Elastic Collision

  • Thread starter KoaDcT
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Homework Statement


The problem is pretty simple, however I don't understand which value to use after using quad formula to solve. See below.

V1(initial) = 2.5 m/s
V2(Initial) = -5.9 m/s

A pool ball moving with a speed of 2.5 m/s makes an elastic head-on collision with an identical ball traveling in the opposite direction with a speed of 5.9 m/s. Find the velocities of the balls after the collision.

Homework Equations


ΣP(initial) = ΣP(final)
M1V1 + M2V2 = M1V1' + M2V2'
M1 = M2
2.5 + (-5.9) = V1' + V2'

V2' = -V1' - 3.4

Elastic collision: KE(initial) = KE(final)
(Mass cancels out)
V12+V22 = V1'2+V2'2


The Attempt at a Solution


From above, I got:
0 = 2V1'2 - 6.8V1' - 29.5
by substituting "V2' = -V1' - 3.4" for V2'

After applying quadratic equation I got:
V1' = -2.5 m/s, 5.9 m/s

My problem is I do not know which one is the correct answer. My answer key says V1' is equal to -2.5 m/s, but I have no idea why.
Any help is appreciated!
 

Answers and Replies

  • #2
ehild
Homework Helper
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Homework Statement


The problem is pretty simple, however I don't understand which value to use after using quad formula to solve. See below.

V1(initial) = 2.5 m/s
V2(Initial) = -5.9 m/s

A pool ball moving with a speed of 2.5 m/s makes an elastic head-on collision with an identical ball traveling in the opposite direction with a speed of 5.9 m/s. Find the velocities of the balls after the collision.

Homework Equations


ΣP(initial) = ΣP(final)
M1V1 + M2V2 = M1V1' + M2V2'
M1 = M2
2.5 + (-5.9) = V1' + V2'

V2' = -V1' - 3.4

Elastic collision: KE(initial) = KE(final)
(Mass cancels out)
V12+V22 = V1'2+V2'2


The Attempt at a Solution


From above, I got:
0 = 2V1'2 - 6.8V1' - 29.5
by substituting "V2' = -V1' - 3.4" for V2'

After applying quadratic equation I got:
V1' = -2.5 m/s, 5.9 m/s

My problem is I do not know which one is the correct answer. My answer key says V1' is equal to -2.5 m/s, but I have no idea why.
Any help is appreciated!
Check the signs of the solutions. The answer key can be wrong.
 
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  • #3
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I don't really understand what you said in the brackets, do you mean if the initial velocity is positive, then the answer must be negative after collision?
I also have another problem with this question. In my solution I substituted "V2' = -V1' - 3.4", so if V1' is equal to -2.5 m/s then V2' should be -0.9 m/s. However the answer is V1' = -2.5 m/s, V2' = +5.9 m/s, which makes no sense.
 
  • #4
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Ah, I found my mistake right after I posted. Now my final answer is V1' = -5.9 m/s, and V2' = 2.5 m/s. I think that is the right answer.
Thanks for the help :)
 
  • #5
ehild
Homework Helper
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Ah, I found my mistake right after I posted. Now my final answer is V1' = -5.9 m/s, and V2' = 2.5 m/s. I think that is the right answer.
Thanks for the help :)
The balls just exchanged velocity due to the collision. The other solution v1'=2.5 m/s, v2'=-5.9 m/s means that nothing has changed, both balls keep moving with the original velocity. That means they did not collide. But the equations for conservation of momentum and conservation of energy are valid also in the case if the balls just pass each other and do not collide.
 

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