Can anyone help me with physics?

Question

This is a college physics problem. I could use any help anyone can give me on this problem. I am stuck and cannot get to the right answers that are in the back of the book :(

Two carts of equal mass, m = 0.250 kg, are placed on a frictionless track that has a light spring of force constant k = 44.0 N/m attached to one end of it, as in the figure I linked below. The red cart is given an initial velocity of v0 (v knot) = 2.40 m/s to the right, and the blue cart is initially at rest.

http://www.webassign.net/sercp/p6-60.gif

a) If the carts collide elastically, find the velocity of the carts (both blue and red) just after the first collision.

b) If the carts collide elastically, find the maximum compression in the spring.

Answer 1

There is conservation of momentum: m1*v0 = (m1v1+m2v2); m1 = m2, so m*v0 = m(v1+v2), therefore v1+v2 = v; The kinetic energy before collision is .5m*v0^2; the kinetic energy after collision is .5*m*v1^2 + .5*m*v2^2, so v0^2 = v1^2 + v2^2. v1 = v0 - v2

v0^2 = (v0-v2)^2 + v2^2
v0^2 =v0^2 - 2v0*v2 + v2^2 +v2^2
0 =-2*v0*v2 + 2*v2^2
0 = -v0 + v2
v2 = v0 then v1 = 0

EDIT: Actually this has two solutions, v2=v0 or v2=0; If v2=0 then v1=v0, meaning cart 1 moves through cart 2, which cannot happen.

The energy of the second cart is .5*m*v0^2. The energy in compressing the spring is ∫F*dx = ∫kx*dx = (k/2)*s^2 where s is the spring compression.

Therefore (k/2)*s^2 = .5*m*v0^2

s^2 = (2/k)*.5*m*v0^2

s = v0*√(m/k)