- #1

Fatentity

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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!