Posted by Help! on .
Two identical pucks are on an air table. Puck A has an initial velocity of 2.0m/s in the +xdirection. Puck B is at rest. Puck A collides elastically with Puck B and A moves off at 1.0m/s at an angle of 60 degrees above the xaxis. What is the speed and direction of puck B after the collision?

Physics 
drwls,
You have two unknowns: the final x and y components of the velocity of Puck B: Vbx and Vby. The two momentum equations (x and y) will let you solve for each. You should not have to use the energy conservation equation at all; they have provided more information than you need. Mathematically speakking, the problem is overdetermined
M*Vo = M*Va*cos60 + M*Vbx
0 = M*Va*sin60 M*Vby
2 = 1/2 + Vbx
Vbx = 3/2
Vby = sqrt3/2
Vbx^2 + Vby^2 = 9/4 + 3/4 = 3
After collision
Vb^2 + Va^2 = 4 = Vo^2
so kinetic energy really is conserved. 
Physics 
Anonymous,
What kind of drug is this guy on. How would anyone understand that answer

Physics 
Anonymous,
You literally explained nothing. Your numbers just spill onto the page with no explanation of where they came from