# Addition of masses in relativistic collision

#### Toby_phys

Really basic question:

a particle, moving at speed u (u is fast enough for relativistic effects) with rest mass m0 collides with a stationary particle with rest mass m0. They coalesce to form a new particle of mass M (observer fame, not rest mass M) and move at speed v. find v in terms of y (y is gamma factor).

Conservation of momentum:

m0y(u)u=M0y(v)v

Is M(the rest mass of the 2 particles): 2m0 (just adding the rest masses and multiplying by the new gamma factor)

or

m0y(u) +m0 (adding up the masses as they were before the collision)

Thank you

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#### PeroK

Homework Helper
Gold Member
2018 Award
Really basic question:

a particle, moving at speed u (u is fast enough for relativistic effects) with rest mass m0 collides with a stationary particle with rest mass m0. They coalesce to form a new particle of mass M (observer fame, not rest mass M) and move at speed v. find v in terms of y (y is gamma factor).

Conservation of momentum:

m0y(u)u=M0y(v)v

Is M(the rest mass of the 2 particles): 2m0 (just adding the rest masses and multiplying by the new gamma factor)

or

m0y(u) +m0 (adding up the masses as they were before the collision)

Thank you
Where are you learning SR? You don't see to have much idea about the basic concepts.

#### Ray Vickson

Homework Helper
Dearly Missed
Really basic question:

a particle, moving at speed u (u is fast enough for relativistic effects) with rest mass m0 collides with a stationary particle with rest mass m0. They coalesce to form a new particle of mass M (observer fame, not rest mass M) and move at speed v. find v in terms of y (y is gamma factor).

Conservation of momentum:

m0y(u)u=M0y(v)v

Is M(the rest mass of the 2 particles): 2m0 (just adding the rest masses and multiplying by the new gamma factor)

or

m0y(u) +m0 (adding up the masses as they were before the collision)

Thank you
These problems are always easier and less confusing if you analyze them in the center-of-momentum (CM) frame.

In a frame whose origin moves with velocity $(v,0)$ you can work out the initial (pre-collision) momenta of the two particles, and equate the total momentum to 0; that tells you the value of $v$ that gives you the CM frame. So, in the CM frame the total momentum before and after the collision is zero. Conservation of energy says that the post-collision mass $M_0$ of the coalesced particle is given by $M_0 c^2 = m_0 c^2 \gamma(u_1) + m_0 c^2 \gamma(-v)$, where $u_1$ and $-v$ are the pre-collision velocities of particles 1 and 2 in the CM frame. That is, $M_0 = m_0 \gamma(u_1) + m_0 \gamma(-v).$

After you have determined $M_0$ you can transform back to the lab frame to find the final momentum of the new particle.

"Addition of masses in relativistic collision"

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