# Silly momentum problem

1. Aug 4, 2008

### ddphuyal

A particle with rest mass m and momentum mc/2 collides with a particle of the same
rest mass that is initially at rest. After the collision, the original two particles have
disappeared. Two other particles, each with rest mass m', are observed to leave the region
of the collision at equal angles of 30° with respect to the direction of the original moving
particle, as shown below.

a) What is the speed of the original moving particle?
here is what i have

1. mc/2 = m'v1 cos(30) + m'v2 cos(30) - for momentum conserved in x direction
2. 0 = m'v1 sin(30) - m'v2 sin(30) - for momentum conserved in y direction

and since this is an elastic collision, Kinetic Energy is also conserved

3. 1/2mVi^2 = 1/2m'v1^2 + 1/2 m'v2^2

from equation 2, i get v1 = v2. i plus this into equation 1, and get v2 = mc/(2*sqrt(3)*m')
thus giving me initial velocity from 3 to be

mVi^2 = m'(v1^2 + v2^2) => substitute from above for v2, and for v1, since v1 = v2
mVi^2 = m'(2m^2c^2/12m'^2)
Vi^2 = mc^2/((sqrt(6)m')....anywho, the answer for Vi in the answer book of this practice test is c/sqrt(5)...please shed some light on this. i've tried and tried, and thus i seek help. thank you

2. Aug 4, 2008

### alphysicist

Hi ddphuyal,

The overall problem with your solution is that you're not treating the problem relativisitically. The formula for the momentum components that you are using $p=mv$ does not apply here. (If it was true for this problem, since you know the initial particle has a momentum of mc/2, you could say the speed was just c/2.)

What is the formula for relativistic momentum? Once you have that, you will be able to determine the initial speed.