StudentAidUnit 11: Special Theory of Relativity — Numericals
11th Class Physics · Unit 11: Special Theory of Relativity
11.1.An electron is accelerated to a speed of 0.995 c which passes down an evacuated tube 500 m long. How long will the tube appear to the electron.
Given
Length of the tube
ell0 = 500 m
Speed of the electron
v = 0.995c
Formula
Length contraction
ell = ell0 × sqrt{1 - v2c2
Working
Substituting values
ell = 500 m × sqrt{1 - (0.995)2}
= 500 m × sqrt{1 - 0.990025}
= 500 m × sqrt{0.009975}
= 500 m × 0.099874
Result
Contracted length
ell ≈ 49.94 m
11.2.A neutron, being not a stable particle, disintegrates in 20 minutes on the average. How long will it seem to exist if it shoots out from a nucleus with a speed of 0.8 c?
Given
Proper time
t0 = 20 minutes = 1200 s
Speed
v = 0.8c
Find
t = ?
Formula
Time dilation
t = frac{t0}{sqrt{1 - v2c2
Working
Substituting values
t = frac{1200}{sqrt{1 - (0.8)2}
= frac{1200}{sqrt{1 - 0.64}
= frac{1200}{sqrt{0.36}
Result
Relativistic time
= 2000 s
Converting to minutes
t = 2000 s / 60 = 33.33 minutes
Final result
t ≈ 33.33 minutes
11.3.A spaceship is measured 100 m long while it is at rest with respect to an observer, if this spaceship now flies by the observer with a speed of 0.99 c. what length will the observer find for the spaceship?
Given
Proper length
ell0 = 100 m
Speed
v = 0.99c
Find
ell = ?
Formula
Length contraction
ell = ell0 × sqrt{1 - v2c2
Working
Substituting values
ell = 100 m × sqrt{1 - (0.99)2}
= 100 m × sqrt{1 - 0.9801}
= 100 m × sqrt{0.0199}
= 100 m × 0.141
Result
Observed length
ell ≈ 14.1 m
11.4.The rest mass of an electron is 9.11 x 10^{-31} kg. Calculate the corresponding rest-mass energy.
Given
Rest mass of an electron
m = 9.11 × 10-31 kg
Speed of light
c = 3 × 108 m/s
Find
E = ?
Formula
Mass-energy relation
E = mc2
Working
Substituting values
E = (9.11 × 10-31 kg) × (3 × 108 m/s)2
= (9.11 × 10-31) × (9 × 1016) J
= 81.99 × 10-15 J
Converting to electron volts (1 eV = 1.602 × 10^{-19} J)
E ≈ frac{81.99 × 10-15{1.602 × 10-19 eV
≈ 0.511 × 106 eV
Result
Rest-mass energy
≈ 0.511 MeV
11.5.An electron is accelerated to a speed v = 0.85 c. Calculate its total energy and kinetic energy in electron volt.
Given
Speed of electron
v = 0.85c
Rest mass
m0 = 9.11 × 10-31 kg
Speed of light
c = 3.0 × 108 m/s
Rest mass energy
E0 = 0.511 MeV
Formula
Total energy
E = frac{E0}{sqrt{1 - left(vcright)2}
Working
Substituting values
E = frac{0.511}{sqrt{1 - (0.85)2}
= frac{0.511}{sqrt{1 - 0.7225}
= frac{0.511}{sqrt{0.2775}
Result
Total energy
= 0.97 MeV
Formula
Kinetic energy
K.E = E - E0
= 0.97 - 0.511
Result
Kinetic energy
= 0.459 MeV
11.6.At what speed, would the mass of a proton in a particle accelerator be tripled?
Given
Rest mass of proton
m0 = m
Relativistic mass
m = 3m0
Speed of light
c = 3.0 × 108 m/s
Find
v = ?
Formula
Relativistic mass
m = frac{m0}{sqrt{1 - left(vcright)2}
3m0 = frac{m0}{sqrt{1 - left(vcright)2}
3 = frac{1}{sqrt{1 - left(vcright)2}
sqrt{1 - left(vcright)2} = 13
1 - left(vcright)2 = 19
left(vcright)2 = 1 - 19 = 89
Result
Speed
vc = sqrt{89 = 0.943
Final result
v = 0.943c
11.7.The period of pendulum is measured to be 3 s in an inertial frame of reference. What will be the period measured by an observer in a spaceship with a constant speed of 0.95 c with respect to the pendulum?
Given
Proper time
t0 = 3 s
Speed
v = 0.95c
Speed of light
c = 3.0 × 108 m/s
Find
t = ?
Formula
Time dilation
t = frac{t0}{sqrt{1 - left(vcright)2}
Working
Substituting values
t = frac{3}{sqrt{1 - (0.95)2}
= frac{3}{sqrt{1 - 0.9025}
= frac{3}{sqrt{0.0975}
Result
Relativistic period
= 9.61 s
11.8.Hypothetically, if a ball of mass 0.5 kg is projected with a velocity of 0.9 c. what will be its mass in flight?
Given
Rest mass
m0 = 0.5 kg
Velocity
v = 0.9c
Speed of light
c = 3.0 × 108 m/s
Find
m = ?
Formula
Relativistic mass
m = frac{m0}{sqrt{1 - left(vcright)2}
Working
Substituting values
m = frac{0.5}{sqrt{1 - (0.9)2}
= frac{0.5}{sqrt{1 - 0.81}
= frac{0.5}{sqrt{0.19}
= 1.147 kg
Result
Relativistic mass
m = 1.15 kg