Inelastic collisions with constant momentum

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The discussion centers on the calculation of kinetic energy in inelastic collisions, specifically questioning why the expected answer of 1/2 mv² is not achieved. Participants agree that the book's phrasing about "identical objects" may be misleading, as only objects of the same mass are necessary for the calculations. The conversation highlights the importance of clarity in terminology, suggesting that "identical objects" could imply additional conditions like equal charges. The confusion stems from the interpretation of the problem, particularly regarding the reference to "the same two objects" from a previous question. Overall, the dialogue emphasizes the need for precise language in physics texts to avoid misunderstandings.
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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?
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Hi,

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

##\ ##
 
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.
 
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.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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