If a proton rebounds at 90 degrees Momentum and elastic collision problem.

In summary: I think I have to write out the first few steps of things like this to get the hang of it. And I have to get familiar with more identities. I think I'll go read that page of them again, and I'll have a look for a list of them to print out.In summary, the conversation discusses the law of conservation of linear momentum and the calculation of the velocity and direction of motion of a recoil oxygen nucleus after colliding with a proton. The conversation uses equations for conservation of momentum and kinetic energy to solve for the unknowns of velocity and direction. The final solutions are a direction of 43.2 degrees and a velocity of 2.56*10^6 ms^-1 for the oxygen nucleus.
  • #1
tleave2000
8
0

Homework Statement


a) State the law of conservation of linear momentum.

b) A proton of mass 1.6*10-27kg traveling with a velocity of 3*107ms-1 collides with a nucleus of an oxygen atom of mass 2.56*10-26kg (which may be assumed to be at rest initially) and rebounds in a direction at 90 degrees to its incident path. Calculate the velocity and direction of motion of the recoil oxygen nucleus, assuming the collision is elastic and neglecting the relativistic increase of mass.

Homework Equations


[tex]KE = \frac{1}{2}mv^2[/tex]

[tex]m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f}[/tex]

The Attempt at a Solution


a) The total linear momentum of a closed system is a constant.

b) At first I tried assuming that the tangent at the point of collision was at 45 degrees. This kinda works out but is a little off. I think that's because a 45 degree tangent would cause the proton to rebound at an angle of 90 degrees if the proton had an infinitely smaller mass than the nucleus. Because the proton has a small mass compared to the nucleus, 45 degrees gives a rough approximation. But it's not the answer.
Next I had a look around the net and got the impression that I might be able to use a simultaneous equation using conservation of momentum and conservation of kinetic energy. The fact they specify that you can assume it's an elastic collision (ie kinetic energy is preserved) also suggests that KE=1/2mv2 comes into it. I feel like I kinda have some of the right ideas now, but I really don't know how to put them all together. Any suggestions would be very much appreciated.
 
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  • #2
tleave2000 said:
Next I had a look around the net and got the impression that I might be able to use a simultaneous equation using conservation of momentum and conservation of kinetic energy. The fact they specify that you can assume it's an elastic collision (ie kinetic energy is preserved) also suggests that KE=1/2mv2 comes into it. I feel like I kinda have some of the right ideas now, but I really don't know how to put them all together. Any suggestions would be very much appreciated.

You have the right idea.

Let the speed of the proton be v1 and v2 before and after collision, Vo the speed of the oxygen nucleus after collision, and θ the direction of its motion with respect to the x-axis which is the direction of the initial velocity of the proton. Using conservation of linear momentum along the x and y axes, you get two equations. Using conservation of kinetic energy, you get another equation.

So there are three unknowns, v2, Vo and θ, and there are three equations.
 
  • #3
Thanks that was a lot of help. I drew this diagram.
https://www.physicsforums.com/attachment.php?attachmentid=18186&d=1238173851
Momentum first:
  • 1. vertical: [tex]0 = m_pv_2 + m_ov_osin\theta[/tex]
  • 2. horizontal: [tex]m_pv_1 = m_ov_ocos\theta[/tex]
Kinetic Energy:
  • 3. [tex]\frac{1}{2}m_pv_1^2=\frac{1}{2}m_pv_2^2+\frac{1}{2}m_ov_o^2[/tex]
Eliminating [tex]v_o[/tex] between equations 1 and 2:
  • I rearranged equation 1 to get:
    1.1 [tex]-m_pv_2=m_ov_osin\theta[/tex]

  • Then divided 1.1 by 2:
    [tex]\frac{-m_pv_2}{m_pv_1}=\frac{m_ov_osin\theta}{m_ov_ocos\theta}[/tex]

  • which simplifies to:

    4. [tex]-\frac{v_2}{v_1}=tan\theta[/tex]
Next to eliminate [tex]v_o[/tex] between equations 2 and 3:

  • From equation 2:
    2.1 [tex]v_o=\frac{m_pv_1}{m_ocos\theta}[/tex]

  • Substitute equation 2.1 into equation 3 for [tex]v_o[/tex]:
    [tex]\frac{1}{2}m_pv_1^2=\frac{1}{2}m_pv_2^2+\frac{1}{2}m_o\left(\frac{m_pv_1}{m_ocos\theta}\right)^2[/tex]

  • Simplify to get:
    5.[tex] m_pv_1^2=m_pv_2^2+\frac{m_p^2v_1^2}{m_ocos^2\theta}[/tex]

Then eliminate [tex]v_2[/tex] or [tex]\theta[/tex] between equations 4 and 5. I'll try to eliminate [tex]v_2[/tex]:
  • From equation 4:
    4.1 [tex]v_2 = -v_1tan\theta[/tex]
  • Substitute equation 4.1 into equation 5 for [tex]v_2[/tex]:
    [tex] m_pv_1^2=m_p\left(-v_1tan\theta\right)^2+\frac{m_p^2v_1^2}{m_ocos^2\theta}[/tex]
  • Simplifies to:
    [tex] m_pv_1^2=m_pv_1^2tan^2\theta+\frac{m_p^2v_1^2}{m_ocos^2\theta}[/tex]

Yikes. To me this looks odd but hoping I've done it right so far, I'll divide through by [tex]m_pv_1^2[/tex] giving: (continued in next post, sorry for length)
 
  • #4
[tex]1=tan^2\theta+\frac{m_p}{m_ocos^2\theta}[/tex]

Could multiply through by [tex]cos^2\theta[/tex] to get:
[tex]cos^2\theta=cos^2\theta tan^2\theta+\frac{m_p}{m_o}[/tex]
That's equivalent to:
[tex]cos^2\theta=sin^2\theta+\frac{m_p}{m_o}[/tex]

And you might have guessed that I really don't know what to do with this equation. I know I have to isolate [tex]\theta[/tex], but I don't know how. Is this a case to use some trigonometric identity or other? I'm not too well up on those. Equally I might have gotten some algebra wrong somewhere(s) along the way, I checked it all but still. Thanks again for the help so far, and again any help with this bit, or pointing out any mistakes I've made so far would be very much appreciated.
 
  • #5
The algebra is OK, but very long.

Use [itex]cos^2\theta-sin^2\theta=cos2\theta[/itex] to get θ. You could have got this from eqns 1 and 2 directly.

Eliminate θ by squaring and adding eqns 1 and 2. Use the result to eliminate v2 from eqn 3. You get v0.
 
  • #6
Thanks alot, that's great.
[tex]\theta=43.2^\circ[/tex]
[tex]v_o=2.56*10^6ms^{-1}[/tex]
I see why your way would have been quicker and how I could have spotted it by looking more carefully at the unknowns in the original equations.
 

1. What is a proton rebound at 90 degrees?

A proton rebound at 90 degrees refers to the situation where a proton undergoes a collision with another particle or object and bounces off at a 90 degree angle from its original direction of motion.

2. What is momentum in a collision?

Momentum in a collision is a measure of the quantity of motion of an object. It is defined as the product of an object's mass and velocity. In other words, it is the amount of force that an object possesses when it is in motion.

3. What is an elastic collision?

An elastic collision is a type of collision where the total kinetic energy of the objects involved in the collision remains constant. This means that no energy is lost or converted into other forms, such as heat or sound, during the collision.

4. How is momentum conserved in an elastic collision?

In an elastic collision, momentum is conserved because the total momentum of the objects before the collision is equal to the total momentum after the collision. This means that the combined momentum of the objects involved in the collision remains the same.

5. What factors affect the rebound angle of a proton in an elastic collision?

The rebound angle of a proton in an elastic collision is affected by the mass and velocity of the proton, as well as the mass and velocity of the object it collides with. The angle of impact and the properties of the surface also play a role in determining the rebound angle.

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