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$$\alpha$$cos$$\beta$$

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 )