freight car-caboose collision finding mass without velocities

hangten1039

A 34000-kg railroad freight car collides with a stationary caboose car. They couple together, and 23 percent of the initial kinetic energy is dissipated as heat, sound, vibrations, and so on. What is the mass of the caboose?



m1v1= (m1+m2)vf
KEbefore= .23KEcollision

I don't know how to set up the last equation so that velocities cancel. I am not sure how to set up the kinetice energy equation, please help! thank you!

Doc Al

OK.
Rethink this one more carefully.

What's the definition of KE? Write vf in terms of v1 (using your first equation).

hangten1039

so would i get for kinetic equation:
.5m1(v1^2)= .23(.5)(m1+m2)(m1v1/(m1+m2))

hangten1039

nevermind i would actually get: .5(m1)((m1+m2)(vf)/m1) = .23(.5)(m1+m2)vf

Doc Al

Two problems:
(1) You forgot to square the velocity on the right hand side.
(2) The final KE is less. If 23 % of the initial KE is "lost", what's left?

hangten1039

I'm not sure I understand what you are explaining in part 2..

so my equation would be .5m1(v1^2)= .23(.5)(m1+m2)(m1v1^2/(m1+m2))

hangten1039

I still have initial velocity as unknown or final velocity, so i'm not sure how to get rid of that

Doc Al

You still have the square wrong on the RHS. The entire vf term must be squared.

Your equation (once you correct the above error) says that the initial KE = 23% of the final KE. Does that make sense?

Doc Al

Once you get the KE equation correct, the velocity cancels.

hangten1039

so actually I would have to take 23% of the initial kinetic energy

Doc Al

You start with one dollar, but then lose 23 cents. What's left?

So the final KE is what percentage of the initial?

hangten1039

the final is 77% of the initial

Doc Al

Good. Use that to write your KE equation.