Angle and speed of can after collision

[gap0063]

Homework Statement

An m2 = 1.2 kg can of soup is thrown upward with a velocity of v2 = 4.6 m/s. It is immediately struck from the side by an m1 = 0.63 kg rock traveling at v1 = 7.9 m/s.
The rock ricochets off at an angle of α = 65◦ with a velocity of v3 = 5.5 m/s.

(a)What is the angle of the can’s motion after the collision?
Answer in units of ◦.

(b)With what speed does the can move immediately after the collision?
Answer in units of m/s.

Homework Equations

P_x=m₁v₃cos\alpha+m₂v₄cos\beta
P_y=m₁v₃sin\alpha+m₂v₄sin\beta

where P_x= m₁v₁
and P_y=0

The Attempt at a Solution

so then I solved for v₄= -m₁v₃sin\alpha/m₂sin\beta

then I plugged v₄ back into the first equation and I think this is where I messed up, probably in the algebra because then:

m₁v₁= m₁v₃cos\alpha+m₂(m₁v₃sin\alpha/m₂sin\beta)cos\beta

m₁v₁= m₁v₃cos\alpha-(m₁v₃sin\alphacos\beta

then I plugged numbers in:

4.977= 3.465 cos 65\circ-3.14036 cot \beta
3.14036cot\beta=-3.51263
\beta=137.203\circ

as for part (b) I don't know how to solve without part (a)

 

[gap0063]

Please, anyone?

Am I even on the right track?

 

[tiny-tim]

hi gap0063!

An m2 = 1.2 kg can of soup is thrown upward with a velocity of v2 = 4.6 m/s. P_x=m₁v₃cos\alpha+m₂v₄cos\beta
P_y=m₁v₃sin\alpha+m₂v₄sin\beta

where P_x= m₁v₁
and P_y=0

no, your P_y should be m₂v₂, shouldn't it?

(and that's messed up your calculation of v₄ at the start )