1. The problem statement, all variables and given/known data A block of mass m is attached to a spring with a force constant k, as in the above diagram. Initially, the spring is compressed a distance x from the equilibrium and the block is held at rest. Another block, of mass 2m, is placed a distance x/2 from the equilibrium as shown. After the spring is released, the blocks collide inelastically and slide together. How far (Δx) would the blocks slide beyond the collision point? Neglect friction between the blocks and the horizontal surface. 2. Relevant equations U_spring = 1/2*k*x^2 K = 1/2*m*v^2 F = m*a v_f^2 = v_0^2+2*a*x 3. The attempt at a solution I used conservation of energy to find the velocity where the two blocks collide at x/2. I'm not sure about this though, because I don't think that conservation of momentum applies and I used 3m for the mass (assuming that the blocks had collided). Then I used N2L to find the acceleration based on the force (-kx) and mass (3m). I plugged that into the kinematics equation for final velocity and solved for delta x. I'm really not having any luck. E_i = 1/2*k*x^2 = E_f = 1/2*m*v^2 + 1/2*k*x_f^2 at x/2 where m collides with 2m: 1/2*k*x^2 = 1/2*3m*v^2 + 1/8*k*x^2 k*x^2 = 3m*v^2 + 1/4*k*x^2 3/4*k*x^2 = 3*m*v^2 yielding v = sqrt(k*x^2/(4*m)) by newtons second law a = -k*x/(3*m) and using kinematics (v_f^2 = v_0^2+2*a*x) 0 = k*x^2/(4*m) - 2*(k*x/(3*m))*delta(x) x/4 = 2/3*delta(x) delta(x) = 3/8*x Am I way off base with this approach?