# Relativistic momentum

## Homework Statement

A particle with mass m has speed 0.802c relative to inertial frame S. The particle collides with an identical particle at rest relative to frame S. Relative to S and in terms of c, what is the speed of a frame S' in which the total momentum of these particles is zero? This frame is called the centre of momentum frame.

## The Attempt at a Solution

If particle 2 is at rest in the S frame, then in the S' frame particle 2 must have the same velocity as the S' frame relative to the S frame, but I'm not sure how to apply this.

## Answers and Replies

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hilbert2
Science Advisor
Gold Member
Do you know how to form the four-momentum vector for a massive particle in one spatial dimension, and how to write Lorentz transformations as matrices? In this case the vector has effectively only two components and the transformations are 2x2 matrices. You will find help here: http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html

You just have to form an equation for the value of $\gamma$ in the Lorentz transformation that makes the momenta equal in magnitude but opposite in sign.

PeroK
Science Advisor
Homework Helper
Gold Member

## Homework Statement

A particle with mass m has speed 0.802c relative to inertial frame S. The particle collides with an identical particle at rest relative to frame S. Relative to S and in terms of c, what is the speed of a frame S' in which the total momentum of these particles is zero? This frame is called the centre of momentum frame.

## The Attempt at a Solution

If particle 2 is at rest in the S frame, then in the S' frame particle 2 must have the same velocity as the S' frame relative to the S frame, but I'm not sure how to apply this.
Do you know how to transform velocities from one frame to another?

@PeroK are you referring to using the equation
v = (u + v')/(1 + uv'/c2)

PeroK
Science Advisor
Homework Helper
Gold Member
@PeroK are you referring to using the equation
v = (u + v')/(1 + uv'/c2)
Yes, you can use that equation. I assume you haven't covered energy-momentum transformations yet? If not, then the above formula is the way to go.