How Is Kinetic Energy Converted to Internal Energy in a Hammer-Spike Collision?

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SUMMARY

The discussion focuses on the conversion of kinetic energy to internal energy during a hammer-spike collision, specifically involving a 2.50 kg hammer striking a 0.5 kg spike at 65 m/s. It is established that one-third of the initial kinetic energy (KEi) is converted to internal energy (Ui). The equation used is ΔU_internal = 1/3 KEi, where KEi is calculated using the formula 1/2 mv². The participants clarify that potential energy (PE) is negligible in this scenario, and the primary focus is on the kinetic energy of the hammer.

PREREQUISITES
  • Understanding of kinetic energy calculations (KE = 1/2 mv²)
  • Familiarity with the concept of internal energy (Ui)
  • Knowledge of conservation of momentum principles
  • Basic grasp of potential energy concepts (PE = mgh)
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  • Study the principles of energy conservation in collisions
  • Learn about the relationship between kinetic energy and internal energy
  • Explore the conservation of momentum in elastic and inelastic collisions
  • Investigate real-world applications of kinetic energy conversion in engineering
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Physics students, mechanical engineers, and anyone interested in understanding energy transformations during collisions.

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A worker drives .5 kg spike into a rail tie with 2.50 kg hammer. Hits spike with 65 m/s. 1/3 kinetic energy converted to internal energy. Find increase in total internal energy.

The problem I have having with this problem is that I don't know how to deal with the masses. I know PEi + KEi + Ui = PEf + KEf + Uf. I am having trouble reasoning which ones equal zero.
I'm grateful for any help.
 
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You are looking for the change in Ui. What does the problem tell you about the change in Ui in relation to KE?
 
Ui is 1/3 KEi, right? And KEi is 1/2mv2. But that mass would only be for the hammer. And would the PEf be zero?
 
The problem really doesn't specify that there is a change in potential energy so PEf would be the same as PEi

As far as which kinetic energy to use, my guess would be the hammer unless the problem states otherwise.
 
Would the initial potential energy mgh, have a mass of the spike? Since no distance is mentioned, I'm assuming it is neglible. And then the KEi would be 1/2mv2 with the mass of hammer. + Ui. And that equals mgh with mass of spike? If so, there is no potential energy. Then + .5mv2 with a new velocity using collsion but with mass of...what? Wait a minute. All of these masses has to be the same, doesn't it? I can't go switching between spike and hammer, can I?
 
When dealing with gravitational potential energy you are really only concerned about the change in potential energy because your height is assigned from an arbitrary point.

Unless I am missing something here, the problem is pretty straightforward. It tells you that \Delta U_{internal} = \frac {1}{3} KE. Unless otherwise specified it would be logical to assume that the kinetic energy referred to in the problem is that of the hammer.

It's possible that the KE they're referring to is that of the spike after it's been hit, in which case you would have to use conservation of momentum to find the spike's velocity, and thus kinetic energy.

Can you copy the problem word for word?
 
Okay, I get this problem now. I was making it way more complicated. The spike can be ignored until the final mass and velocity combined witht eh hammer. Thanks for your help, what you said made exact sense. Thanks!
 

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