1. Dec 18, 2008

### roxxyroxx

1. The problem statement, all variables and given/known data

A 1.1 kg cart and a 2.2 kg cart are at rest with a compressed string between them. the force constant of the spring is 600 N/m. when the spring is released, the carts rapidly separate. how far must the spring have been compressed for the 2.2 kg cart to end up moving at 1.5 m/s?
(involves both momentum and energy)

2. Relevant equations

p = mv
Ek = 0.5mv2
Ep = 0.5kx2

3. The attempt at a solution

i dont even know where to start !!

2. Dec 18, 2008

### Staff: Mentor

Figure out the total KE of the carts. Hint: What's the momentum of the system?

3. Dec 18, 2008

### roxxyroxx

ok so i got:
0.5(1.1)(3^2) = 4.95 J for Cart 1
0.5(2.2)(1.5^2) = 2.475 J for Cart 2

4. Dec 18, 2008

### jgens

Using what you know about the system, apply the law of conservation of energy now.

5. Dec 18, 2008

### roxxyroxx

ok so
0.5mv2
= 0.5(7.425)(1.5)2
=8.35

but the answer should be 0.16 m.. did i do something wrong or is there another setp now?

6. Dec 18, 2008

### jgens

Well, your total kinetic energy is the sum of the kinetic energy of each kart, so KE = 7.425. You know that the resultant KE must equal the initial potential energy so KE = .5(k)(x^2).

7. Dec 18, 2008

### roxxyroxx

great thanks !

8. Dec 18, 2008

### jgens

You're welcome.