## Homework Statement

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)

p = mv
Ek = 0.5mv2
Ep = 0.5kx2

## The Attempt at a Solution

i dont even know where to start !!

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

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

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

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

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

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?

jgens
Gold Member
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).

great thanks !

jgens
Gold Member
You're welcome.