Speed and Mechanical Energy Lost in Railroad Car Collision

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SUMMARY

The discussion centers on a railroad car collision involving a single car of mass 2.49e4 kg moving at 3.89 m/s colliding with three additional cars, each of the same mass and moving at 1.95 m/s. The final speed of the coupled cars after the collision is calculated using the conservation of momentum formula: 2.49e4 * 3.89 + 3 * 2.49e4 * 1.95 = 4 * 4.29e4 * V, resulting in a final speed of 2.92 m/s. The mechanical energy lost during the collision can be determined by comparing the initial and final kinetic energies.

PREREQUISITES
  • Understanding of conservation of momentum
  • Basic knowledge of kinetic energy calculations
  • Familiarity with algebraic manipulation of equations
  • Concept of inelastic collisions
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  • Calculate the initial and final kinetic energies of the railroad cars
  • Explore the concept of inelastic collisions in physics
  • Study the implications of momentum conservation in multi-body collisions
  • Learn about energy loss in collisions and its real-world applications
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TrippingBilly
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A railroad car of mass 2.49 104 kg, is moving with a speed of 3.89 m/s. It collides and couples with three other coupled railroad cars, each of the same mass as the single car and moving in the same direction with an initial speed of 1.95 m/s.
(a) What is the speed of the four cars after the collision?
(b) How much mechanical energy is lost in the collision?

Sorry, but I have no idea about this problem :(
 
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Hint: What is conserved in any collision?
 
Momentum, so...

2.49e4 * 3.89 + 3*2.49e4 *1.95 = 4*4.29e4 * V

?
 
TrippingBilly said:
Momentum, so...

2.49e4 * 3.89 + 3*2.49e4 *1.95 = 4*4.29e4 * V
Right. Looks good, except for a typo. Once you solve for the final speed, you can figure out the initial and final kinetic energies and compare them.
 

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