Solving Physics Puzzle: Hockey Puck Collision

AI Thread Summary
The discussion revolves around solving a physics problem involving the collision of two hockey pucks, where puck A strikes puck B at rest. The key points include the conservation of momentum in both x and y directions, leading to two equations that can be solved for the speeds of the pucks after the collision. The user initially struggled with using component velocities instead of the total velocities, complicating their calculations. After receiving guidance, they realized their mistake and acknowledged that the problem was simpler than they had perceived. The conversation emphasizes the importance of correctly applying conservation principles in collision problems.
RussG
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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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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!
 
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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