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Approaching a nucleus, physics question

I have been having issues with this question https://isaacphysics.org/questions/closest_approach_nucleus?board=9dbf7646-23a2-4670-9f5a-1d475c0e5f97&stage=all
To start with, I conserved momentum, however, I am not sure if there is a flaw in this my working for this part is: 4m*v_0=201m+v' and from there you get the speed after the collision, although I am pretty sure it is incorrect as somehow I should take into account the closest approach, as it will not be when all the kinetic energy has transformed into potential, as both particles are moving. Then, I calculated the difference in KE for the alpha particle, although maybe I should do it like: KEafter(uranium)+KEafter(alpha)-KE before(alpha). And then equate that to the potential energy and get the radius? I would be thankful for any help.
Your working out for momentum is perfecto and your speed after 'collision' should be correct (I can let you know if you post it)

Ill clarify what energy is being conserved

initially theres just KE from the alpha particle
then theres KE from both the alpha particle AND the nucleus PLUS the electric potential energy.

your current plan is also perfecto (although not sure why you said uranium instead of gold nucleus) - so yeah pretty much equate the electric potential energy and use the equations given in the hint to find 'r'.
hope that helps.
If you have any problems post your working and ill check what went wrong
Original post by mosaurlodon
Your working out for momentum is perfecto and your speed after 'collision' should be correct (I can let you know if you post it)
Ill clarify what energy is being conserved
initially theres just KE from the alpha particle
then theres KE from both the alpha particle AND the nucleus PLUS the electric potential energy.
your current plan is also perfecto (although not sure why you said uranium instead of gold nucleus) - so yeah pretty much equate the electric potential energy and use the equations given in the hint to find 'r'.
hope that helps.
If you have any problems post your working and ill check what went wrong

Ok let me follow up your advice and I will tell you how it went
Original post by mosaurlodon
Your working out for momentum is perfecto and your speed after 'collision' should be correct (I can let you know if you post it)
Ill clarify what energy is being conserved
initially theres just KE from the alpha particle
then theres KE from both the alpha particle AND the nucleus PLUS the electric potential energy.
your current plan is also perfecto (although not sure why you said uranium instead of gold nucleus) - so yeah pretty much equate the electric potential energy and use the equations given in the hint to find 'r'.
hope that helps.
If you have any problems post your working and ill check what went wrong

so I have calculated the initial velocity to be:69651m*s^-1. Then, I have the conservation of energy equation that is, 1/2*(4m)*v´^2+1/2*(197m)*v´^2+(2*79)*(q^2)/(4*pi*epsilon_0*r)=1/2*(4m)*v_0^2. 4m is the mass of the alpha particle. I get the distance of closest approach to be r=8.7157*10^-13m. Is there is a flaw in this method please let me know.
Original post by mosaurlodon
Your working out for momentum is perfecto and your speed after 'collision' should be correct (I can let you know if you post it)
Ill clarify what energy is being conserved
initially theres just KE from the alpha particle
then theres KE from both the alpha particle AND the nucleus PLUS the electric potential energy.
your current plan is also perfecto (although not sure why you said uranium instead of gold nucleus) - so yeah pretty much equate the electric potential energy and use the equations given in the hint to find 'r'.
hope that helps.
If you have any problems post your working and ill check what went wrong
Thanks for the help, I finally got it!! I found the change in KE and then equated that to the potential. It makes sense, as that energy lost must be the energy used in potential.

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