Relativistic Momentum of a particle

In summary, a particle of unknown mass M decays into two particles of known masses m1 = 0.5 GeV/c2 and m2 = 1.0 GeV/c2. The momenta of the two particles are measured to be p1 = 2.0 GeV/c along the positive y-axis and p2 = 1.5 GeV/c along the positive x-axis. Using conservation of momentum and energy, the unknown mass M can be calculated, as well as its speed in the ground frame.
  • #1
~Sam~
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Homework Statement


A particle of unknown mass M decays into two particles of known masses m1 = 0.5 GeV/c2 and m2 = 1.0 GeV/c2, whose momenta are measured to be p1 = 2.0 GeV/c along the positive y-axis and p2 = 1.5 GeV/c along the positive x-axis. Find the unknown mass M and its speed.


Homework Equations



p=1/sqrt(1-v^2/c^2)(mu)

The Attempt at a Solution



I tried summing the momentas, via pythagoras theorem. after that I'm not sure.
 
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  • #2
What's the general principle which governs how you can find the answer? I.e. for a given mass 'M,' why wouldn't m1 = 500000 GeV and m2 = 100000 Gev?

Hint: what if you think about the decay reaction as a collision?
 
  • #3
Ahh I got it...using conservation of energy.
 
  • #4
use conservation of momentum and energy to get the velocity in ground frame
 
  • #5


I can provide a response to this question by using the principle of conservation of momentum and the equation for relativistic momentum. The total momentum of the system before and after the decay must be equal. Therefore, we can write the equation:

p1 + p2 = p

Where p is the momentum of the unknown particle M. We can also write the momentum of the two particles in terms of their masses and velocities:

p1 = m1v1
p2 = m2v2

Substituting these equations into the first one, we get:

m1v1 + m2v2 = p

Using the equation for relativistic momentum, we can write:

p = 1/sqrt(1-v^2/c^2)(m1u1 + m2u2)

Where v is the velocity of the unknown particle M and u1 and u2 are the velocities of the two particles after the decay. Since we know the values of m1, m2, p1, and p2, we can solve for u1 and u2:

u1 = p1/m1 = 2.0 GeV/c / 0.5 GeV/c^2 = 4 c
u2 = p2/m2 = 1.5 GeV/c / 1.0 GeV/c^2 = 1.5 c

Substituting these values into the equation for p, we get:

p = 1/sqrt(1-v^2/c^2)(0.5*4c + 1.0*1.5c) = 1.5/sqrt(1-v^2/c^2)c

Now, we can solve for v by equating this expression to the total momentum of the system:

p1 + p2 = p = 1.5/sqrt(1-v^2/c^2)c

Solving for v, we get:

v = 0.8c

Finally, we can use the equation for relativistic energy to solve for the mass of the unknown particle M:

E = mc^2 = sqrt(p^2c^2 + m^2c^4)

Substituting the values of p and v, we get:

M = sqrt((1.5/sqrt(1-0.8^2)c)^2 + m^2c^4)/c^2

Solving for M
 

What is relativistic momentum?

Relativistic momentum is a concept in physics that takes into account the effects of special relativity on the momentum of a moving particle. It is based on the idea that the mass of a particle increases as its velocity approaches the speed of light.

How is relativistic momentum calculated?

The formula for relativistic momentum is p = mv/sqrt(1-v^2/c^2), where p is the momentum, m is the mass of the particle, v is its velocity, and c is the speed of light. This formula takes into account the effects of special relativity and allows for the calculation of momentum at high speeds.

What is the difference between relativistic and classical momentum?

Classical momentum is based on Newton's laws of motion and assumes that the mass of a particle remains constant, regardless of its velocity. Relativistic momentum, on the other hand, takes into account the increase in mass as a particle's velocity approaches the speed of light. This allows for more accurate calculations at high speeds.

Why is relativistic momentum important?

Relativistic momentum is important because it helps us understand the behavior of particles at high speeds, such as those in particle accelerators or in space. It also allows for accurate calculations and predictions in these scenarios.

What are some real-world applications of relativistic momentum?

Relativistic momentum is used in various fields, including particle physics, astrophysics, and engineering. It is essential for understanding the behavior of particles in particle accelerators and for predicting the movement of objects in space, such as satellites and spacecraft. It also has applications in nuclear reactors and nuclear medicine.

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