Conservation of Linear Momentum (man on a moving railcar)

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Homework Help Overview

The problem involves a man running on a moving railroad flatcar, requiring an analysis of conservation of linear momentum. The scenario includes a flatcar moving at a specified speed while the man runs in the opposite direction, prompting questions about the resulting speed of the flatcar after the man changes position.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the initial and final momentum calculations, questioning the setup of the equations used. Some participants suggest re-evaluating the initial conditions and the relative speeds involved in the momentum conservation equations.

Discussion Status

The discussion includes various attempts to calculate the final speed of the flatcar, with participants providing different equations and interpretations of the momentum conservation principle. There is no explicit consensus on the correct approach or answer, but some guidance on considering the system as a whole has been offered.

Contextual Notes

Participants note discrepancies in the calculations and assumptions about the initial conditions, such as the velocities of the man and flatcar before and after the man runs. There is an emphasis on the need to clarify the definitions of momentum and the system being analyzed.

Vertiviper
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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
 
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What is the answer?
 
Abdul Quadeer said:
What is the answer?

The answer is supposed to be 14.4 m/s.
 
-46.00 m/s relative to the flatcar final speed, so check Pf=m1(-46.0m/s+ 17m/s) + m2 (Vi+17m/s).
 
Last edited:
Vertiviper said:

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
 
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.
 

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