The discussion is ongoing, with some participants providing affirmations of the original poster's calculations while others express a desire for more validation. There is an exploration of different methods to approach the problem, indicating a lack of consensus on the correctness of the solution.

Contextual Notes

Participants note the possibility of multiple methods to solve the problem, which may lead to different interpretations of the results. The original poster's calculations are based on specific assumptions regarding the masses and initial conditions of the collision.

proxy98
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1-D Perfectly Elastic Collison PLEASE CHECK! THANKS

Homework Statement


Mass m1 is moving to the right at a velocity of 17.6m/s. Suddenly it strikes a stationary ball. The stationary ball has a mass of .685kg. The collision is perfectly elastic and 1 dimensional. The collision forces m2 to move to the right at 11.1m/s (v2prime).

Total momentum = 5.5 kg m/s ??
m1 = .3155kg ??
v1prime= -6.5m/s ??

Homework Equations



momentum before = momentum after
m1v1 = m1v1prime + m2v2prime

v2prime = (2m1)/(m1+m2) (v1) + (m2-m1)/(m1+m2) (v2)

v1prime = (m1-m2)/(m1+m2) (v1) + (2m2)/(m1+m2) (v2)

The Attempt at a Solution

11.1 = (2*m1)/(m1+.685) ( 17.6) + 0 (cancels cause m2 is stationary
11.1(m1+.685) = 35.2m
11.1m1 + 7.6035 = 35.2m
24.1m = 7.6035
m1= .3155 kg

v1prime = (.316-.685)/(.316+.685)(17.6) + 0 (v2 = 0)
v1prime = -6.50 m/s Left.

momentum before = momentum after
(17.6)(.316) = (.316)(-6.5) + (.685)(11.1)
5.55 kgm/s


Thanks for taking your time to help ! I just want to make sure this is right.
 
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yeah, it's right. to check just see if the momentum is the same before & after and check if the kinetic energy is the same before & after. they are, so you did it right.
 


sorry to ask this, but are you 100% sure? Maybe I can get some more people to assure this answer is correct, as there are multiple other ways of doing the same problem.
 


proxy98 said:
sorry to ask this, but are you 100% sure? Maybe I can get some more people to assure this answer is correct, as there are multiple other ways of doing the same problem.

Yes. there is only one solution to the problem, so if you plug it into find the correct results, it must be right.
 

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