Conservation of energy - elastic potential energy and k.e.

AI Thread Summary
The discussion revolves around the application of the conservation of energy principle in a physics problem involving a 2 kg particle and elastic potential energy (E.P.E). A participant questions why the solution provided does not account for the E.P.E of the string after the collision, suggesting that tension should still be present. Another participant agrees with this concern and prompts a comparison of the participant's answer with the book's answer to identify any misunderstandings. The conversation also touches on the formatting of the solution, indicating a discrepancy between the problem statement and the solution's presentation. The thread emphasizes the importance of accurately accounting for all forms of energy in such problems.
Janiceleong26
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1. Homework Statement
image.jpg

For part (iii) , I used the principle of conservation of energy,
K.E of the 2 kg particle after collision + E.P.E = K.E of the 2 kg particle at the furthest distance away from A + E.PE,

But the solution for this question did not include the E.P.E of the string after collision (in bold), why? Shouldn't there still be tension in the string?
 
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Janiceleong26 said:
1. Homework Statement
View attachment 97599
For part (iii) , I used the principle of conservation of energy,
K.E of the 2 kg particle after collision + E.P.E = K.E of the 2 kg particle at the furthest distance away from A + E.PE,

But the solution for this question did not include the E.P.E of the string after collision (in bold), why? Shouldn't there still be tension in the string?
Based on the information you have provided, I would agree with you. But this leaves the possibility that you have misunderstood something in the solution provided and therefore not portrayed it accurately.
As a way of checking, what answer did you get and what answer does the book get?
 
haruspex said:
Based on the information you have provided, I would agree with you. But this leaves the possibility that you have misunderstood something in the solution provided and therefore not portrayed it accurately.
As a way of checking, what answer did you get and what answer does the book get?
image.jpg


I did this :
1/2 (2)(0)2 +1/2 (72) (x)2 = 1/2 (2) (4)2 + 1/2 (72)(1)2
 
Janiceleong26 said:
I did this :
1/2 (2)(0)2 +1/2 (72) (x)2 = 1/2 (2) (4)2 + 1/2 (72)(1)2
I agree with your equation.
I note that the wrong solution is monospaced type, as from an old typewriter, whereas the problem statement is a page from a book.
 
Yeah it is
Thanks for your help !
 
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