Elastic collision between pool balls

[John004]

Homework Statement

A super cue ball is made of the same material with the target ball (radius r) but slightly larger: r_c = (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?

Homework Equations

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

 

[ehild]

Homework Statement

A super cue ball is made of the same material with the target ball (radius r) but slightly larger: r_c = (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?

Homework Equations

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.

 

[John004]

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.

 

[ehild]

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.

 

[John004]

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<u_1x,u_1y> = m₁<v_1x, v_1y> + m₂<v_2x, v_2y>

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 = u_1x 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?

 

[ehild]

I end up with

(√3)/6 v₂ + v₁cos(φ) = v₀ where v₀ = u_1x these are the magnitudes of the velocities

v₂ = -(6/√3)v₁sin(φ)

Check the equations, there are mistakes in both.

and then the energy conservation expression

3v₀² = 3v₁² + v₂²

does this look ok so far?

The energy equation is OK.

 

[John004]

Check the equations, there are mistakes in both.

The energy equation is OK.

m₁<u_1x, 0> = m₁<v_1x, v_1y> + m₂<v_2x, v_2y>

<m₁u_1x, 0> = <m₁v_1x, m₁v_1y> + <m₂v_2x, m₂v_2y>

<m₁u_1x, 0> = < m₁v_1x + m₂v_2x, m₁v_1y + m₂v_2y>

therefore,

m₁v_1x + m₂v_2x = m₁v₀

m₁v_1y = -m₂v_2y

3v₁cos(φ) + ½v₂ = 3v₀

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

 

[ehild]

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.