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# Isaac Physics closest approach problem watch

1. I've been stuck on this question for a while now, calculate distance of closest approach between an aplha particle and a gold nucleus. the alpha particle has inital velocity of 3.5x10^6 m/s.
2. (Original post by danielh12)
I've been stuck on this question for a while now, calculate distance of closest approach between an aplha particle and a gold nucleus. the alpha particle has inital velocity of 3.5x10^6 m/s.
I think the question asked

An α
-particle with velocity 3.5×106ms−1
strikes a block of gold, atomic number 79
, mass number 197
. Find the distance of closest possible approach between the α
-particle and a gold nucleus, assuming that only Coulomb's law holds over such distances. Assume that mp=mn=1.67×10−27kg
.
So there's several important things you've edited out... that's not going to help people on here understand the question.

It's asking for closest possible approach which occurs when the alpha travels straight towards the centre of the nucleus and is then repelled straight back out down the same path it came in on.
The point of closest approach occurs when the initial KE has entirely been converted to Electric Potential Energy - paths that deflect by other angles don't approach the nucleus as closely, less of the original KE is converted to EPE.
3. I got 8.9x10^-13 and it said its wrong and to consider momentum and energy conservation
4. (Original post by danielh12)
I got 8.9x10^-13 and it said its wrong and to consider momentum and energy conservation
Joinedup had already "elaborated" the conceptual understanding of the problem, but it seems that you did not pay attention to the explanation.

You cannot use conservation of momentum. You should think about it why.

You can post what you had done to let others know what you had done wrong, so that they can advise accordingly.
5. (Original post by danielh12)
I got 8.9x10^-13 and it said its wrong and to consider momentum and energy conservation

I seem to know what is the problem.

Your answer 8.9 × 10−13 m suggests that you had considered the conservation of energy only by assuming that the gold nucleus is stationary. So the momentum of the system is NOT conserved.

If you really apply both conservation of energy and momentum, you would get 9.1 × 10−13 m.
At the distance of closest approach, both alpha particle and gold nucleus would move at the same velocity.
Apply this to both the conservation of momentum and energy would give 9.1 × 10−13 m.
6. Thanks, I understand were I went wrong thanks for your help

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