Momentum: Glancing collisions in two dimensions

[anti404]

Homework Statement

a ball of mass 2kg is traveling North with a Velocity 6m/s when it collides with an identical, stationary ball. after the collision, one of the ball moves with a velocity of 2m/s at an angle of 30deg east of north. find Velocity of the other ball.
m1=2kg
m2=m1
v1=6m/s
v1'=2m/s
alpha=30deg
v2'=?
theta=?

Homework Equations

P(initial)=P(final) <--conservation of momentum
P=mv

The Attempt at a Solution

split the (conserved) momentum into x and y components.
Px: 0+0=m1v1'sin(alpha)-m1v2'sin(theta)
Py: m1v1=m1v1'cos(alpha)+m1v2'cos(theta)

after that, I'm lost. I honestly have no idea where to go from there. I think maybe you're supposed to solve for... theta? and then plug that in. but again, I don't know how to do so. we just barely[and when I mean barely, I mean for about 1 minute at the end of class] covered this today, and our book details only inelastic collisions in two dimensions, which is of little help here, as it is not stated to be elastic or inelastic. help would be very, VERY much appreciated.

 

[rl.bhat]

Px: 0+0=m1v1'sin(alpha)-m1v2'sin(theta)
Py: m1v1=m1v1'cos(alpha)+m1v2'cos(theta)
Rewrite these equations as
m1v1'sin(alpha) = m1v2'sin(theta)...(1)
m1v1- m1v1'cos(alpha) = m1v2'cos(theta)...(2)
Cancel m1 form both sides.Substitute the values of v1' and α.
Find v2' from eq.(1) and substitute in eq.(2) and solve for θ.

 

[Lachlan1]

remember that, like in a snooker game, these balls paths after the collision will be 90 degrees apart.

 

[anti404]

Px: 0+0=m1v1'sin(alpha)-m1v2'sin(theta)
Py: m1v1=m1v1'cos(alpha)+m1v2'cos(theta)
Rewrite these equations as
m1v1'sin(alpha) = m1v2'sin(theta)...(1)
m1v1- m1v1'cos(alpha) = m1v2'cos(theta)...(2)
Cancel m1 form both sides.Substitute the values of v1' and α.
Find v2' from eq.(1) and substitute in eq.(2) and solve for θ.

this is what I kept trying to do, but I don't know how to solve for v2' since you're missing angle theta. because what I'd get was m1v1'sin(alpha)/sin(theta)=v2'. and I don't know how to solve from that point on.

Lachlan1: would that mean that you'd just add 90deg to alpha and you'd get theta...?

 

[rl.bhat]

v1'sin(alpha) = v2'sin(theta)...(1)
v1- v1'cos(alpha) = v2'cos(theta). (2)
From eq.(1) and (2) you will get
v1'sin(alpha)/[v1- v1'cos(alpha)/ = v2'sin(theta)/v2'cos(theta)...
Substitute the values of v1. v1' and α. Simplify and find tanθ. From that find θ.

 

[anti404]

v1'sin(alpha) = v2'sin(theta)...(1)
v1- v1'cos(alpha) = v2'cos(theta). (2)
From eq.(1) and (2) you will get
v1'sin(alpha)/[v1- v1'cos(alpha)/ = v2'sin(theta)/v2'cos(theta)...
Substitute the values of v1. v1' and α. Simplify and find tanθ. From that find θ.

ah, okay. for whatever reason I kept getting cos/sin, which confused me. thanks!