Conservation of Energy question :cry:

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SUMMARY

The discussion revolves around calculating the closest approach of a proton to an oxygen nucleus, given its initial velocity of 3.2E4 m/s and a distance of 2 nm from the nucleus. The correct closest approach is determined to be 0.57 nm, although the user initially calculated 1.04 nm due to an error in copying the velocity. The relevant equations used include kinetic energy (Ek = 0.5 mv^2) and potential energy (U = (kqQ)/(r)). The user ultimately identified a mistake in their notes that led to the incorrect calculation.

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Homework Statement


A proton 2nm away from an oxygen nucleus is launched towards the nucleus with a velocity of 3.2E4 m/s. What is the closest approach?


Homework Equations


# of protons in oxy = 8
Ek =.5 mv^2
U = (kqQ)/(r)


The Attempt at a Solution


I know the answer to this question is 0.57 nm, and I also got this answer when I did this question very first time (that was like last week) however, now whenever I try this question I get 1.04 nm.

I am using this equation
change in Ek = work

-0.5mv^2 = - (9E9 x 1.6E-19^2 x 8) (1/Rf - 1/Ri)

0.5(1.67E-27)(3.2E4)^2/(9E9 x 1.6E-19^2 x 8) + 1/2nm = 1/Rf
Rf = 1.04E-9


I know I am making some mistake, but just cannot figure it out:cry:
 
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[tex]E_{k} = qV = q\frac{KQ}{r}[/tex]
find r
q = proton (p) (move)
Q = charge against proton (8p)

in work or energy calculation not use negative
 
Last edited:
after spending like more than 2 hours, I figured out that when I was copying this question from the black board, I copied the value of v wrong, and I didn't make correction in my notes though wrote the correct answer.
So, that screwed up everthing.

Anyways, thanks a lot.
 

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