What Is the Speed and Direction of the Second Hockey Puck After Collision?

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In an elastic collision between two equal-mass hockey pucks, one puck initially moves at 0.48 m/s and collides with a stationary puck. After the collision, the first puck deflects at 39° and moves at 0.35 m/s. The user attempts to calculate the final speed and direction of the second puck but arrives at an incorrect result of -0.21 m/s. The user seeks assistance to resolve the discrepancies in their calculations. Clarification on the correct application of momentum conservation principles is needed to determine the second puck's speed and direction accurately.
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Hello,

Q.) A hockey puck moving at 0.48 m/s collides elastically with another puck that was at rest. The pucks have equal mass. The first puck is deflected 39° to the right and moves off at 0.35 m/s. Find the speed and direction of the second puck after the collision. Make sure the speed is in m/s and direction in terms of degrees

A.) The way i tried solving this problem is below:

m1=m2

Hockey Puck #1:
p=p(final)-p(initial)
=m1(vfinal-vinitial)


Hockey Puck #2:
=m2(vfinal-vinitial)


p1=p2

m1(v1final-v1initial)=m2(v2final-v2initial)

Both the masses cancel out... we have the following:
v1initial=.48
v1final= .35cos39
v2initial=0

Plug in formula and get v2final=-.21 since it's going to the left...but this answer is wrong.

I tried 39° for the second puck since they both weight the same but this is wrong as well.


Please i don't know why I'm getting this wrong. Any help will be appeciated :smile:
 
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