A muon is traveling at 0.4 c relative to a laboratory frame of reference. The sp

Question

A muon is traveling at 0.4c relative to a laboratory frame of reference. The speed of the muon is doubled to 0.8c.

Part A

What happens to the momentum of the muon in the laboratory frame of reference?

Part B

What happens to the kinetic energy of the muon in the laboratory frame of reference?

Part C

What happens to the total energy of the muon in the laboratory frame of reference?

Explain your answers, thanks.

Explanation / Answer

A.)

given that relativistic momentum is p = gamma * m0 * v, where gamma is
1/sqrt(1 - (v^2/c^2)), m0 = rest mass, v = velocity
gamma is the proportion of the relativistic mass to the rest mass.

at 0.4c

P = 1/sqrt(1 - (v^2/c^2))  * m0 * v

P = 1.091 * m0 * 0.4c = 0.436c *m0

at 0.8c

P = 1/sqrt(1 - (v^2/c^2))  * m0 * v

P = 1.666 * m0 * 0.8c = 1.333c *m0

So it become roughly thrice of 0.4c
Momentum at 0.8c would be roughly 3 times that at 0.4c.

so it become more than double in laboratory frame of refrence.

B) K.E = 0.5 mv^2

given that relativistic momentum is p = gamma * m0 * v, where gamma is
1/sqrt(1 - (v^2/c^2)), m0 = rest mass, v = velocity
gamma is the proportion of the relativistic mass to the rest mass.

kinetic energy at 0.8c would be more than 4 times that at 0.4c.

kinetic energy become more than quadruples in laboratory frame of refrence

C.)

similarly energy become more than double in the laboratry frame of refrence

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