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What is the resulting increase in the speed of the flatcar?

  • Thread starter norcal
  • Start date
  • #1
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Homework Statement



A man (weighing 915 N) stands on a long railroad flatcar (weighing 2805 N) as it rolls at 18.0 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 40.00 m/s relative to the flatcar. What is the resulting increase in the speed of the flatcar?

Homework Equations



Pi=Pf
m1v1=m2v2

The Attempt at a Solution



(915+2805)(18)=(915)(-40+18)+(2805)(v+18)
v=13.05 m/s

Well this is the wrong answer and I have no clue why. I know that the fact that the weights are given and not mass is not an issue since W=ma and a would cancel out on both sides so I cannot figure out why this answer is wrong. Anything else that I can try???
 

Answers and Replies

  • #2
93
1
Hm
well
my physics teacher always says when you are dealing with speed or velocity, you always use energy.
So, since the net vertical change is 0, change in potential energy is negligible.
So, start with this equation
PEi + KEi = PEf + KEf
You can rearrange to get -(deltaPE) = deltaKE. We know that deltaPE = 0, so we have this formula
deltaKE=0
KEf - KEi = 0
KEf = summation of final kinetic energy
KEi = summartion of initial kinetic energy

you can find m because w = mg and you are given w and g is known.
Once you find m, use the above KE equations and solve for v final of flatcar, and that should give you the right answer.
 
  • #3
340
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why do you have (v+18)?
 
  • #4
340
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pakmingki- be careful here. Total mechanical energy is not conserved here.
 
  • #5
93
1
pakmingki- be careful here. Total mechanical energy is not conserved here.
Why wouldn't it be? The only force that could possibly act against the system is friction, which is negligible.
 
  • #6
19
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I have v+18 because its asking for the resulting increase in speed. do you have a suggestion of doing it another way???
 
  • #7
19
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KE is not involved in this problem, only conservation of momentum.
 
  • #8
340
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Why wouldn't it be? The only force that could possibly act against the system is friction, which is negligible.
Looks inelastic to me....

but it is late cause I dont see anything wrong with the work shown above. :zzz:
 
  • #9
93
1
KE is not involved in this problem, only conservation of momentum.
well,
using KE could be a different approach to the problem, and if there is mass and velocity involved, it could very well be a KE problem as much as a P problem.
 
  • #10
340
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KE is not conserved in inelastic collisions.
 
  • #11
19
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maybe so but in the section of this book it says not to use KE as KE is conserved. Thus, using KE for this problem would be ridiculous.
 
  • #12
93
1
if its an inelastic collision, you can use the inelastic collision equation.
Vfinal = sum P / sum m
where Vfinal is the velocity of the system, which is the man and the train.
 

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