- #1

Eve Torchic

- 4

- 0

If the rocks stick together after the collision, what is their velocity (in the reference frame of the star) immediately after the collision? Consider the direction away from the star to be the positive direction.

**Attempt at solution:**

p for momentum

v for velocity

m for mass

Momentum:

P

_{AB}= P

_{A}+P

_{B}

ϒ

_{AB}m

_{AB}v

_{AB}= ϒ

_{A}m

_{A}v

_{A}+ ϒ

_{B}m

_{B}v

_{B}

Energy:

E

_{Total}= E

_{A}+ E

_{B}

ϒ

_{AB}m

_{AB}c^2 = ϒ

_{A}m

_{A}c^2 + ϒ

_{B}m

_{B}c^2

divide c^2

ϒ

_{AB}m

_{AB}= ϒ

_{A}m

_{A}+ ϒ

_{B}m

_{B}

When dividing momentum by energy, the mass and Lorentz factors cancel to leave the velocity of AB

v

_{AB}= (ϒ

_{A}m

_{A}v

_{A}+ ϒ

_{B}m

_{B}v

_{B}) / (ϒ

_{A}m

_{A}+ ϒ

_{B}m

_{B})

substituting the numbers and

m

_{A}= 1 kg

m

_{B}= 1 kg

v

_{A}= 0.800c

v

_{B}= 0.750c

I got

v

_{AB}= 0.776c

but the answer seems to be wrong. I have no idea where to start in answering this question.

Edit: I tried changing the gamma symbols to make them the same, hope it worked.