# Elastic collision between pool balls

• John004
In summary: It is OK. But do not forget, that you get negative angle φ. The cue ball moves under the x-axis after collision.
John004

## Homework Statement

A super cue ball is made of the same material with the target ball (radius r) but slightly larger: rc = (3)^(1/3)*r The cue ball collides with the target ball on a frictionless table, as shown below, with initial speed of v0. The collision is not head-on, as shown below.

a) What are the velocities of the two balls after the collision? You need to determine both the speeds and directions!

b) In the CM frame, what is the target ball’s direction, relative to the x-axis?

ρ = m/V
ρ = 4/3 π r3

## The Attempt at a Solution

[/B]
so for the masses,

m1 = ρV1 , m2 = ρV2

taking the ratio of those guys gives m1 = 3 m2

if i use conservation of linear momentum and kinetic energy conservation, i end up with two sets of equations but one is a vector equation and the other is not, so i don't know how to even start this problem

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John004 said:

## Homework Statement

A super cue ball is made of the same material with the target ball (radius r) but slightly larger: rc = (3)^(1/3)*r The cue ball collides with the target ball on a frictionless table, as shown below, with initial speed of v0. The collision is not head-on, as shown below.

a) What are the velocities of the two balls after the collision? You need to determine both the speeds and directions!

b) In the CM frame, what is the target ball’s direction, relative to the x-axis?

ρ = m/V
ρ = 4/3 π r3

## The Attempt at a Solution

[/B]
so for the masses,

m1 = ρV1 , m2 = ρV2

taking the ratio of those guys gives m1 = 3 m2

if i use conservation of linear momentum and kinetic energy conservation, i end up with two sets of equations but one is a vector equation and the other is not, so i don't know how to even start this problem
There should be a figure to the problem. Attach it please!
The vector equation can be written as two scalar equations for the components of the linear momentum. So you have three equations.

ehild said:
There should be a figure to the problem. Attach it please!
The vector equation can be written as two scalar equations for the components of the linear momentum. So you have three equations.
Oh right! its there now.

John004 said:
Oh right! its there now.
OK. So the velocity of the target ball encloses the angle pi/3 with the original velocity of the cue ball after the collision. The direction of velocity of the cue ball is unknown after collision. And energy is conserved. Write the equations for the components of the momenta, and also for the conservation of energy.

ehild said:
OK. So the velocity of the target ball encloses the angle pi/3 with the original velocity of the cue ball after the collision. The direction of velocity of the cue ball is unknown after collision. And energy is conserved. Write the equations for the components of the momenta, and also for the conservation of energy.
Ok so this is what I've done so far

m1<u1x,u1y> = m1<v1x, v1y> + m2<v2x, v2y>

where u1 is the velocity of mass 1 before the collision, v1 is the velocity of mass 1 afterwards, and similarly for mass 2.

I end up with

(√3)/6 v2 + v1cos(φ) = v0 where v0 = u1x these are the magnitudes of the velocities

v2 = -(6/√3)v1sin(φ)

and then the energy conservation expression

3v02 = 3v12 + v22

does this look ok so far?

John004 said:
I end up with

(√3)/6 v2 + v1cos(φ) = v0 where v0 = u1x these are the magnitudes of the velocities

v2 = -(6/√3)v1sin(φ)
Check the equations, there are mistakes in both.
John004 said:
and then the energy conservation expression

3v02 = 3v12 + v22

does this look ok so far?
The energy equation is OK.

ehild said:
Check the equations, there are mistakes in both.

The energy equation is OK.

m1<u1x, 0> = m1<v1x, v1y> + m2<v2x, v2y>

<m1u1x, 0> = <m1v1x, m1v1y> + <m2v2x, m2v2y>

<m1u1x, 0> = < m1v1x + m2v2x, m1v1y + m2v2y>

therefore,

m1v1x + m2v2x = m1v0

m1v1y = -m2v2y

3v1cos(φ) + ½v2 = 3v0

v2 = -(6/√3)v1sin(φ) still got the same thing for this one

John004 said:
3v1cos(φ) + ½v2 = 3v0

v2 = -(6/√3)v1sin(φ) still got the same thing for this one
It is OK. But do not forget, that you get negative angle φ. The cue ball moves under the x-axis after collision.
Isolate cos(φ) and sin(φ), square and add the equations, so as φ eliminates.

## 1. What is an elastic collision between pool balls?

An elastic collision between pool balls is a type of collision where the total kinetic energy of the system is conserved. This means that the total energy of the two colliding pool balls before and after the collision remains the same.

## 2. Why is an elastic collision between pool balls important?

Elastic collisions are important in physics because they follow the fundamental laws of conservation of energy and momentum. This means that they can be used to predict the behavior of objects in motion and understand the physical principles at work.

## 3. How is an elastic collision between pool balls different from an inelastic collision?

In an inelastic collision, some of the kinetic energy is lost and converted into other forms of energy, such as heat or sound. This means that the total energy of the system is not conserved. In an elastic collision, however, no energy is lost and the total energy remains the same.

## 4. What factors affect the outcome of an elastic collision between pool balls?

The outcome of an elastic collision between pool balls is affected by the masses and velocities of the colliding balls. The angle and direction of the collision, as well as the elasticity of the balls and the surface they are colliding on, also play a role in the outcome.

## 5. How does the elasticity of pool balls affect the outcome of a collision?

The elasticity of pool balls is a measure of how much they can deform and then return to their original shape after a collision. The more elastic the balls are, the less energy will be lost in the collision and the more closely the final velocities of the balls will match the initial velocities.

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