Hockey Puck Collision--momentum problem

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SUMMARY

The discussion centers on a physics problem involving the conservation of momentum during a collision between two hockey pucks. Puck A, moving at 10 m/s, collides with stationary puck B, resulting in puck A scattering at an angle of 40 degrees and puck B at 50 degrees. The key equations used include the momentum conservation equation, mAv1A + mBv1B = mAv2A + mBv2B. The solution approach involves breaking down the velocities into x and y components, and the assumption of equal masses for both pucks simplifies the calculations, allowing for the cancellation of mass terms.

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Homework Statement


The problem reads:
"Hockey puck A sliding on ice with velocity 10 m/s in the x direction collides at a glancing blow with a second stationary puck B. Puck A scatters at an angle of θA= 40 degrees above the x axis, while B slides away at an angle ofθB= 50 degrees below the x axis. (a) Use momentum conservation to find the final speeds vA and vB of the two pucks, respectively. What is the change in kinetic energy? (b) What are the two scattering angles according to an observer in the center of mass frame of reference?"

Homework Equations


mAv1A + mBv1B = mAv2A + mBv2B

The Attempt at a Solution


Ok, so far I divided the problem into x and y components, where the final velocities in the x direction would be v2Acos40 and v2Bcos50 and the final velocities in the y direction would be v2Asin40 and -v2Bsin50. I feel like there are too many unknowns, we aren't given the masses of the two pucks, and I have absolutely no clue how to do this problem without the masses of the pucks. I have 4 unknowns (final velocities and the masses) and only two equations. Am I missing something? How do I approach this problem??
 
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Assume the pucks have equal mass.
 
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TSny said:
Assume the pucks have equal mass.
Then the masses will cancel out?? BRILLIANT, that makes a lot of sense, thank you!
 

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