A man (weighing 746 N) stands on a long railroad flatcar (weighing 3000 N) as it rolls at 16.8 m/s in the positive direction of an x axis, with negligible friction. Then the man runs along the flatcar in the negative x direction at 4.46 m/s relative to the flatcar. What is the resulting increase in the speed of the flatcar? Answer 0.888 m/s Relevant Equations m1u1 + m2u2 = m1v1 + m2v2 F = ma In my attempt I saw that I would probably have to work with the equation m1u1 + m2u2 = m1v1 + m2v2 since I was asked to find the difference in velocities. Knowing that i solved for the masses of each object. Man (Object 1) F = 746 N m = 746/9.8 = 76.12 kg Flatcar (Object 2) F = 3000 N m = 3000/9.8 = 306.12 kg After that I rearranged the conservation of momentum equation to fit the problem stated above. Since, in the beginning, they can be treated as one object the equation is, u(m1 + m2) = m1(u-v1) + m2v2 I also noted that the velocity of the man was relative to the flatcar so I took the difference of the two velocities and assigned it to the man. After having this I solved for v2 and plugged in the knowns. v2 = [u(m1 + m2) - m1(u-v1)]/m2 v2 = [16.8(76.12 + 306.12) - 76.12(16.8 - 4.46)]/306.12 v2 = 17.909 m/s However, when I go and calculate the difference I get... well, something that is not the answer! Δv = vf - vi = 17.909 - 16.8 = 1.109 m/s. What did I do wrong? Did I not understand what the question was asking of me? Thank you for taking the time to read and review my question.