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

[tleave2000]

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

KE = \frac{1}{2}mv^2

m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f}

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.

 

[Shooting Star]

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.

 

[tleave2000]

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: 0 = m_pv_2 + m_ov_osin\theta
• 2. horizontal: m_pv_1 = m_ov_ocos\theta

Kinetic Energy:

• 3. \frac{1}{2}m_pv_1^2=\frac{1}{2}m_pv_2^2+\frac{1}{2}m_ov_o^2

Eliminating v_o between equations 1 and 2:

• I rearranged equation 1 to get:
  1.1 -m_pv_2=m_ov_osin\theta
  
  
• Then divided 1.1 by 2:
  \frac{-m_pv_2}{m_pv_1}=\frac{m_ov_osin\theta}{m_ov_ocos\theta}
  
  
• which simplifies to:
  
  4. -\frac{v_2}{v_1}=tan\theta

Next to eliminate v_o between equations 2 and 3:

• From equation 2:
  2.1 v_o=\frac{m_pv_1}{m_ocos\theta}
  
  
• Substitute equation 2.1 into equation 3 for v_o:
  \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
  
  
• Simplify to get:
  5.m_pv_1^2=m_pv_2^2+\frac{m_p^2v_1^2}{m_ocos^2\theta}

Then eliminate v_2 or \theta between equations 4 and 5. I'll try to eliminate v_2:

• From equation 4:
  4.1 v_2 = -v_1tan\theta
  
• Substitute equation 4.1 into equation 5 for v_2:
  m_pv_1^2=m_p\left(-v_1tan\theta\right)^2+\frac{m_p^2v_1^2}{m_ocos^2\theta}
  
• Simplifies to:
  m_pv_1^2=m_pv_1^2tan^2\theta+\frac{m_p^2v_1^2}{m_ocos^2\theta}

Yikes. To me this looks odd but hoping I've done it right so far, I'll divide through by m_pv_1^2 giving: (continued in next post, sorry for length)

 

[tleave2000]

1=tan^2\theta+\frac{m_p}{m_ocos^2\theta}

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

And you might have guessed that I really don't know what to do with this equation. I know I have to isolate \theta, 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.

 

[Shooting Star]

The algebra is OK, but very long.

Use cos^2\theta-sin^2\theta=cos2\theta 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.

 

[tleave2000]

Thanks a lot, that's great.
\theta=43.2^\circ
v_o=2.56*10^6ms^{-1}
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.