Hello, I've recently came across a question that I have two solutions for. Though I don't see how either of them can be wrong, I'm getting two different answers :( Can someone explain me why this is happening (maybe I'm doing something wrong?) and which solution is better to use. 1. The problem statement, all variables and given/known data During a free dance program in figure skating, Phrank (m = 71kg) glides at a 2.1m/s to a stationary Phyllis (m= 52kg) and hangs on. How far will the pair slide after the "collision" if coefficient of kinetic friction (μk) between their skates and the ice is 0.052? 2. Relevant equations m1v1 1+m2v2 2=(m1+m2)vf (Inelastic collision, since Phrank grabs on) Solution 1: F⋅Δt=Δp =m(vf−vi) (Impulse) Δd = 1/2(vi+vf)Δt Solution 2: W=F⋅d Wnet=ΔK 3. The attempt at a solution Solution 1: Use the inelastic collision formula to solve for the final velocity (vf). m1v1 + m2v2 = (m1+m2)vf (71kg)(2.1m/s) + (52kg)(0) = (71kg + 52kg)vf vf = 1.2122m/s Plug that into the impulse forumla. F⋅Δt=Δp =m(vf−vi) Δt = Δp/F Δt = m(vf - vi)/μk⋅m⋅g (masses cancel out) Δt = vf - vi/μk⋅g Δt = (0 - 2.1m/s)/0.052⋅9.81 m/s2 Δt = 2.3763s Plug the time into the the second kinematics equation. Δd = 1/2(vi+vf)Δt Δd = 1/2(2.1m/s + 1.2122m/s)(2.3763s) Δd = 1.44 m Solution 2: Use the inelastic collision formula to solve for the final velocity (vf). m1v1 + m2v2 = (m1+m2)vf (71kg)(2.1m/s) + (52kg)(0) = (71kg + 52kg)vf vf = 1.2122m/s Combine the two work equations. W=F⋅d Wnet=ΔK F⋅d=ΔK Δd = ΔK/F Δd = 1/2(m1 + m2)vf2 - 1/2(m1)(v1)2 / μk⋅m⋅g Its messy, so I'll skip plugging in all the values. Δd = -66.185/-62.74476 Δd = 1.0548m Both solutions seem reasonable to me, but why am I getting two different answers? Which solution is more accurate?