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## Homework Statement

a) State the law of conservation of linear momentum.

b) A proton of mass 1.6*10

^{-27}kg traveling with a velocity of 3*10

^{7}ms

^{-1}collides with a nucleus of an oxygen atom of mass 2.56*10

^{-26}kg (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=1/2mv

^{2}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.