What was the speed before the collision?

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SUMMARY

The speed of puck A before the collision was determined to be 0.79 m/s based on the conservation of momentum principle. The initial momentum (Pi) was calculated using the mass of puck A (0.250 kg) and the unknown velocity (V), while the final momentum (Pf) was computed using the post-collision velocities of both pucks. The calculations confirmed that the direction of puck A's velocity after the collision must be negative, which is crucial for accurate momentum conservation. The discrepancy in earlier calculations was resolved by correctly applying the negative sign to puck A's post-collision velocity.

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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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