Conservation of Linear Momentum (man on a moving railcar)

[Vertiviper]

Homework Statement

A man (weighing 915 N) stands on a long railroad flatcar (weighing 2005 N) as it rolls at 17.0 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 46.00 m/s relative to the flatcar. What is the resulting increase in the speed of the flatcar?

Homework Equations

Pi=Pf

The Attempt at a Solution

Pi=m1v1+m2v2
Pi= 5065.49

Pf=m1(-46.0m/s+ 17m/s) + m2 (Vi+17m/s)
V=18.26 -incorrect

 

[zorro]

What is the answer?

 

[Vertiviper]

What is the answer?

The answer is supposed to be 14.4 m/s.

 

[Quinzio]

-46.00 m/s relative to the flatcar final speed, so check Pf=m1(-46.0m/s+ 17m/s) + m2 (Vi+17m/s).

 

[HallsofIvy]

Homework Statement

A man (weighing 915 N) stands on a long railroad flatcar (weighing 2005 N) as it rolls at 17.0 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 46.00 m/s relative to the flatcar. What is the resulting increase in the speed of the flatcar?

Homework Equations

Pi=Pf

The Attempt at a Solution

Pi=m1v1+m2v2
Pi= 5065.49

This is incorrect. Initially, the man was standing on the rail car so both had the same velocity, 17 m/s. The total momentum was (915)(17)+ (2005)(17)= 49640 Nm/s.

Pf=m1(-46.0m/s+ 17m/s) + m2 (Vi+17m/s)
V=18.26 -incorrect

 

[zorro]

Consider the man and the flat car as your system. No external force acts on it in the horizontal direction, conserving the momentum.
Let M and m represent the mass of flatcar and man resp. Let 'v' be the final velocity of flatcar w.r.t GROUND.
Initial momentum = (M+m) x 17
Final momentum= Mv + m(v-46) [Taking the direction of 'v' along positive x-axis]

Solving this you will get v=31.4m/s
You can find increase in velocity from this.