Special Relativity - Relativistic Energy

A lump of clay whose rest mass is 4 kilograms is travelling at three-fifths the speed of light when it collides head-on with an identical lump going the opposite direction at the same speed. If the two lumps stick together and no energy is radiated away, what is the mass of the composite lump?

A. 4 kg
B. 6.4 kg
C. 8 kg
D. 10 kg
E. 13.3 kg

(GR9677 #36)

Solution:

No energy is radiated away = energy is conserved

E_rel(a) + E_rel(b)  = E_rest(a,b)

γ_am_ac² + γ_bm_bc²  = Mc²

Given:
m_a = m_b = m₀ = 4 kg
|v_a| = |v_b| = ³/₅ c

γ = 1/(1 − ^v2/_c²)^½ 
γ = 1/(1 − ⁹/₂₅)^½ = 1/(¹⁶/₂₅)^½ = 1/(⁴/5)^½ = ⁵/₄
γ_a = γ_b = γ = ⁵/₄  

γ_am_ac² + γ_bm_bc²  = Mc²
M = 2γm₀ = 2(⁵/₄ )(4) = 10 kg

Answer: D

Special Relativity - Relativistic Energy

What is the speed of a particle having a momentum of 5 MeV/c and a total relativistic energy of 10 MeV?

A. c
B. 0.75 c
C. 1/√3 c
D. ½ c
E.  ¼ c

(GR9677 #38)

Solution:

p = γm₀v
E = γm₀c²

p/E = v/c²
v = pc²/E = (5 MeV/c) c²/10 MeV = ½ c

Answer: D

Special Relativity - Momentum

A monoenergetic beam consists of unstable particles with total energies 100 times their rest energy. If the particles have rest mass m, their momentum is most nearly 

A. mc
B. 10 mc
C. 70 mc
D. 100 mc
E. 10⁴ mc

(GR9277 #70)

Solution:

E = γE₀ = 100E₀ → γ = 100
p = γm₀c = 100 mc

Answer: D

Special Relativity - Relativistic Energy

A free electron (rest mass m_e = 0.5 MeV/c²) has a total energy of 1.5 MeV. Its momentum p in units of MeV/c is about

A. 0.86
B. 1.0
C. 1.4
D. 1.5
E. 2.0

(GR9277 #85)

Solution:

E² = p²c² + m₀²c⁴

1.5² = p²c² + (0.5/c²)²c⁴

 p²c² = 1.5² − 0.5² = 2

p = ^√2/_c ≈ ^1.4/_c

Answer: C

Special Relativity - Relativistic Properties

If a newly discovered particle X moves with a speed equal to the speed of light in vacuum, then which of the following must be true?

A. The rest mass of X is zero
B. The spin of X equals the spin of a photon
C. The charge of X is carried on its surface
D. X does not spin
E. X cannot be detected

(GR8677 #68)

Solution:

Relativistic Energy: E² = p²c² + m₀²c⁴

If the particle moves with v = c, de Broglie relation: E = hf = pc

E² = E² + m₀²c⁴ 
m₀²c⁴ = E² − E² = 0
m₀ = 0

Answer: A

Special Relativity - Relativistic Momentum

If the total energy of a particle of mass m is equal to twice its rest energy, then the magnitude of the particle’s relativistic momentum is

A. mc/2
B. mc/√2
C. mc
D. √3mc
E. 2mc

(GR0177 #32)

Solution:

Rest Energy: E₀ = m₀c²

Relativistic Energy:  E² = p²c² + m₀²c⁴

Given: m₀ = m and E = 2E₀

p²c² + m²c⁴ = 4E₀² = 4m²c⁴
p² = 3m²c²
p = √3mc

Answer: D

Special Relativity - Rest Mass

A particle leaving a cyclotron has a total relativistic energy of 10 GeV and a relativistic momentum of 8 GeV/c. What is the rest mass of this particle? 

A. 0.25 GeV/c²
B. 1.20 GeV/c²
C. 2.00 GeV/c²
D. 6.00 GeV/c²
E. 16.0 GeV/c²

(GR0177 #79)

Solution:

E² = p²c² + m₀²c⁴

Given: E = 10 GeV and p = 8 GeV/c

100 = (64/c²)c² + m₀²c⁴
m₀²c⁴ = 100 − 64 = 36
m₀ = 6 GeV/c²

Answer: D

Special Relativity - Relativistic Energy

A photon strikes an electron of mass m that is initially at rest, creating an electron-positron pair. The photon is destroyed and the positron and two electrons move off at equal speed along the initial direction of the photon. The energy of the photon was

A. mc²
B. 2mc²
C. 3mc²
D. 4mc²
E. 5mc²

(GR0177 #99)

Solution

Relativistic Energy: 

E = (p²c² + m₀²c⁴)^1/2

Photon has no mass, m₀ = 0, therefore,

E_(photon) = (p²c² )^1/2 = pc

Initially, electron is at rest, thus momentum is zero, therefore,

 E_(electron) = (m₀²c⁴)^1/2 = m_ec²

 
Total initial energy of the system,

E_i = E_(photon) + E_(electron) = pc + m_ec²

After collision, 3 particles (2 electrons and a positron) move off at equal speed, thus each has p/3 momentum.

Electron and positron has the same mass.

Thus, each particle has energy,

E_{f (electron)} = E_{f (positron)} = [(p/3)²c² + m_e²c⁴]^1/2

Total final energy of the 3 particles, E_f  = 3 [(p/3)²c² + m_e²c⁴]^1/2

Conservation of energy, E_i = E_f 

pc + m_ec²  = 3 [(p/3)²c² + m_e²c⁴]^1/2

(pc + m_ec²)² = 9 [(p/3)²c² + m_e²c⁴]

p²c² + m_e²c⁴ + 2pm_ec³ = 9(p/3)²c²  + 9m_e²c⁴

p²c² + m_e²c⁴ + 2pm_ec³ = p²c²  + 9m_e²c⁴

m_e²c⁴ + 2pm_ec³ = 9m_e²c⁴

2pm_ec³ = 8m_e²c⁴

p = 4m_ec

Energy, E = pc = 4m_ec²

Answer: D