2D momentum of two objects next to a wall

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SUMMARY

The discussion focuses on calculating the post-collision velocity of a stationary puck (Puck 2) after an inelastic collision with a moving puck (Puck 1) in a two-dimensional scenario. Puck 1, with a mass of 0.1 kg and an initial velocity of 15 m/s in the y-direction, reverses direction post-collision with a velocity of 10 m/s. The mass of Puck 2 is 0.5 kg, and the collision dynamics are analyzed using the momentum conservation principle, specifically the equation P=mv, and the Pythagorean theorem to determine the angle of deflection.

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Homework Statement


We have two pucks next to a wall. Puck 1 has velocity v0=15 m/s in the direction y (as seen on the picture). Puck 2 is stationary. After the collision puck 1 moves in the opposite direction with velocity v1=10 m/s and puck 2 moves with unknown velocity in the direction x. The radius of puck 1 is r1=3cm and of puck 2 r2=6cm. The mass of puck 1 is m1=0.1kg and of puck 2 m2=0.5kg. The collision is not elastic.

What is the velocity of puck 2 after the collision?

http://img134.imageshack.us/img134/8082/breznaslova.jpg

Homework Equations


Equation of momentum P=mv
Pythagorean theorem c2=a2+b2

The Attempt at a Solution


First I calculated the angle using the radiuses of the pucks and I got Cos([tex]\alpha[/tex])=[tex]\frac{r<sub>2</sub>-r<sub>1</sub>}{r<sub>2</sub>+r<sub>1</sub>}[/tex]
I got that [tex]\alpha[/tex] = 70.5 degrees.

Then I went on to figure out momentums. I figured out this:
http://img256.imageshack.us/img256/8120/breznaslova2.jpg
The angle between P0 and P2 is [tex]\alpha[/tex].
Then I went on to make the P1 the hypotenus and the opposite is P1y which is the same as P=m1v1. From that I calculated the angle and then the adjacent P1x. Am I right to think that the velocity I am looking for is vx=[tex]\frac{P<sub>1x</sub>}{m<sub>2</sub>}[/tex].

Thanks for any assistance.
 
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