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

1. Mar 26, 2009

### tleave2000

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
$$KE = \frac{1}{2}mv^2$$

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

3. 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.

2. Mar 26, 2009

### Shooting Star

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. Mar 27, 2009

### tleave2000

Thanks that was alot 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)

4. Mar 27, 2009

### 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.

5. Mar 27, 2009

### 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.

6. Mar 28, 2009

### tleave2000

Thanks alot, 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.