Inelastic collisions with constant momentum

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SUMMARY

The discussion centers on the concept of inelastic collisions, specifically addressing the calculation of kinetic energy before a collision involving identical objects. Participants confirm that the kinetic energy should be expressed as mv² rather than 1/2 mv², highlighting an error in the referenced textbook. The conversation emphasizes that the term "identical objects" is a simplification, as only the mass needs to be the same for the kinematic analysis to hold true.

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  • Understanding of inelastic collisions in physics
  • Familiarity with kinetic energy calculations
  • Basic knowledge of mass and momentum conservation
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  • Review the principles of inelastic collisions in classical mechanics
  • Study the conservation of momentum in different collision scenarios
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haha0p1
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Homework Statement
The total momentum before the collision in an inelastic collisions is 0, but the total kinetic energy before the collision is 1/2mv². Calculate how the total kinetic energy before collision is 1/2mv².
Relevant Equations
Ek=1/2mv²
Kinetic energy before collision =1/2 mv² + 1/2 mv² = mv² (since energy is a scalar quantity, the direction does not matter). Kindly tell why am I not getting the required answer i.e: 1/2 mv². Am I doing the calculation wrong?
IMG_20230102_154827.jpg
 
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Hi,

I agree with your calculation. An unfortunate error in the book.

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It's interesting that the book specifies the "same object" or "identical" objects, where all that is required kinematically is objects of the same mass!
 
PeroK said:
It's interesting that the book specifies the "same object" or "identical" objects, where all that is required kinematically is objects of the same mass!
Truee
 
PeroK said:
It's interesting that the book specifies the "same object" or "identical" objects, where all that is required kinematically is objects of the same mass!
I think it's economy of words. "Identical objects" is shorter than "objects of the same mass" and conveys the idea of symmetry. "Identical" becomes relatively conciser when the masses also carry equal charges.
 
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PeroK said:
It's interesting that the book specifies the "same object"
No, it says "the same two objects". Presumably the same two as in the preceding question.
 

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