Kinetic energy of 2 cart system

AI Thread Summary
The discussion focuses on calculating the translational kinetic energy of a two-cart system before collision, with one cart weighing 371 kg moving at 1.90 m/s and the other weighing 495 kg at 2.09 m/s. The kinetic energy formula K=1/2mv^2 is highlighted for these calculations. Participants clarify that the total translational kinetic energy is the sum of the individual kinetic energies of both carts, rather than averaging their speeds. Additionally, it is noted that during a collision, not all kinetic energy may be converted into other forms due to momentum conservation, which affects the final velocity of the combined object. Understanding these principles is essential for solving the problem effectively.
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Homework Statement


On cart with a mass of 371 kg and another cart with a mass of 495 kg are rolling toward each other. Just before they collide, the 495 kg cart has a speed of 2.09 m/s, and the 371 kg cart has a speed of 1.90 m/s. We will consider the friction between the cart wheels and track negligible for this problem. Evaluate the translational kinetic energy and convertible kinetic energy of the two-cart system before they collide.

Homework Equations


K=1/2mv^2

The Attempt at a Solution


Do you find the kinetic energy for both of the carts? or average the speeds? Also what's the different between translational kinetic energy and convertible kinetic energy
 
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The total translational KE of a system is the sum of the KE's of all its moving parts.

Now consider a scenario in which there is a two-body collision. Even if the collision is perfectly inelastic it is not always possible to convert all the KE into some other form of energy (like heat or sound or potential energy). This is because conservation of momentum will dictate what the final velocity of the combined object will be post-collision. If there's motion remaining then there's KE remaining, too. So that portion of the original KE could not be converted.
 
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Thank you so much! This helps a lot!
 
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