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Collisions of hockey pucks

  • Thread starter Tabe
  • Start date
  • #1
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Okay, here's the problem I'm working on...
A .50 kg hockey puck moving at a speed of 5.70 m/s collides with a stationary hockey puck of equal mass. After the collisions, the first puck moves off in a direction 60 degrees to the left of its original direction, while the second puck moves off at a right angle to the first puck. What is the speed of the second puck after the collision?
I know that I have to solve for the components, but I'm not sure which numbers to use to solve for it.
 

Answers and Replies

  • #2
quasar987
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mmh... These are the "numbers":

initial horizontal momentum = (0.50 kg)(5.70)

initial vertical momentum = 0.
 
  • #3
dextercioby
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1)Make a diagram of the particles which collide in which put the momentum vectors.Be careful with the angles.
2)Chose a coordinate system with the center in the collision point and 2 orthonorlmal axis.
3)Write down the conservation of momentum in vector form and conservation of KE.
4) Project the vector eq.written at 3) on the axis of coordinates.
5)Find the "speed of the second puck after the collision".Actually from 4) u find its components on the 2 axis of coordinates.Apply Pythagora's theorem to find the modulus.

Daniel.
 

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