A simple problem of conservation of momentum

In summary, after the collision the momentum in system S' is greater than the sum of the masses of the original particles.
  • #1
epovo
114
21
Trying to understand the conservation of relativistic momentum, I thought of this problem.
It's very simple, and possibly my mistake is embarrassing so I apologize in advance :blushing:

Two particles of identical mass m and speeds v and -v in some frame S collide inelastically and they stop. The relativistic momentum in S is zero all the time.
Now, in system S' moving at -v, the initial momentum should be

p = m v' / √(1-v'2/c2) [1]

where v' is the velocity of the other particle in S'

v' = 2v / (1+v2/c2) [2]

if I substitute [2] into [1] I can get the momentum before impact as a function of v. After some manipulation I get

p = 2mv / (1 - v2/c2)

But the momentum in S' after the impact is that of a particle of mass 2m moving with speed v. So its momentum should be

p = 2mv / √(1 - v2/c2))

so - where did I go wrong?
Thank you
 
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  • #2
The easiest way to handle these types of problems is through the conservation of the four-momentum. Here is a wikipedia page on four-momentum: http://en.wikipedia.org/wiki/Four-momentum

Using units where c=1 the four-momentum of the particles in S are:
##P_1=(\gamma m, \gamma m v, 0, 0)##
##P_2=(\gamma m, -\gamma m v, 0, 0)##
where ##\gamma=1/\sqrt{1-v^2}##

The four-momentum is conserved, so after the collision we have:
##P_3=P_1+P_2=(\gamma 2 m,0,0,0)##
Note that the mass after the collision is ##m_3=|P_3|=\gamma 2 m## so the mass of the combined particle is greater than the sum of the masses of the original particles! This is the key part that you missed by doing it the long way rather than using four-vectors.

Now, in frame S' we have:
##P_1=((2\gamma^2-1)m, 2mv\gamma^2,0,0)##
##P_2=(m,0,0,0)##
##P_3=(2 m \gamma^2, 2 m v \gamma^2,0,0)##
Note that the rest mass after the collision is again ##m_3=|P_3|=\gamma 2 m##
 
  • #3
Of course! I forgot about the increased rest mass because of the energy-mass conservation principle... That's where my missing γ went!
I will start using four-momentum but actually this mistake has helped me understand things better.
Thank you very much
 

FAQ: A simple problem of conservation of momentum

1. What is conservation of momentum?

Conservation of momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant over time, regardless of any external forces acting on the system.

2. How is momentum defined?

Momentum is defined as the product of an object's mass and velocity. It is a vector quantity, meaning it has both magnitude and direction. The SI unit for momentum is kilogram meters per second (kg m/s).

3. What is an example of a simple problem of conservation of momentum?

An example of a simple problem of conservation of momentum would be a collision between two objects, such as billiard balls or cars. In this scenario, the total momentum before the collision would be equal to the total momentum after the collision, as long as there are no external forces acting on the system.

4. How does the law of conservation of momentum apply to real-world situations?

The law of conservation of momentum applies to real-world situations by helping us understand and predict the motion of objects in various situations, such as collisions, explosions, and rocket launches. It is also the basis for many engineering and technological applications, such as airbags in cars and rocket propulsion systems.

5. What are some limitations of the conservation of momentum principle?

The conservation of momentum principle is only applicable to closed systems, meaning there are no external forces acting on the system. This can be difficult to achieve in real-world situations, as there are almost always some external forces present. Additionally, the law of conservation of momentum does not take into account friction, which can have a significant impact on the motion of objects in real-world scenarios.

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