- #1
Eve Torchic
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- 0
Two chunks of rock, each having a mass of 1.00 kg, collide in space. Just before the collision, an observer at rest in the reference frame of a nearby star determines that rock A is moving toward the star at 0.800 c and rock B is moving away from the star at 0.750 c .
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:
PAB = PA+PB
ϒABmABvAB = ϒAmAvA + ϒBmBvB
Energy:
ETotal = EA + EB
ϒABmABc^2 = ϒAmAc^2 + ϒBmBc^2
divide c^2
ϒABmAB = ϒAmA + ϒBmB
When dividing momentum by energy, the mass and Lorentz factors cancel to leave the velocity of AB
vAB = (ϒAmAvA + ϒBmBvB) / (ϒAmA + ϒBmB)
substituting the numbers and
mA = 1 kg
mB = 1 kg
vA = 0.800c
vB = 0.750c
I got
vAB = 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.
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:
PAB = PA+PB
ϒABmABvAB = ϒAmAvA + ϒBmBvB
Energy:
ETotal = EA + EB
ϒABmABc^2 = ϒAmAc^2 + ϒBmBc^2
divide c^2
ϒABmAB = ϒAmA + ϒBmB
When dividing momentum by energy, the mass and Lorentz factors cancel to leave the velocity of AB
vAB = (ϒAmAvA + ϒBmBvB) / (ϒAmA + ϒBmB)
substituting the numbers and
mA = 1 kg
mB = 1 kg
vA = 0.800c
vB = 0.750c
I got
vAB = 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.