Elastic Billiard Ball Collision

In summary: So if you solve for the velocities in the CM system, you'll solve for the velocities in the lab system.
  • #1
iandelaney
4
0

Homework Statement


A cue ball with speed U hits a stationary red ball of equal mass. The collision is elastic (ie no energy is converted into other forms). After the collision the cue ball is moving at an angle θ to its original path. Find the final speed of the cue ball.


Homework Equations



conservation of momentum: mu=mv1mv2

conservation of energy: 1/2mu2 = 1/2mv12 + 1/2mv22

The Attempt at a Solution



As I understand it this should be a fairly standard problem but I have been struggling to find any help on the tinterweb hence my posting :-)

I know that from the conservation of momentum that the velocities of the two balls after the collision will be equal to the velocity of the cue ball initially. I would then like to substitute u=v1+v2 into the conservation of energy to obtain (v1+v2)2 = v12+v22 which is only the case when v1 and v2 are zero. I was wondering if anyone could spot my mistake?

thanks in advance
Ian
 
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  • #2
The velocities are vectors, remember that. You can get rid of the velocity of the red ball using conservation momentum, plug that into the equation of conservation of energy. You get the angle dependence from the dot product between the two velocities.
 
  • #3
When I do that I obviously find that v2=u-v1 and when I sub this into the conservation of energy equation it still comes out as either v1=u or v1=0, and I am assuming that this can't be the case as it states in the question that the ball continues to move after the collision at an angle θ :s
 
Last edited:
  • #4
iandelaney said:

Homework Statement


A cue ball with speed U hits a stationary red ball of equal mass. The collision is elastic (ie no energy is converted into other forms). After the collision the cue ball is moving at an angle θ to its original path. Find the final speed of the cue ball.


Homework Equations



conservation of momentum: mu=mv1mv2

conservation of energy: 1/2mu2 = 1/2mv12 + 1/2mv22

The Attempt at a Solution



As I understand it this should be a fairly standard problem but I have been struggling to find any help on the tinterweb hence my posting :-)

I know that from the conservation of momentum that the velocities of the two balls after the collision will be equal to the velocity of the cue ball initially. I would then like to substitute u=v1+v2 into the conservation of energy to obtain (v1+v2)2 = v12+v22 which is only the case when v1 and v2 are zero. I was wondering if anyone could spot my mistake?

thanks in advance
Ian

Such problems are almost always easier to solve if you first go into the CM (center-of-mass) system, then transform back to the lab system to finally finish the calculation. In the CM system, the total momentum = 0 before (and after) the collision, so if the masses of the balls are equal, their speeds are equal in the CM system before and after the collision. If energy is conserved in the lab system it is also conserved in the CM system, and vice-versa.
 

1. What is an elastic billiard ball collision?

An elastic billiard ball collision is a type of collision that occurs between two billiard balls where the total kinetic energy and momentum of the system are conserved. This means that the balls bounce off each other without losing any energy or changing direction.

2. What factors affect the outcome of an elastic billiard ball collision?

The factors that affect the outcome of an elastic billiard ball collision include the mass and velocity of the balls, as well as the angle and point of impact. The elasticity or "bounciness" of the balls also plays a role in the collision.

3. How is the velocity of the balls calculated after an elastic billiard ball collision?

The velocity of the balls after an elastic billiard ball collision can be calculated using the laws of conservation of momentum and energy. These equations take into account the masses and velocities of the balls before and after the collision, as well as the elasticity of the collision.

4. Can an elastic billiard ball collision be perfectly elastic?

Yes, in theory, an elastic billiard ball collision can be perfectly elastic. This means that the total kinetic energy and momentum of the system are conserved, and the balls bounce off each other without any loss of energy. In real-world scenarios, some energy is always lost due to factors such as friction and deformation of the balls.

5. How is an elastic billiard ball collision different from an inelastic collision?

An elastic billiard ball collision is a type of collision where the total kinetic energy and momentum of the system are conserved, while an inelastic collision is a type of collision where some kinetic energy is lost and the objects stick together after the collision. Inelastic collisions are typically seen in real-world scenarios, while elastic collisions are idealized scenarios used for calculations and simulations.

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