Closest approach of alpha particle - two separate methods

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Two methods for calculating the closest approach of an alpha particle yield different results due to assumptions about the larger nucleus's movement. Method 1 assumes the nucleus is stationary, while Method 2 accounts for its motion, using conservation of linear momentum and energy. The discussion highlights that if the nucleus is significantly larger, Method 1 may suffice, but Method 2 is necessary otherwise. Practicing various problems helps determine when assumptions are justified, with a rule of thumb suggesting that if the energy gained by the larger mass is less than 10%, Method 1 can be used for quick calculations. Understanding these methods is crucial for solving common exam questions effectively.
etotheipi
Homework Statement
Find the distance of closest approach of an alpha particle and nucleus (of arbitrary atomic number Z)
Relevant Equations
##E_{k}##, relative velocities
I've found two methods for doing this problem and they give different answers.

Method 1: Assume the larger nucleus does not move, and simply equate energies before the collision and at the point of closest approach:$$\frac{1}{2} m v^{2} = \frac{qQ}{4\pi\epsilon_{0}r}$$

Method 2: Assume the larger nucleus is capable of moving; the distance of closest approach occurs when the relative velocity of the two particles is zero. From this, we can use CLM and then do another energy calculation:

Conservation of Linear Momentum:
$$m_{\alpha}v_{\alpha} = (m_{\alpha} + M_{nucleus})v_{common}$$Energy:
$$\frac{1}{2} m_{\alpha} v_{\alpha}^{2} = \frac{1}{2} (m_{\alpha} + M_{nucleus})v_{common}^{2} + \frac{qQ}{4\pi\epsilon_{0}r}$$It makes sense that the two methods give different answers for ##r##, however this is a very common exam question, and it is never usually stated whether we permit the larger nucleus to move or not. So, I was wondering which approach I should use for this and similar problems. Thank you!
 
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Unless you can assume the nucleus is large enough that its movement is negligible you must use method 2. In the case where the nucleus is very large, methods 1 and 2 should give approximately the same answer.
 
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PeroK said:
Unless you can assume the nucleus is large enough that its movement is negligible you must use method 2. In the case where the nucleus is very large, methods 1 and 2 should give approximately the same answer.

Thank you, this is essentially what I was thinking. It can be quite hard though to determine whether the assumption is justified or not. For instance, in the one below, I'd assume since silver is about 30x as massive as helium we could get away with the simpler method.

I suppose the only way to get a better feel for when certain assumptions are justified is just by doing lots of practice problems.

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Here's an idea. In this collision, the energy gained by the large mass is approx ##\frac{m}{M}## of the energy of the system. Is that significant? In the above problem, this ratio is about ##4/50## = 8%.

So, do you need the answer to within 10%? If not, you can do a quick calculation by method 1.
 
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PeroK said:
Here's an idea. In this collision, the energy gained by the large mass is approx ##\frac{m}{M}## of the energy of the system. Is that significant? In the above problem, this ratio is about ##4/50## = 8%.

So, do you need the answer to within 10%? If not, you can do a quick calculation by method 1.

That's a nice rule of thumb!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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