Inelastic collisions with constant momentum

AI Thread Summary
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.
haha0p1
Messages
46
Reaction score
9
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
 
Last edited by a moderator:
Physics news on Phys.org
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top