Conservation of Momentum/Linear Momentum

  • Thread starter norcal
  • Start date
  • #1
norcal
19
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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

The Attempt at a Solution



(915+2805)g(18)=(915)g(-40+18)+(2805)g(v+18)
v=-1.3 m/s

But this is the wrong answer...what am I doing wrong?
 

Answers and Replies

  • #2
mjsd
Homework Helper
726
3
firstly, weights were given in Newton already, no needs to add an extra "g". secondly, if you do want to use momentum (somehow), you want the "mass" of the objects.
 
  • #3
jay ambekar
26
0
usually i get ma equations in variables n thn substitute the values...suppose the man moves at a speed w towards left and the platform recoils at a speed V towards right...hence the speed of the man relative to the platform is V+w=v...or w= v-V...
 
  • #4
jay ambekar
26
0
taking the man n the platform to b inna system, there is no external horizontal force on the system, the linear momentum remians constant. thus
(M+m)18 = MV- mv...now solve n check
 
  • #5
jay ambekar
26
0
sorry, the eqn above shud hav been (M+m)18 = MV- mw
 
  • #6
jay ambekar
26
0
ooookay...the catch is , it will recoil wit V+18 and not just V...ill just try n let u know
 
  • #7
norcal
19
0
If You look at what I posted above that is the equation that I used and it did not work...(M+m)18=M(v+18)+m(v+18) and it did not give me the right answer.
 

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