Block 2 (mass 1.10 kg) is at rest on a frictionless surface and touching the end of an unstretched spring of spring constant 144 N/m. The other end of the spring is fixed to a wall. Block 1 (mass 1.70 kg), traveling at speed v1 = 3.60 m/s, collides with block 2, and the two blocks stick together. When the blocks momentarily stop, by what distance is the spring compressed? Answer 0.305 m Relavant Equations Fsp = -kx ______________________________ How do I solve for this? I'm given k but what is Fsp? Is there any other way to solve for Fsp? After scratching my head for a good while I tried something out of pure desperation, despite my gut feeling that it was wrong. I said, m1v1^2 + m2v2^2 = m1u1^2 + m2u2^2 m1v1^2 = m2u2^2 u2^2 = (m1v1^2)/m2 u2 = sqrt((m1v1^2)/m2) u2 = sqrt((1.7(3.6)^2)/1.10) = 4.4753 Being unsure about the relationship I said u2 = Fsp. Therefore, Fsp = -kx 4.4753 = -144x 4.4753/-144 = x -0.03107 = x Now, I am way off with this attempt but I'm not quite sure how I'm suppose to approach this problem. Can somebody help explain to me what, and most importantly why, I have to do to solve this problem? I'm having difficulty with this section so baby steps is appreciated :). Thank you for taking the time to review my question.