Mechanical Energy loss due to NC Forces: two railroad cars collide.

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SUMMARY

The discussion focuses on calculating the thermal energy produced during a collision between two railroad cars, each with a mass of 4800 kg, traveling at 80 km/h. The initial kinetic energy is calculated using the formula \(E_k = \frac{1}{2}mv^2\), resulting in an energy output of 2.4 x 10^6 J when considering both cars. The error in the initial calculation stemmed from only accounting for one car's kinetic energy, neglecting the contribution of the second car. The correct approach involves summing the kinetic energies of both cars to determine the total thermal energy produced.

PREREQUISITES
  • Understanding of kinetic energy calculations
  • Familiarity with the conservation of energy principle
  • Basic knowledge of physics concepts related to collisions
  • Ability to convert units (e.g., km/h to m/s)
NEXT STEPS
  • Study the conservation of momentum in collisions
  • Learn about energy transformations in inelastic collisions
  • Explore the concept of thermal energy and its calculations
  • Investigate real-world applications of collision physics in engineering
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Physics students, educators, and engineers interested in understanding energy loss during collisions and the principles of kinetic energy transformation.

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[SOLVED-Thanks AL]Mechanical Energy loss due to NC Forces: two railroad cars collide.

SOLVED! - Thanks Al![/color]

Homework Statement



Two railroad cars each of mass 4800 kg and traveling @ 80 km/h collide head on and come to rest. How much thermal energy is produced in this collision?

Homework Equations



[tex]E_k + E_p = E^{'}_k + E^{'}_p + Q[/tex]

The Attempt at a Solution



This problem does not use momentum. Energy considerations only.

I treat one car as a single entity.

Energy before = energy after.

There was only an initial kinetic energy at the beginning and no kinetic energy afterwards. so i assume all the energy was transformed to Q.

[tex]\frac{1}{2}mv^2 = Q[/tex]

after finding 80 km/h = 22.2222... m/s

i solved and got 1.2 x 10^6 J, the answer is double that, 2.4 x 10^6 J, where did i go wrong?
 
Last edited:
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There are two moving trains. :wink:
 
OH! I only calculated the amount of thermal energy one train would release, and not the other. So the answer would be Q of both trains (KE of train one + the KE of train two) as it's released simultaneously ?? Ugh... i thought i'd isolate only one train because I wasn't dealing with momentum.. got my head stuck in dynamics :) Thanks a lot Doc!
 

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