Conservation of momentum

  • Thread starter RussG
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  • #1
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Ok, this is probably the simplest problem I was assigned out of an intro physics course and simply cannot figure out how to go about it.

Hockey Puck B at rest is struck by puck A, which was traveling at 40.0m/s in the +x direction. After they hit, A is deflected 30 degrees above the x axis, and Puck B acquires a velocity at 45 degrees below the x axis. The pucks have the same mass and friction forces aren't considered.

a) The speed of each puck after collision
b) what fraction of the original kinetic energy of puck A dissipates during the collision

I have the solutions, but can't figure out how to solve for them. Yes, I suck at physics :( Any help would be appreciated.

Since the message says solutions can't be solved for us, just a hint on where to start would be useful. I'm just ending up with ridiculous equations with sin/cos trying to account for the conservation of momentum.
 
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Answers and Replies

  • #2
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Those equations might not be so ridicuous after all. The total momentum is 40*m in the positive x-direction. Keep that in mind.

So the total momentum in the y-direction is 0. The pucks do get a speed in the y-direction though, so these must be equal and opposite to make the total momentum 0:
Va*sin(30 degrees) *m = - Vb*sin(45 degrees)*m (the positive y direction is above the x-axis). So here's equation [1]. (Just forget about the mass btw. It can be annihilated from the equation)

Now in the x-direction there is also conservation of momentum, meaning it has to be 40 m/s * m
Now we achieve:
Va*cos(30 degrees)*m + Vb*cos(45 degrees)*m = 40 *m (Just take out the m again.)
This leaves you with 2 equations and 2 unknowns. Do some basic maths and you'll be able to solve it. Good luck!
 
  • #3
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Hey - thanks a lot for the help. Those are the same equations as what I was getting albeit one difference, and looking at the problem I thought "wow, this one's easy." yet I couldn't get any solutions. I see what I was doing wrong now. Instead of the velocities, Va and Vb, I was using their component velocities with the trig functions, like Vaysin30 Vbysin45 etc. Needless to say that made things complicated when trying to equate the two. Looking back, I have no clue why I was doing that, maybe some inner desire to make things harder than they really are. ;) Thanks for the clarification.
 

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