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

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The discussion focuses on solving a physics problem involving the elastic collision of two hockey pucks of equal mass. The first puck, moving at 0.48 m/s, collides with a stationary puck and is deflected at 39° to the right with a final speed of 0.35 m/s. The user attempts to apply the conservation of momentum principle but incorrectly calculates the final speed and direction of the second puck, resulting in confusion regarding the correct application of the equations involved.

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~*Noor*~
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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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