Speed and direction of clay balls after inelastic collision

[Nulligan]

Here's my problem guys, I've been trying to puzzle this out for a while. Would really appreciate any help at all, thanks!

Homework Statement

A 50.0 g ball of clay traveling east at 5.50 m/s collides and sticks together with a 40.0 g ball of clay traveling north at 6.00 m/s.

What is the speed of the resulting ball of clay?

2. Homework Equations and 3. The Attempt at a Solution

My work so far:
Let Ball traveling East be 1
Let ball traveling North be 2

Known
M1 = 0.05 kg
M2 = 0.04
M1 + M2 = 0.09

(Vix)1 = 5.5 m/s; (Viy)1 = 0 m/s
(Vix)2 = 0; (Viy)2 = 6 m/s

(Vfx) = 5.5 + 0 = 5.5 m/s
(Vfy) = 0 + 6 = 6 m/s


So the X components are equal to Vf* (as in, the vector V)Cos(Theta)

And the Y components are equal to Vf*Sin(Theta)

=> (M1 + M2)Vfx = (M1 + M2)VfCos(Theta) = M1(Vix)1 + M2(Vix)2 {which equals zero} = M1(Vix)1

(M1 + M2)Vfy = (M1 + M2)VfSin(Theta) = M1(Viy)1{which equals zero} + M2(Viy)2 = M2(Viy)2

 

[Nulligan]

Nevermind, I worked it out in the end. For anyone on the same problem, just get the momenta of the two balls, and then divide the momenta by the final mass (0.09kg) to get the velocities in the X and Y directions. Then simply use the baldy Greek's theorem (Pythagoras).


x momentum = (0.05kg)(5.5 m/s) = 0.275 J
y momentum = (0.04kg)(6 m/s) = 0.24 J
(0.24 J) / (0.09 kg) = FInal y velocity = 2.667 m/s
(0.275 J) / (0.09 kg) = Final x velocity = 3.056 m/s
Final Speed = sqrt(Vfx^2 + Vfy^2) = 4.056 m/s
Theta = arctan(2.667 / 3.056) = 41.1 deg