# Proton collision

1. Sep 24, 2011

### iscofield

A proton travelling at 3*10^7 m/s collides with the nucleus of a stationary oxygen atom and rebounds in a direction at 90 degree to its original path, calculate the velocity and the direction of the oxygen nucleus assuming the that the collision was perfectly elastic and ignoring relativistic effects .
Mass of proton =1.6*10^-27kg and mass of oxygen atom =2.56*10^-26kg . Thnks to you in advance.

2. Sep 24, 2011

### danielakkerma

Re: collision

Hi there!
Welcome to the forum...
Surely, you've heard of the conservation of momentum & kinetic energy; Have you tried applying those principles here? It makes for very, easily solving in this case...
Daniel

3. Sep 24, 2011

### iscofield

Re: collision

can u pls try it out and explain how

4. Sep 24, 2011

### danielakkerma

Re: collision

Of course,
Look, first we need to solve any such problem analytically, by taking the mass of the proton as m1 and its speed as v1, the oxygen nucleus at m2 and v2 as its velocity after the collison.
Let's assume they were both on the x axes.
Therefore:
$\Large p_x = p'_x \Longrightarrow m_1v_1 = m_2v_{2,x}$
There's also conservation on the y axes:
Note that since the new velocity of the proton is now at 90 degrees to the original path, it's now travelling on the y axis.
$\Large p_y = p'_y \Longrightarrow 0 = m_1v'_1+m_2v_{2,y}$
Finally the conservation of kinetic energy:
$\Large m_1{v_1}^2 = m_1{v'_1}^2+m_2({v_{2,x}}^2+{v_{2,y}}^2).$
From the first equation you can extract v_2x, from the second, v_2y, and then find v'_1..
And that's it!!!

5. Sep 24, 2011

### danielakkerma

Re: collision

Check your answers, to see that you should get
$\large v_2 = \frac{\alpha v_1\sqrt{2}}{\sqrt{1+2\alpha}}, \alpha=\frac{m_1}{m_2};$
The direction with the horizontal is given by:
$\tan{\theta} = \frac{v_{2, y}}{v_{2, x}}$
Which is easily discovered from the former...
Give it a go!
Daniel

6. Sep 25, 2011

### iscofield

thnks alot that was really helping.