- #1

aamirza

- 1

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## Homework Statement

Consider the following head-on elastic collision. Particle 1 has rest mass 2m

_{o}, and particle 2 has rest mass m

_{o}. Before the collision, particle 1 movies toward particle 2, which is initially at rest, with speed u (= 0.600c ). After the collision each particle moves in the forward direction with speeds of u

_{1}and u

_{2}, respectively.

a) Apply the laws of conservation of total energy (or, equivalently, of relativistic mass) and of relativistic momentum to this collision and solve the resulting equation to find u

_{1}and u

_{2}(the resulting speeds of the two particles).

## Homework Equations

Well, I know the mass is conserved in elastic collisions.

##m_1\gamma_1 + m_2\gamma_2 = m_1\gamma_{1}^{'} + m_2\gamma_{2}^{'} ##

where gamma is the Lorrentz factor. I also know energy is conserved, (E

_{i}= E

_{f}, where E = M

_{o}c

^{2}+ (M - M

_{o})c

^{2}where M

_{o}is the rest mass and M is the relativistic mass), but that basically reduces to the same thing as the mass equation.

I also know the equation for the conservation of momentum,

##p_1 + p_2 = p_{1}^{'} + p_{2}^{'} ##

where

##p = M_o\gamma u ##

M

_{o}is the rest mass and u is the speed.

## The Attempt at a Solution

So first I plugged in the values of the masses and speeds into the mass-conservation equation and got

##2m_o\gamma_1 + m_o\gamma_2 = 2m_o\gamma_1^{'} + m_o\gamma_2^{'} ##

##3.5 - 2\gamma_1^{'} = \gamma_2^{'} ##

but when I got around to plugging the numbers into the momentum equation, I got

## 2m_o\gamma_1u_1 = 2m_o\gamma_{1}^{'}u_{1}^{'} + m_o\gamma_{2}^{'}u_{2}^{'} ##

## 2\gamma_1u_1 = 2\gamma_{1}^{'}u_{1}^{'} + \gamma_{2}^{'}u_{2}^{'} ##

The problem arises when, no matter what I substitute in for any of the Lorrentz factors (gammas), I always get two unkonws in the equation, u

_{1}prime and u

_{2}prime. I can't think of a way to isolate just for one. Any help?