Calculating Thermal Energy from a Head-On Collision

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In a head-on collision between two railroad cars, each weighing 6500 kg and traveling at 86 km/h, the calculation of thermal energy produced requires converting the speed to meters per second, which is approximately 24 m/s, not 2.389 m/s. The kinetic energy (KE) of both cars must be considered, as they both contribute to the total energy converted into thermal energy upon collision. The correct formula for kinetic energy is KE = 1/2 mv^2, and the total kinetic energy before the collision can be calculated by summing the kinetic energies of both cars. The discussion emphasizes the importance of accurate unit conversion and considering the contributions from both vehicles in the calculation. Understanding these principles is crucial for accurately determining the thermal energy generated in such collisions.
crisicola
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I've got a question on thermal energy:

Two railroad cars, each of mass 6500 kg and traveling 86 km/h, collide head-on and come to rest. How much thermal energy is produced in this collision?

Converted 86 km/h to m/s = 2.389 m/s
I assumed that KE converted all into thermal energy:
KE = 1/2 mv^2, but it doesn't turn out right.

Thank you beforehand for any help!
 
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Check your conversion. Also, did you include the kinetic energy of both cars?
 
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