What was the speed before the collision?

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The discussion revolves around calculating the speed of puck A before a collision with puck B on a frictionless air table. Puck A has a mass of 0.250 kg and moves to the right, while puck B, with a mass of 0.350 kg, is initially at rest. After the collision, puck A moves left at 0.120 m/s, and puck B moves right at 0.650 m/s. The initial momentum equation is set up, but the direction of puck A's velocity after the collision must be considered as negative. The participant successfully resolves the problem after correcting their approach and confirming their answer with a teacher.
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Homework Statement


On a frictionless, horizontal air table, puck A (with a mass 0.250 kg) is moving to the right toward Puck B (with mass 0.350 kg), which initially at rest. After the collsion, puck A has a velocity of 0.120 m/s to the left, and puck B has a velocity of 0.650 m/s to the right.

Question A: What was the speed before the collision?
Question B: Calculate the change in the total kinetic energy of the system that occurs during the collision.


Homework Equations


*P = Momentum*
P = mv
P(total) = Pa + Pb
U = 1/2 K X^2.
Kinetic energy = 1/2 (m1 + m2)V^2.


The Attempt at a Solution


(Note: I haven't tried working out the Kinetic energy problem yet...so don't worry about that. Just trying to solve this part for now).

I think it's the fact that puck B is at rest which is throwing me off.
I know the answer is V = 0.79 m/s for Puck A before collision *Answer is in the back of the book*, but I can not figure out how to get this.

I know Pi = (0.25kg) V + (0.35kg) V = ?
And Pf = (0.25kg) (0.12 m/s) + (0.35 kg) (0.65 m/s) = 0.2575 Kg*m/s.

I've tried setting them equal to each other and canceling from both sides. Which gave me V = 0.12 m/s + 0.65 m/s => V = 0.77 m/s. But that's off by 0.02

I tried solving it a few other ways, but that gave me = 0.91 m/s. And I tried another way and got 1.03 m/s. So none of that was close.

I tried working it backwards an using the answer plugging it into Pi giving me:

Pi = (0.25kg) (0.79 m/s) + (0.35kg) (V) = 0.1975 Kg*m/s.
Pf = (0.25kg) (0.12 m/s) + (0.35 kg) (0.65 m/s) = 0.2575 Kg*m/s.
Setting them equal to each other:
0.1975 Kg*m/s + 0.35Kg (V) = 0.2575 Kg*m/s
and getting to

(0.35kg)(V) = 0.06 Kg*m/s.

This is far as I can go. In what I'm missing have some relation to maybe that the square root of 0.35 is almost 0.06?

Or is there a reason my one answer that V = 0.77 m/s was off by just 0.02 m/s, and it's something I'm just not seeing?

Any and all help is appreciated =]. Thank you in advance!
 
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Fatentity said:
I know Pi = (0.25kg) V + (0.35kg) V = ?
Almost. What's the initial velocity of puck B? It's not the same as puck A.
And Pf = (0.25kg) (0.12 m/s) + (0.35 kg) (0.65 m/s) = 0.2575 Kg*m/s.
Careful. Direction matters. The fact that after the collision puck A is moving to the left means that you need to give its velocity a negative sign.

I've tried setting them equal to each other
Fix the above and try it again.
 
Thank you! Fixed what you said and tried again in class and asked the teacher if I had the right answer last week and I did, so it worked. Much appreciated Doc Al :).
 
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