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

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.