Elastic collision speed and time problem

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SUMMARY

The discussion centers on the calculation of velocities after a perfectly elastic collision between two ice pucks: a 0.45 kg puck moving east at 3 m/s and a 0.9 kg puck. The correct final velocities are determined to be 2 m/s east for the 0.9 kg puck and 1 m/s west for the 0.45 kg puck. The momentum equation used is 0.45*3 = 0.9v1 + 0.45v2, which simplifies to 3 = 2v1 + v2. The confusion arises from the interpretation of the velocities, but the values ultimately align with the conservation of momentum.

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A 0.45 kg ice puck, moving east with a speed of 3 m/s, has a head-on collision with a 0.9 kg ice puck. Assuming a perfectly elastic collision, what will be the speed and direction of each object?

Alright, so the first time i go through this problem i solve it, no problem. the answer is 1 m/s west and 2 m/s east and that's also the answer in the book. but then i go back and i realize something weird. i set up the momentum equation like normal so that it's
0.45*3 = 0.9v1 + 0.45 v2
and then you can make it
0.45*3 = 0.45*2v1 + 0.45 v2
divide both sides by 0.45
3 = 2v1 + v2
and you get that but the values determined for the speeds don't match up with that. What's going on?
 
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fizzzzzzzzzzzy said:
A 0.45 kg ice puck, moving east with a speed of 3 m/s, has a head-on collision with a 0.9 kg ice puck. Assuming a perfectly elastic collision, what will be the speed and direction of each object?

Alright, so the first time i go through this problem i solve it, no problem. the answer is 1 m/s west and 2 m/s east and that's also the answer in the book. but then i go back and i realize something weird. i set up the momentum equation like normal so that it's
0.45*3 = 0.9v1 + 0.45 v2
and then you can make it
0.45*3 = 0.45*2v1 + 0.45 v2
divide both sides by 0.45
3 = 2v1 + v2
and you get that but the values determined for the speeds don't match up with that. What's going on?

they match. If v1 is 2m/s east and v2 is 1m/s west... v1 = 2. v2 =-1.

3 = 2(2) + (-1)
 

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