Thursday, January 21, 2010

Train A has a mass 4500kg successfully couples with Train b which has a mass of 7500kg. Train A is moving at?

6.0 m/s before the collision and Train B is initially at rest. What is the final velocity of the coupled trains? What is the initial KE of the system? Final KE of the coupled train? How much KE is lost? What fraction of initial KE is lost?Train A has a mass 4500kg successfully couples with Train b which has a mass of 7500kg. Train A is moving at?
MaVo+MbVo=Vf(Ma+Mb) Try that.


KE=1/2mv^2 Final KE





Ma Mass of Object A


Mb Mass of Object B


Vo Initial Velocity of the relative train


Vf Final velocity


KE kinetic energy





Solve.Train A has a mass 4500kg successfully couples with Train b which has a mass of 7500kg. Train A is moving at?
Since the two objects are joined together after the collision, this is a ';perfectly inelastic'; collision, so total momentum will be conserved, but total kinetic energy won't.





First, figure out the total initial kinetic energy and momentum





Train A has 4500kg * 6 m/s of momentum, or 27000 kg-m/s


Train A has 1/2 * 4500kg * (6 m/s)^2 of kinetic energy, or 81000 J





Train B initially has no momentum or kinetic energy, since it's not moving.





The final momentum of the two joined trains will still be 27000 kg-m/s, so divide that by their total mass to get the final velocity:





27000/(4500+7500) = 2.25 m/s





The final kinetic energy of the joined trains is:





1/2 * (4500+7500) * (2.25)^2 = 30375 J





So you've lost (81000J-30375J) of kinetic energy, or 50625J.





50625J is 62.5% of 81000J, so you've lost 62.5% of the initial kinetic energy.
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