Conservation of Momentum involving Vf, elastic collisions

In summary: This would cause the second ball to move in a circle with a radius equal to the distance the first ball had traveled. This contradicts the textbook's statement that the second ball moves in a straight line.But there is anyway another problem here. It does not say exactly how the first ball was moving. If it had traveled any distance it would surely have been rolling, so its total KE would have been 17J. If the collision lost no mechanical energy, the second ball must have set off with the same rotation rate, but since that would have come about by static friction the second ball's rotation would be backspin. This would cause the second ball to move in a circle with a radius equal to the distance the first
  • #1
ericcy
19
1
Homework Statement
A billiard ball of mass 0.155kg 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.
Relevant Equations
Pt=Pt`
m1v1+m2v2=m1v1`+m2v2`
I tried solving it using this method and I got 12.5m/s, and assumed the collision was elastic.

The answer is actually 6.32m/s [41.5 degrees counterclockwise from the original direction of the first ball]; the collision is not elastic: Ek = 12.1J Ek`= 10.2J

I have absolutely no idea how the textbook could even get this answer. If you could explain the steps and why that would be greatly appreciated. Thanks!
 
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  • #2
Please show the details of your calculation. If the collision is head-on, there is no scattering angle involved. Where did the 41.5 deg. counterclockwise come from?
 
  • #3
That was the textbooks answer, I have no idea how they got that. The collision is head-on. I used m1v1=m2v2` and subbed in all the values except for v2` and got 12.5m/s for m2. This means that when I solved for the total energies before and after the collision to check if it was elastic, they were both the same (12.1=12.1). Would you say my answer is right?
 
  • #4
ericcy said:
That was the textbooks answer, I have no idea how they got that. The collision is head-on. I used m1v1=m2v2` and subbed in all the values except for v2` and got 12.5m/s for m2. This means that when I solved for the total energies before and after the collision to check if it was elastic, they were both the same (12.1=12.1). Would you say my answer is right?
Yes. When balls of identical mass collide and the collision is elastic, they simply exchange velocities. This is the case here. I have no idea where the textbook got that answer either. Are there more parts to this problem? Maybe the answer belongs to a different part.
 
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  • #5
kuruman said:
Yes. When balls of identical mass collide and the collision is elastic, they simply exchange velocities. This is the case here. I have no idea where the textbook got that answer either. Are there more parts to this problem? Maybe the answer belongs to a different part.
Nope. Just that question. That's what made me so confused because I had literally no idea whatsoever how they could have gotten that answer. Must have been a mix up. Thanks for the confirmation on the answer :)
 
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  • #6
You were rightfully confused. The answer appears to belong to a two-dimensional collision problem. :oldsmile:
 
  • #7
Curiously, the 12.1J does match the initial linear KE in the question. I think a lot of these textbook errors come about because someone adapted an existing question and failed to update the answer.

But there is anyway another problem here. It does not say exactly how the first ball was moving. If it had traveled any distance it would surely have been rolling, so its total KE would have been 17J. If the collision lost no mechanical energy, the second ball must have set off with the same rotation rate, but since that would have come about by static friction the second ball's rotation would be backspin.
 

What is the conservation of momentum?

The conservation of momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant, regardless of any external forces acting on the system. This means that in a closed system, the total amount of momentum before and after an event remains the same.

How is momentum calculated?

Momentum is calculated by multiplying the mass and velocity of an object. The formula for momentum is: p = mv, where p is momentum, m is mass, and v is velocity.

What is an elastic collision?

An elastic collision is a type of collision between two objects where both objects bounce off each other and no kinetic energy is lost during the collision. This means that the total momentum of the system before and after the collision remains the same.

What is the equation for conservation of momentum in an elastic collision?

In an elastic collision, the equation for conservation of momentum is: m1v1i + m2v2i = m1v1f + m2v2f, where m1 and m2 are the masses of the objects, v1i and v2i are the initial velocities of the objects, and v1f and v2f are the final velocities of the objects.

How can conservation of momentum be applied in real-life situations?

The principle of conservation of momentum can be applied in many real-life situations, such as in car accidents, sports, and rocket propulsion. It is also used in the design of safety features in vehicles to reduce the impact of collisions. In sports, it is used to explain the motion of objects, such as a basketball being bounced off the ground. In rocket propulsion, it is used to explain how a rocket propels forward by expelling exhaust gases in the opposite direction.

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