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

  • #1
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?
 

Answers and Replies

  • #2
haruspex
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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?
 
  • #3
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
 
  • #4
haruspex
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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.
 
  • #5
Yeah it is
Thanks for your help !
 

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