How Do You Calculate the Mass of a Stationary Train After a Collision?

[EthanB]

Sorry that I don't have LaTeX, I'll do my best without it:

A train with known mass collides with a stationary train and the two link together. If 27% of the initial kinetic energy is dissipated, what is the mass of the stationary train? We are given no velocities.

M1 = mass of moving train (known)
M2 = mass of stationary train
Vi = initial velocity of moving train
Vf = final velocity of the 2-train system
Ki = initial kinetic energy
Kf = final kinetic energy
Wnc = work done by non-conservative forces
Pi = initial momentum
Pf = final momentum

Using conservation of momentum:
Pi = Pf
(M1)(Vi) = (M1+M2)(Vf)

Using conservation of energy:
Ki + Wnc = Kf
(1/2)(M1)(Vi)^2 - (.27)(1/2)(M1)(Vi)^2 = (1/2)(M1+M2)(Vf)^2
(.73)(M1)(Vi)^2 = (M1+M2)(Vf)^2

...2 equations, 3 unknowns. What am I forgetting?

 

[OlderDan]

Sorry that I don't have LaTeX, I'll do my best without it:

A train with known mass collides with a stationary train and the two link together. If 27% of the initial kinetic energy is dissipated, what is the mass of the stationary train? We are given no velocities.

M1 = mass of moving train (known)
M2 = mass of stationary train
Vi = initial velocity of moving train
Vf = final velocity of the 2-train system
Ki = initial kinetic energy
Kf = final kinetic energy
Wnc = work done by non-conservative forces
Pi = initial momentum
Pf = final momentum

Using conservation of momentum:
Pi = Pf
(M1)(Vi) = (M1+M2)(Vf)

Using conservation of energy:
Ki + Wnc = Kf
(1/2)(M1)(Vi)^2 - (.27)(1/2)(M1)(Vi)^2 = (1/2)(M1+M2)(Vf)^2
(.73)(M1)(Vi)^2 = (M1+M2)(Vf)^2

...2 equations, 3 unknowns. What am I forgetting?

That you were not asked to find the velocities. Perhaps they are not unique. Does that give you any ideas?

 

[EthanB]

That you were not asked to find the velocities. Perhaps they are not unique. Does that give you any ideas?

I can't see that it does. I realize the velocity of the trains after impact is the same, that they become a system. I've already taken that into account, though.

Help?

 

[OlderDan]

I can't see that it does. I realize the velocity of the trains after impact is the same, that they become a system. I've already taken that into account, though.

Help?

Try dividing your two equations

 

[EthanB]

Righteous, dude. Can't believe that one slipped me by. I get so stuck on comparing numbers of equations and unknowns that I miss the little things. Thanks much.

 

[dextercioby]

Sorry that I don't have LaTeX, I'll do my best without it:

But we do have LaTex. All you need next time is write the code between [ tex ] [ /tex ] tags and it comes out nicely.

Daniel.

 

[EthanB]

Oh, great. Where can I go to find out how to write the code?

 

[dextercioby]

https://www.physicsforums.com/showthread.php?t=8997"

Daniel.