Conservation of Momentum/Linear Momentum

[norcal]

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?

 

[mjsd]

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.

 

[jay ambekar]

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...

 

[jay ambekar]

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

 

[jay ambekar]

sorry, the eqn above shud hav been (M+m)18 = MV- mw

 

[jay ambekar]

ooookay...the catch is , it will recoil wit V+18 and not just V...ill just try n let u know

 

[norcal]

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.