1. The problem statement, all variables and given/known data A 5.0 kg block with a speed of 8.0 m/s travels 2.0 m along a horizontal surface where it makes a head-on, perfectly elastic collision with a 15.0 kg block which is at rest. The coecient of kinetic friction between both blocks and the surface is 0.35. How far does the 15.0 kg block travel before coming to rest? ANS. 1.79m 2. Relevant equations Elastic Collision: K_0 = K m1iv1i = m1v1 + m2v2 Ff = μmg 3. The attempt at a solution First it travels 2.0 m with friction: K_o - Wf = K .5mv^2 - 2μmg = .5mv^2 v_0^2 - 4μg = v^2 v^2 = 50.28 v = 7.1 m/s Now cons. of momentum: 5(7.1) + 0 = 5(v1) + 15(v2) 7.1 = v1 + 3v2 KE: 1/2mv_0^2 + 0 = 1/2m1v1^2 + 1/2m2v2^2 160 = 2.5v1^2 + 7.5v2^2 64 = v1^2 + 3v2^2 Solving for v1 by substituting then quad formula gives: v2 = -.29 m/s This does not make any sense?