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Elastic collision in two dimensions

  • Thread starter linnus
  • Start date
23
0
1. Homework Statement

On a frictionless surface, a 0.35 kg puck moves horizontally to the right (at an angle of 0°) and a speed of 2.3 m/s. It collides with a 0.23 kg puck that is stationary. After the collision, the puck that was initially moving has a speed of 2.0 m/s and is moving at an angle of −32°. What is the velocity of the other puck after the collision?

2. Homework Equations
Momentum before= momentum after
kinetic energy before= kinetic energy after
AB=AB cos (x)


3. The Attempt at a Solution

From Momentum before= momentum after, I got
M1=first mass
Vi1=initial velocity of the first mass
Vf1=final velocity of the first mass...and I think you get the rest
M1(Vi1-Vf1)=(M2)(Vf2)

From the kinetic energy before= kinetic energy after, i got
M1(Vi1^2-V1f^2)=(M2)(V2f^2)

Now i'm kinda lost on what to do...some advice?
 
Last edited:
18
0
ok so you have one puck moving straight to the right initially. this has a certain momentum. this certain momentum is equal to the momentum of both of the pucks after the collision (which you already stated)

initial momentum = .35 x 2.3 = .805 to the right (0 degrees)

if you know that one will move at -32 degrees with a velocity of 2 so determine its momentum. all you have to do now is find the other momentum vector that, when added to this one, will equal .805 at 0 degrees.
 
Last edited:
23
0
i got it, thanks
 

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