# Homework Help: Relativistic Collision - Momentum and Energy

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1. Mar 31, 2017

### Eve Torchic

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.

2. Mar 31, 2017

### Orodruin

Staff Emeritus
The masses are moving in opposite directions.

3. Mar 31, 2017

### Eve Torchic

oh, but is the method correct?

edit: if velocity A is negative the answer would be -0.0627c and its still the wrong answer so there seems to be a problem with my method

Last edited: Mar 31, 2017
4. Mar 31, 2017

### vela

Staff Emeritus

5. Mar 31, 2017

### Eve Torchic

The homework site I'm using says I got the answer wrong.

6. Mar 31, 2017

### robphy

If you got the answer "wrong", it might be that it's the positive value 0.0627c by letting $v_1=0.8$ and $v_2=-0.75$.

https://www.wolframalpha.com/input/.../(m1*cosh(arctanh(v1))+m2*cosh(arctanh(v2)));

Since the input masses are equal, they drop out of the ratio to find the velocity of the output particle.
Geometrically, it turns out that the output particle is along the Minkowski-angle bisector:
https://www.wolframalpha.com/input/?i=m1:=1;m2:=1;v1:=.8;v2:=-.75;tanh(+(arctanh(v1)+arctanh(v2))/2)

7. Mar 31, 2017

### Eve Torchic

I tried 0.0627 and it still says I have the answer wrong. I only have 2 attempts left. Could it be a rounding problem?

8. Apr 1, 2017

### PeroK

I put your figures into a spreadsheet and found that the answer is sensitive to small changes in the initial velocities. Technically, $0.800c$ could be anywhere in the range $0.7995c - 0.8005c$ and likewise $0.750c$ is in the range $0.7495c - 0.7505c$.

If you do the calculations for the two extremes. First, using $0.8005c$ and $0.7495c$ gives an answer of $0.064c$. And, using $0.7995c$ and $0.7505c$ gives an answer of $0.061c$.

Technically, therefore, the answer is in the range $0.061c - 0.064c$, hence $0.06c$ is the best you can say.

That said, it's entirely possible that the expected answer is $0.063c$.