Was the collision elastic or not?

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Homework Help Overview

The discussion revolves around a physics problem involving a head-on collision between two identical billiard balls, focusing on whether the collision is elastic or inelastic. The original poster presents calculations related to momentum and kinetic energy before and after the collision.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the conservation of momentum and kinetic energy, questioning the assumptions made about the collision type. The original poster calculates velocities and kinetic energies, while others challenge the textbook's answer and discuss the implications of their calculations.

Discussion Status

The discussion is active, with participants expressing differing views on the nature of the collision. Some suggest that the collision is elastic based on their interpretations of the calculations, while others question the assumptions regarding energy conservation and the setup of the problem.

Contextual Notes

Participants note the head-on nature of the collision, which simplifies the analysis by eliminating angular considerations. There is also mention of potential confusion regarding the textbook's answer and the calculations presented by the original poster.

jaron
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Homework Statement


a billiard ball of mass of 0.115kg moves with a velocity of 12.5m/s toward a stationary billiard ball of identical mass and strikes it in a head on collision. the first billiard ball comes to a complete stop. determine whether the collision was elastic.


Homework Equations


Ek1 + Ek2 = Ek1' + Ek2'

p=mv


The Attempt at a Solution


first i calculate the velocity of the steel ball after being struck by the billiard ball...
mv + mv = mv' + mv'
0.155kg(12.5m/s) = (0.155kg + 0.155kg)v
1.94 = 0.31v
v = 1.94/0.31
v = 6.26m/s

then i calculate the total kinetic energy before the collision (only the billiard ball because the steel ball was idle)

Ek = 1/2mv^2
Ek = 1/2(0.155)(12.5)^2
Ek = 12.1J

then i calculate the total kinetic energy after the collision

Ek + Ek = Ek' + Ek'
12.1 + 1/2(0.155)(0)^2 = 1/2(0.155)(0)^2 + 1/2(0.155)v^2
12.1 = .078v^2
v^2 = 12.1/0.078
v^2 = 155
v = 12.46m/s (but this doesn't make sense...)

so maybe this:
Ek = 1/2(0.155 + 0.155)(6.26)^2
Ek = 0.155(39.20)
Ek = 6.1J

- the answer in the back of my textbook is this:
v=6.32m/s[41.5' counterclockwise from the original direction of the first ball]; the collision is not elastic: Ek = 12.1J; Ek' = 10.2J

i got the first 2 correct but the third one i don't understand. as well.. i am not sure how to know whether something is considered an elastic collision or not)
 
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Excuse me,
But I have to disagree with the answer in your textbook.
assume system is isolated , no external forces, thus momentum conserved,
m*V+m*U=m*V'+m*U'
U=0 ,V'=0
we get that U'=V, 1 ball stops and the other one gets the speed of the other one.
thus Ek(initial)= mV^2/2 , Ek(final)=m*U'^2/2
because U'=V
Ek initial =Ek final...
Am I understanding Your question wrong??Do they scatter with an angle?? but it's head to head collision?
 
ok so are you agreeing with my answers?
and yes it was a head on collision, so there are no angles to worry about.
 
jaron said:
ok so are you agreeing with my answers?
and yes it was a head on collision, so there are no angles to worry about.

Well given what Dweirdo says, it does look like the final velocity of the other ball is 12.5m/s, which would imply that the collision is elastic.

0.155kg(12.5m/s) = (0.155kg + 0.155kg)v


you also used this to find v. This implies that you assumed the collision was completely inelastic.
 

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