Calculate Change in Velocity of Flatcar When Man Runs Opposite Direction

[DavidAp]

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.

 

[hage567]

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.

You are on the right track but the red part is not quite correct. Remember that once the man starts moving, the flatcar is no longer going at 16.8 m/s. How does that change your equation?