Elastic Collision of a gold nucleus

In summary: Hi there Rutherford,Your formula sheet has a weird way of writing the formula and they had m2 on there, that's why instead of having 2 in my numerator, i have 100. You guys saved me from using that formula on the exam! Thanks a lot!Thanks for letting us know!
  • #1
habibclan
55
0

Homework Statement


A positively charged alpha particle is fired, in a direct line, at a positively charged gold nucleus:

o--> O

o: alpha, O: golden nucleus.

Because of their mutual repulsion, the alpha particle does not actually hit the nuleus: it comes to rest, for a moment, some distance from the nucleus and then recedes. The alpha particle may be considered to have undergone an elastic "collision" with the golden nucleus.The golden nucleus has a mass of about 50 times the alpha particle. Compared to the gold nucleus, after the collision, the alpha particle has

A. more momentum but less kinetic energy
B. more momentum and more kinetic energy.
C. less momentum and less kinetic energy
D. less momentum but more kinetic energy
E. the same momentum but more kinetic energy.


Homework Equations



Elastic collision: object 2 initially at rest. vi is the initial velocity of the alpha particle. The post-collision velocities are:
v1= (m1 - m2)vi / (m1+ m2)
v2= (2*m2)vi / (m1+ m2)

The Attempt at a Solution



This problem can be solved in two ways: either by using logic and doing no calculations or doing the calculations based on the elastic collision formulas. Logically, I know that the initial momentum is to the right, therefore post collision the overall momentum will be to the right. So the gold nucleus will definitely have more momentum. The question now remains about the kinetic energy. Logically, one would think that even though the net momentum is to the right, the velocity of the gold nucleus < velocity of alpha, due to the huge mass difference, therefore the KE of the alpha particle would be greater. However, to back this up, I wanted to prove it mathematically but it doesn't work out that way.

Let the initial velocity of the alpha particle be vi.

Velocity of gold nucleus post-elastic-collision:
v2= (2*m2)vi / (m1+ m2)
v2= (2*50m)vi / (m+ 50m)
v2= (100m) vi/ (51m) = 100v/51

Velocity of alpha nucleus post-elastic-collision:
v1= (m1 - m2)vi / (m1+ m2)
v1= (m - 50m)vi / (m+ 50m)
v1= (-49m)vi / (51m)= -49/51 v


From this I can see that the velocity of the alpha particle is much less than the velocity of the gold nucleus, which having the greater mass as well, will have the greater momentum. How is it that the math doesn't work out but logically answer D makes sense? Thanks in advance!
 
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  • #2
D doesn't make sense. Total kinetic energy is conserved since the collision is elastic. The alpha comes in with all of the kinetic energy and the gold nucleus with none. The gold nucleus leaves with some kinetic energy so that of the alpha MUST decrease. Momentum is a little trickier since it has a sign. But still, can you rethink this?
 
Last edited:
  • #3
This question is from a past midterm and the answer given in the answer key is D. I've thought of it in terms of momentum and since velocity is squared in the KE expression, larger velocity of the alpha particle gives it a greater KE. Besides this, I don't know what to conclude from this problem. Logic makes sense to me, but the math isn't working out, when technically it should be. :S I'm confused!
 
  • #4
habibclan said:
Compared to the gold nucleus, after the collision, the alpha particle has

D. less momentum but more kinetic energy

I think you're misreading the question.

D says that momentum(alpha) < momentum(gold) and KE(alpha) > KE(gold). :wink:
 
  • #5
hehe. i do get that. the only discrepency is that the formulas give a greater velocity for the gold nucleus. Therefore, making it have a higher KE. Or should I not use the formula over here and it's not a REAL elastic collision.
 
  • #6
Hi habibclan,

The equation you have for v2 in your relevant equations section is not correct. Do you see what it needs to be?
 
  • #7
tiny-tim said:
I think you're misreading the question.

D says that momentum(alpha) < momentum(gold) and KE(alpha) > KE(gold). :wink:

I was definitely reading it wrong. And that's with absolute values on the momentum. habibclan, check your math. I'm getting v alpha=(-49/51)vi (just as you did) but for v gold I get (2/51)vi. Not (100/51)vi.
 
  • #8
Dick said:
I was definitely reading it wrong. And that's with absolute values on the momentum. habibclan, check your math. I'm getting v alpha=(-49/51)vi (just as you did) but for v gold I get (2/51)vi. Not (100/51)vi.

My formula sheet has a weird way of writing the formula and they had m2 on there, that's why instead of having 2 in my numerator, i have 100. You guys saved me from using that formula on the exam! Thanks a lot!

[Now my logical answer matches the math too! Yay!]
 
  • #9

Related to Elastic Collision of a gold nucleus

1. What is an elastic collision?

An elastic collision is a type of collision in which the total kinetic energy of the system is conserved. This means that the sum of the kinetic energies of the objects before the collision is equal to the sum of the kinetic energies after the collision.

2. How does a gold nucleus participate in an elastic collision?

A gold nucleus is a subatomic particle found in the center of an atom. In an elastic collision, the gold nucleus would act as one of the objects involved in the collision, either colliding with another gold nucleus or with another object.

3. What factors affect the outcome of an elastic collision involving a gold nucleus?

The outcome of an elastic collision involving a gold nucleus can be affected by factors such as the mass and velocity of the gold nucleus, as well as the mass and velocity of the object it is colliding with.

4. How does the conservation of momentum apply to an elastic collision of a gold nucleus?

The conservation of momentum states that the total momentum of a system before a collision is equal to the total momentum after the collision. In the case of an elastic collision of a gold nucleus, the momentum of the gold nucleus before the collision will be equal to the momentum of the gold nucleus after the collision.

5. What are some real-world applications of elastic collisions involving gold nuclei?

Elastic collisions involving gold nuclei can be observed in particle accelerators, where gold nuclei are accelerated to high speeds and then collide with other particles. This can help scientists study the properties and behavior of subatomic particles. Additionally, elastic collisions of gold nuclei can also be seen in natural phenomena, such as the fusion reactions that occur in stars.

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