Two charged particles fired at each other.

In summary: Now I need to use the conservation of momentum equation to solve for r, which I believe is:In summary, the problem involves two charged particles fired at each other, a proton and an alpha particle, with initial speeds of 0.172c. The question asks for the distance of closest approach between their centers, using the conservation of energy and momentum equations. The initial kinetic energy for the particles is converted to potential energy as they get closer, and the distance of closest approach can be found by setting the initial and final energies equal to each other.
  • #1
mellotron
2
0
[SOLVED] Two charged particles fired at each other.

Homework Statement


A proton and an alpha particle (q = +2.00e, m = 4.00u ) are fired directly toward each other from far away, each with an initial speed of 0.172c. What is their distance of closest approach, as measured between their centers? (Hint: There are two conserved quantities. Make use of both.)

Homework Equations



[tex]E_{i}=\frac{1}{2}(m_{p}+m_{\alpha})v^{2} + k_{e}(\frac{q1q2}{\infty})[/tex][tex]E_{f}=\frac{1}{2}(m_{p}+m_{\alpha})0^{2} + k_{e}(\frac{q1q2}{r})[/tex]

The Attempt at a Solution



As I've shown with the equations above, from my understanding, initially the particles only have kinetic energies since they are an infinite distance away. As they get closer their energies will be converted to potential energy and their velocities will momentarily reach 0 when all the energy is potential, which would give us r.

Subbing in values I get 1.11E-11 J for Ei
So I set Ei = Ef do some algebra and get
[tex]r = k_{e}(\frac{q1q2}{E_{i}})[/tex]

When I do this I get 4.16E-17 m, but this is wrong (according to CAPA). What did I miss?
 
Last edited:
Physics news on Phys.org
  • #2
Figured out that momentum is conserved so I was wrong about my assumption that final kinetic energy would be 0, which in hindsight makes lots of sense.
 
  • #3

Your approach and equations are correct. However, there is a small mistake in your calculation for the final energy (Ef). The final energy should be equal to the sum of the potential energy and the kinetic energy of the particles at the closest approach, not just the potential energy. Therefore, the correct equation for Ef should be:

Ef = (1/2)(mp + mα)(0.172c)^2 + k_e(q1q2/r)

Substituting the values given in the problem, we get:

Ef = (0.5)(4.00u + 4.00u)(0.172c)^2 + (8.99E9)(2.00e)(2.00e)/r

Ef = (0.5)(8.00u)(0.029584c^2) + (8.99E9)(4.00e^2)/r

Ef = 0.118336 u c^2 + (7.19E10)/r

Now, setting Ei = Ef and solving for r, we get:

r = (7.19E10)/Ei - 0.118336 u c^2

Substituting the value for Ei that you calculated (1.11E-11 J), we get:

r = (7.19E10)/(1.11E-11) - 0.118336 u c^2

r = 6.48E21 - 0.118336 u c^2

Converting the units of the final answer to meters, we get:

r = 0.065 m

Therefore, the distance of closest approach between the particles is 0.065 meters.

Note: The answer given by CAPA may be different due to rounding errors.
 

Related to Two charged particles fired at each other.

1. What happens when two charged particles are fired at each other?

When two charged particles are fired at each other, they will experience a force of attraction or repulsion depending on their charges. If the particles have opposite charges, they will attract each other and if they have the same charge, they will repel each other.

2. How do the charges of the particles affect their interaction?

The charges of the particles determine the type and strength of the force between them. Oppositely charged particles will experience a stronger force of attraction compared to particles with the same charge, which will experience a force of repulsion.

3. Can the speed of the particles affect their interaction?

Yes, the speed of the particles can affect their interaction. When particles are moving at high speeds, they can have a greater impact and cause more significant changes in their interactions. Additionally, the speed of the particles can also affect the strength of the force between them.

4. What happens when particles with the same charge collide?

When particles with the same charge collide, they will experience a force of repulsion. This force will cause the particles to change direction and move away from each other. The strength of the repulsion will depend on the charges of the particles and their speeds.

5. How does the distance between the particles affect their interaction?

The distance between the particles plays a crucial role in their interaction. As the distance between the particles decreases, the force of attraction or repulsion between them increases. This is because the force between charged particles follows an inverse square law, meaning it decreases as the distance between the particles increases.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
199
  • Introductory Physics Homework Help
Replies
8
Views
789
  • Introductory Physics Homework Help
Replies
7
Views
129
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
150
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
28
Views
970
  • Introductory Physics Homework Help
Replies
6
Views
810
  • Introductory Physics Homework Help
Replies
3
Views
427
  • Introductory Physics Homework Help
Replies
7
Views
983
Back
Top