Aproton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationary nitrogen nucleus of mass 14m. find the velocities of the proton and the nitrogen nucleus after the collision.

Explanation:

using the law of the conservation of the linear momentum:

where is the inicial momemtum and is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

where m is the mass of the proton and is the velocity of the proton after the collision, is the mass of the nucleus and is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = m + 14m

then, m is cancelated and we have:

270 = +

This is a elastic collision, so the kinetic energy K is conservated. Then:

and

Kf =

then,

=  

here we can cancel the m and get:

=  

now, we have two equations and two incognites:

270 = +  (eq. 1)

=  

in the second equation, we have:

36450 =    (eq. 2)

from this last equation we solve for as:

and replace in the other equation as:

270 = 14

so,

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get: