1. Length Contraction of a Moving Rod
A rod 1 meter long is moving along its length with a velocity 0.6c. Calculate its length as it appears to an observer (a) on the earth (b) moving with the rod itself.
Data:
- Actual (Proper) Length (L0) = 1 m
- Speed (v) = 0.6c
- a) Length observed from Earth (L) = ?
- b) Length for observer moving with the rod (L') = ?
Solution:
(a) The relativistic length contraction is given by: L = L0 × √(1 - v²/c²)
L = 1 × √(1 - (0.6c)²/c²)
L = 1 × √(1 - 0.36)
L = 1 × √(0.64)
L = 1 × 0.8 = 0.8 m
(b) An observer moving with the rod is at rest relative to it, so they measure its proper length: L' = L0 = 1 m
Result: Its length as it appears to an observer (a) on the earth is 0.8 m and (b) moving with the rod itself is 1 m.
2. Required Speed for Target Length Contraction
How fast would a rocket have to go relative to an observer for its length to be contracted to 99% of its length at rest?
Data:
- Length at rest (L0) = L0
- Contracted Length (L) = 99% of L0 = 0.99 L0
- Speed (v) = ?
Solution:
The length contraction formula is: L = L0 × √(1 - v²/c²)
Substitute values: 0.99 L0 = L0 × √(1 - v²/c²)
Dividing by L0: 0.99 = √(1 - v²/c²)
Squaring on both sides (S.O.B.S.): 0.9801 = 1 - v²/c²
Rearranging terms: v²/c² = 1 - 0.9801
v²/c² = 0.0199
Taking square root on both sides: v/c = √(0.0199) ≈ 0.141
v = 0.141c = 0.141 × 3 × 10⁸ m/s ≈ 4.23 × 10⁷ m/s
Result: The rocket must travel at a speed of 0.141c (or 4.23 × 10⁷ m/s) relative to the observer.
3. Time Dilation and Particle Range
A particle with a proper lifetime of 1 µs moves through the laboratory at 2.7 × 10⁸ m/s. (a) What is its lifetime, as measured by observers in the laboratory? (b) What will be the distance traversed by it before disintegrating?
Data:
- Proper lifetime (Δt0) = 1 µs = 1 × 10⁻⁶ s
- Speed (v) = 2.7 × 10⁸ m/s
- Speed of light (c) = 3.0 × 10⁸ m/s (so v/c = 2.7/3.0 = 0.9)
- a) Observed Lifetime (Δt) = ?
- b) Range / Distance (d) = ?
Solution:
(a) The time dilation formula is given by: Δt = Δt0 / √(1 - v²/c²)
Δt = (1 × 10⁻⁶) / √(1 - (0.9)²)
Δt = (1 × 10⁻⁶) / √(1 - 0.81)
Δt = (1 × 10⁻⁶) / √(0.19)
Δt = (1 × 10⁻⁶) / 0.4359 ≈ 2.294 × 10⁻⁶ s = 2.29 µs
(b) The laboratory distance traveled before decaying is: d = v × Δt
d = (2.7 × 10⁸ m/s) × (2.294 × 10⁻⁶ s)
d ≈ 619.4 m (or 621 m using rounded intermediate steps)
Result: (a) Its dilated lifetime in the lab is 2.29 µs, and (b) the distance traversed before disintegrating is approximately 619.4 m.
4. Mass Variation and Speed Limit
At what speed is a particle moving if its relativistic mass is equal to three times its rest mass?
Data:
- Rest Mass = m0
- Relativistic Mass (m) = 3 × m0
- Speed (v) = ?
Solution:
The mass variation expression is: m = m0 / √(1 - v²/c²)
According to the problem criterion: 3m0 = m0 / √(1 - v²/c²)
Dividing by m0: 3 = 1 / √(1 - v²/c²)
Rearranging yields: √(1 - v²/c²) = 1/3
Squaring on both sides (S.O.B.S.): 1 - v²/c² = 1/9
v²/c² = 1 - 1/9 = 8/9
Taking square root on both sides: v/c = √(8/9) = (√8)/3 ≈ 0.943
v = 0.943c = 0.943 × 3 × 10⁸ m/s ≈ 2.83 × 10⁸ m/s
Result: The speed of the particle is equal to 0.943c (or 2.83 × 10⁸ m/s).
5. Total Mass-Energy Conversion
If 4 kg of a substance is fully converted into energy, how much energy is produced?
Data:
- Mass (m) = 4 kg
- Speed of light (c) = 3 × 10⁸ m/s
- Energy produced (E) = ?
Solution:
Einstein's mass-energy equivalence equation is: E = m × c²
E = 4 × (3 × 10⁸)²
E = 4 × (9 × 10¹⁶)
E = 3.6 × 10¹⁷ J
Result: The total energy produced in this full conversion is 3.6 × 10¹⁷ Joules.
6. Rest Energy of an Electron
Calculate the rest energy of an electron in joules and in electron volts.
Data:
- Rest mass of an electron (m0) = 9.11 × 10⁻³¹ kg
- Speed of light (c) = 3 × 10⁸ m/s
- Conversion factor: 1 eV = 1.6 × 10⁻¹⁹ J
- Rest Energy (E0) = ?
Solution:
The rest energy equation is given by: E0 = m0 × c²
E0 = (9.11 × 10⁻³¹) × (3 × 10⁸)²
E0 = 9.11 × 10⁻³¹ × 9 × 10¹⁶
E0 = 8.199 × 10⁻¹⁴ J
To convert the answer into electron volts (eV): E0 (eV) = E0 (J) / (1.6 × 10⁻¹⁹)
E0 (eV) = (8.199 × 10⁻¹⁴) / (1.6 × 10⁻¹⁹)
E0 (eV) ≈ 512,437 eV = 0.512 MeV
Result: The rest energy of an electron is 8.2 × 10⁻¹⁴ J or 0.512 MeV.
7. Relativistic Kinetic Energy
Calculate the K.E. of an electron moving with a velocity of 0.98 times the velocity of light in the laboratory system.
Data:
- Rest mass of an electron (m0) = 9.11 × 10⁻³¹ kg
- Velocity (v) = 0.98c
- Kinetic Energy (K.E.) = ?
Solution:
The relativistic total energy relation states: E = K.E. + E0 ⇒ K.E. = E - E0 = (m - m0)c²
Substitute the mass variation rule: K.E. = m0c² × [1 / √(1 - v²/c²) - 1]
First calculate the Lorentz factor item: 1 / √(1 - (0.98)²) = 1 / √(1 - 0.9604) = 1 / √(0.0396) ≈ 5.0252
Now calculate rest energy value m0c²: (9.11 × 10⁻³¹) × (3 × 10⁸)² = 8.199 × 10⁻¹⁴ J
Substitute back to solve for K.E.: K.E. = 8.199 × 10⁻¹⁴ × [5.0252 - 1]
K.E. = 8.199 × 10⁻¹⁴ × 4.0252
K.E. ≈ 3.30 × 10⁻¹³ J
Result: The relativistic kinetic energy of the electron is 3.30 × 10⁻¹³ Joules.
8. Velocity Where Kinetic Energy Equals Rest Energy
At what velocity does the K.E. of a particle equal its rest energy?
Data:
- Condition: Kinetic Energy (K.E.) = Rest Energy (E0)
- Velocity (v) = ?
Solution:
The formula for total relativistic energy is: E = K.E. + E0
Since K.E. = E0, total energy becomes: E = E0 + E0 = 2E0
Expressing via mass variables: m × c² = 2 × m0 × c² ⇒ m = 2m0
Using the relativistic mass equation: 2m0 = m0 / √(1 - v²/c²)
Dividing out mass variables: √(1 - v²/c²) = 1/2
Squaring on both sides (S.O.B.S.): 1 - v²/c² = 1/4
v²/c² = 1 - 1/4 = 3/4
Taking square root on both sides: v/c = √(3)/2 ≈ 0.866
v = 0.866c = 0.866 × 3 × 10⁸ m/s ≈ 2.60 × 10⁸ m/s
Result: The velocity of the particle must be equal to 0.866c (or 2.60 × 10⁸ m/s).
9. Relativistic Properties Analysis
A particle of rest mass m0 moves with speed c/√2. Calculate its mass, momentum, total energy, and kinetic energy.
Data:
- Rest Mass = m0
- Speed (v) = c / √2
- Properties required: Relativistic mass (m), Momentum (p), Total Energy (E), Kinetic Energy (K.E.)
Solution:
First compute the denominator root parameter: v²/c² = (c/√2)² / c² = 1/2
Thus, the lorentz factor component becomes: √(1 - v²/c²) = √(1 - 1/2) = 1/√2
1) Relativistic Mass: m = m0 / √(1 - v²/c²) = m0 / (1/√2) = √2 m0
2) Momentum: p = m × v = (√2 m0) × (c / √2) = m0c
3) Total Relativistic Energy: E = m × c² = √2 m0c²
4) Kinetic Energy: K.E. = E - E0 = √2 m0c² - m0c² = (√2 - 1)m0c² ≈ 0.414 m0c²
Result: The mass of the particle is √2 m0, Momentum is m0c, Total Energy is √2 m0c², and Kinetic Energy is (√2 - 1)m0c².
10. Spaceflight Time Dilation (Proxima Centauri Voyage)
The nearest star to Earth is Proxima Centauri, 4.3 light-years away. A spaceship with a constant speed of 0.800c relative to the Earth travels toward the star. (a) How much time would elapse on a clock as measured by travelers on the spacecraft? (b) How long does the trip take according to Earth observers?
Data:
- Distance (d) = 4.3 light-years
- Speed (v) = 0.800c
- b) Dilated Trip Time for Earth Observers (Δt) = ?
- a) Proper Trip Time for Spacecraft Travelers (Δt0) = ?
Solution:
(b) We first calculate time according to Earth frame observers using linear kinematics: Δt = Distance / Velocity
Δt = 4.3 light-years / 0.800c = 5.375 years
(a) According to time dilation formulas, the frame measuring proper time interval is the traveling system: Δt = Δt0 / √(1 - v²/c²)
Rearranging terms for proper time: Δt0 = Δt × √(1 - v²/c²)
Δt0 = 5.375 × √(1 - (0.800)²)
Δt0 = 5.375 × √(1 - 0.64) = 5.375 × √(0.36)
Δt0 = 5.375 × 0.6 = 3.225 years
Result: (a) The elapsed time measured by travelers on the spacecraft is 3.225 years, and (b) the trip takes 5.375 years according to Earth observers.
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