Finding angle of billiard ball after collision

[luffy3san]

Homework Statement

A billiard ball of mass 0.162 kg has a speed of 1.86 m/s and collides with the side of the billiard table at an angle of 51.8°. For this collision, the coefficient of restitution is 0.841. What is the angle relative to the side (in degrees) at which the ball moves away from the collision?

m = 0.162 kg
v_i = 1.86 m/s
θ = 51.8°
C_R = 0.841

Homework Equations

C_R = v/u
mv_xi + mv_yi = mv'_xf + mv'_yf

The Attempt at a Solution

C_R = v/u
0.841 = v / 1.86 m/s
v = 1.56 m/s

mv_xi = mv_xf
mvcosθ = mvcosθ'
1.86 * cos 51.8° = v_xf
v_xf = 1.15 m/s

cosθ' = v_xf/v
cosθ' = 1.15/1.56
θ' = cos^-1(1.15/1.56) = 42.5°

 

[rude man]

The Attempt at a Solution

C_R = v/u
0.841 = v / 1.86 m/s
v = 1.56 m/s

mv_xi = mv_xf
mvcosθ = mvcosα

I have a problem with these two. E.g. the second says θ = α and I don't believe the first one either!

Is momentum really conserved in the billiard ball? What about momentum transferred to the Earth via the pool table etc.?

M.E.'s out there, help?

1.86 * cos 51.8° = v_xf
v_xf = 1.15 m/s

cosα = v_xf/v
cosα = 1.15/1.56
α = cos^-1(1.15/1.56) = 42.5°[/QUOTE]

 

[luffy3san]

I have a problem with these two. E.g. the second says θ = α and I don't believe the first one either!

Is momentum really conserved in the billiard ball? What about momentum transferred to the Earth via the pool table etc.?

M.E.'s out there, help?

1.86 * cos 51.8° = v_xf
v_xf = 1.15 m/s

cosα = v_xf/v
cosα = 1.15/1.56
α = cos^-1(1.15/1.56) = 42.5°

Hmm not sure about the momentum transfer to the Earth etc. I checked a couple sites where they said momentum is conserved in the x direction in this situation.

 

[rude man]

Hmm not sure about the momentum transfer to the Earth etc. I checked a couple sites where they said momentum is conserved in the x direction in this situation.

Yeah, that makes sense. Momentum transfer to Earth should be only in the direction perpendicular to the table's side. Good point. Did you pick x as going along the side of the collision?

 

[luffy3san]

Yeah, that makes sense. Momentum transfer to Earth should be only in the direction perpendicular to the table's side. Good point. Did you pick x as going along the side of the collision?

Yes, I think so. I chose the x-axis to be the side of the table. On a x-y coordinate, the ball travels from the 3rd quadrant, hits at the origin, and deflects towards the 4th quadrant. And, I chose the angles to be along the x axis. Don't know if that's the right way.

 

[Andrew Mason]

I think you are supposed to assume that the speed is reduced only in the direction perpendicular to the cushion.

AM

 

[luffy3san]

I think you are supposed to assume that the speed is reduced only in the direction perpendicular to the cushion.

AM

Yeah, that's what I was assuming but i found out my mistake. I noticed that C_R should've been C_R = $\vec{v}$sinθ' / $\vec{u}$sinθ

Thanks rudeman and Andrew for the suggestions!