Relativistic energy of particle of mass

In summary, a particle of mass M decays into two identical particles with mass m = 0.3M. The initial particle has a total energy of 5Mc^2 in the laboratory reference frame. Using the equation Etot=mc^2/√(1-u^2/c^2), the velocities of the decay products in the laboratory frame can be found to be +/- 0.9798c. However, when performing the velocity transformation, the faster particle is moving at 0.99c and the slower particle at 0.83c relative to the laboratory. This discrepancy may be due to a rounding error.
  • #1
w3390
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Homework Statement



A particle of mass M decays into two identical particles each of mass m, where m = 0.3M. Prior to the decay, the particle of mass M has a total energy of 5Mc2 in the laboratory reference frame. The velocities of the decay product are along the direction of motion M. Find the velocities of the decay products in the laboratory reference frame.

Homework Equations



Etot=[tex]\frac{mc^2}{\sqrt{1-\frac{u^2}{c^2}}}[/tex]

The Attempt at a Solution



Since I am given the total energy to be 5Mc^2 in the frame of the laboratory, I plugged it into the above equation and solved for u. I get u to be .9798c. I have also found the velocities of the particles in the frame of the initial particle to be +/-.8c. However, when I do the velocity transformation, I get the faster particle to be moving .99c and the slower particle to be moving .83c relative to the laboratory, but my online homework is saying these answers are wrong. Can anybody help me figure out where I am going wrong? I am comfortable with my velocity transformation mathematics, so I figure the mistake is with the velocity of the particle of mass M.
 
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  • #2
Could just be a rounding error...how many sig-digs are your final results supposed to be?
 
  • #3
You're right. I ended up getting one of the answers wrong because I input .833c instead of .832c when the problem only gave one number with one significant figure. Thanks though.
 

What is the equation for relativistic energy of a particle of mass?

The equation for relativistic energy is E = mc^2, where E is energy, m is mass, and c is the speed of light.

How does the relativistic energy of a particle of mass differ from classical kinetic energy?

The relativistic energy of a particle of mass takes into account the effects of special relativity, including the increase in mass at high speeds. This means that at high speeds, the relativistic energy will be higher than the classical kinetic energy.

What is the significance of the speed of light in the equation for relativistic energy?

The speed of light, denoted by c, is a fundamental constant in physics. It represents the maximum speed at which any object or information can travel in the universe. In the equation for relativistic energy, it shows how mass and energy are interrelated.

Can the relativistic energy of a particle of mass be negative?

No, the relativistic energy of a particle of mass cannot be negative. This is because the equation for relativistic energy does not account for negative mass values, and the speed of light is always a positive value.

How is the concept of relativistic energy used in modern physics?

The concept of relativistic energy is used in many areas of modern physics, such as particle accelerators, nuclear reactions, and cosmology. It is also an important factor in the theory of relativity and the understanding of the behavior of objects at high speeds.

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