GRADE 11 PRACTICE
Physics Problems: Special relativity, Grade 11
Relativistic mass, time and length.
Here the conditions and analysis of the solution can be read without an account. In the simulator, you need to collect the same analyzes yourself: “Given,” formulas, order of steps and answer.
Problem statements
The answer options are there to check against, and the walkthrough opens under each problem.
Problem #54
Fundamentals of Special RelativityHow far will light travel in a vacuum in an hour?
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- t
- 1\,\text{ч}=3600\,\text{с}
- c
- 3\cdot 10^{8}\,\text{м/с}
S - ?
- S = ctS = 3\cdot 10^{8}\cdot 3600S = 1{,}08\cdot 10^{12}\,\text{м} = 1{,}08\cdot 10^{9}\,\text{км}
The speed of light is the maximum speed at which interaction can be transmitted.
AnswerS = 1{,}08\cdot 10^{12}\,\text{м}
Problem #57
Fundamentals of Special RelativityAt what speed of a moving clock does 2 times more time pass between its two signals in the laboratory frame of reference than according to this clock? Inertial reference systems.
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- \tau
- 2\tau_0
- c
- 3\cdot 10^{8}\,\text{м/с}
v - ?
- \tau = \dfrac{\tau_0}{\sqrt{1 - \dfrac{v^{2}}{c^{2}}}}\ \Rightarrow\ 2\tau_0 = \dfrac{\tau_0}{\sqrt{1 - \dfrac{v^{2}}{c^{2}}}}\sqrt{1 - \dfrac{v^{2}}{c^{2}}} = \dfrac{1}{2}
- 1 - \dfrac{v^{2}}{c^{2}} = \dfrac{1}{4}\ \Rightarrow\ v^{2} = 0{,}75c^{2}v = \sqrt{0{,}75}\,c = 0{,}866\cdot 3\cdot 10^{8}v = 2{,}6\cdot 10^{8}\,\tfrac{\text{м}}{\text{с}}
Answerv = 2{,}6\cdot 10^{8}\,\tfrac{\text{м}}{\text{с}}
Problem #58
Fundamentals of Special RelativityBurnt out 100\,\text{г} gunpowder with specific heat of combustion 10^{7}\,\text{Дж/кг}. All released energy is transferred to the bullet. Find the increase in E/c² for the bullet, where E is its total energy. This is the energy equivalent of the energy transferred; The rest mass of the bullet does not change during acceleration.
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- m
- 100\,\text{г}=0{,}1\,\text{кг}
- q
- 10^{7}\,\tfrac{\text{Дж}}{\text{кг}}
- c
- 3\cdot 10^{8}\,\text{м/с}
\Delta m - ?
- \Delta E = Q = mq\Delta E = 0{,}1\cdot 10^{7}\Delta E = 10^{6}\,\text{Дж}
- \Delta m = \dfrac{\Delta E}{c^{2}} = \dfrac{mq}{c^{2}}\Delta m = \dfrac{0{,}1\cdot 10^{7}}{\left(3\cdot 10^{8}\right)^{2}}\Delta m = 1{,}1\cdot 10^{-11}\,\text{кг}
This is the increase in E/c², traditionally called the increase in relativistic mass. An increase in kinetic energy does not change the rest mass of the bullet.
Answer\Delta m = 1{,}1\cdot 10^{-11}\,\text{кг}
Problem #59
Fundamentals of Special RelativityThe first satellite had a mass 80\,\text{кг} and flew at speed 8\,\text{км/с}. By how much did its relativistic mass increase?
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- m
- 80\,\text{кг}
- v
- 8\,\tfrac{\text{км}}{\text{с}}=8000\,\text{м/с}
- c
- 3\cdot 10^{8}\,\text{м/с}
\Delta m - ?
- \Delta E = E_{\text{к}} = \dfrac{mv^{2}}{2}\Delta E = \dfrac{80\cdot 8000^{2}}{2}\Delta E = 2{,}56\cdot 10^{9}\,\text{Дж}
- \Delta m = \dfrac{\Delta E}{c^{2}} = \dfrac{mv^{2}}{2c^{2}}\Delta m = \dfrac{80\cdot 8000^{2}}{2\cdot\left(3\cdot 10^{8}\right)^{2}}\Delta m = 2{,}84\cdot 10^{-8}\,\text{кг} = 2{,}84\cdot 10^{-2}\,\text{мг}
Answer\Delta m = 2{,}84\cdot 10^{-8}\,\text{кг}
Problem #60
Fundamentals of Special RelativityFind the magnitude of the accelerating voltage, after passing which the total energy of the initially resting electron will become 4 times greater than its rest energy.
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- m
- 4m_0
- m_0
- 9{,}1\cdot 10^{-31}\,\text{кг}
- \left|q\right|
- e = 1{,}6\cdot 10^{-19}\,\text{Кл}
- c
- 3\cdot 10^{8}\,\text{м/с}
U - ?
- \Delta m = m - m_0 = 3m_0\Delta m = 3\cdot 9{,}1\cdot 10^{-31}\Delta m = 2{,}73\cdot 10^{-30}\,\text{кг}
- \Delta E = \Delta m\,c^{2}\Delta E = 2{,}73\cdot 10^{-30}\cdot 9\cdot 10^{16}\Delta E = 2{,}457\cdot 10^{-13}\,\text{Дж}
- \Delta E = A = \left|q\right|U\ \Rightarrow\ U = \dfrac{\Delta E}{\left|q\right|}U = \dfrac{2{,}457\cdot 10^{-13}}{1{,}6\cdot 10^{-19}}U = 1{,}54\cdot 10^{6}\,\text{В}
AnswerU = 1{,}54\cdot 10^{6}\,\text{В}
Problem #134
Relativistic effectsObject mass 0{,}001 kg. Find the rest energy at c = 300000000 m/s.
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- m
- 0{,}001\,\text{кг}
- c
- 300000000\,\text{м/с}
E_0 - ?
- E_0=mc^2E_0=0{,}001 \cdot {300000000}^{2}E_0=90000000000000\,\text{Дж}
Rest energy is inherent in the body even without movement.
Answer90000000000000\,\text{Дж}
Problem #135
Relativistic effectsObject mass 0{,}002 kg. Find the rest energy at c = 300000000 m/s.
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- m
- 0{,}002\,\text{кг}
- c
- 300000000\,\text{м/с}
E_0 - ?
- E_0=mc^2E_0=0{,}002 \cdot {300000000}^{2}E_0=180000000000000\,\text{Дж}
Rest energy is inherent in the body even without movement.
Answer180000000000000\,\text{Дж}
Problem #136
Relativistic effectsOwn process time 4 s. System speed is 0{,}6 from the speed of light. Find the duration in the laboratory system.
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- t
- 4\,\text{с}
- b
- 0{,}6\,\text{доля c}
T - ?
- k=\sqrt{1-\beta^2}k=\sqrt{1 - {0{,}6}^{2}}k=0{,}8\,\text{безразм.}
β = v/c is less than one.
- T=\frac{t}{\sqrt{1-\beta^2}}T=\frac{4}{0{,}8}T=5\,\text{с}
Laboratory duration is longer than your own time.
Answer5\,\text{с}
Problem #137
Relativistic effectsOwn process time 6 s. System speed is 0{,}8 from the speed of light. Find the duration in the laboratory system.
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- t
- 6\,\text{с}
- b
- 0{,}8\,\text{доля c}
T - ?
- k=\sqrt{1-\beta^2}k=\sqrt{1 - {0{,}8}^{2}}k=0{,}6\,\text{безразм.}
β = v/c is less than one.
- T=\frac{t}{\sqrt{1-\beta^2}}T=\frac{6}{0{,}6}T=10\,\text{с}
Laboratory duration is longer than your own time.
Answer10\,\text{с}
Problem #138
Relativistic effectsParticle rest energy 3 MeV, the speed is 0{,}8 from the speed of light. Find the kinetic energy.
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- E
- 3\,\text{МэВ}
- b
- 0{,}8\,\text{доля c}
K - ?
- k=\sqrt{1-\beta^2}k=\sqrt{1 - {0{,}8}^{2}}k=0{,}6\,\text{безразм.}
Find the factor in the denominator.
- W=\frac{E_0}{k}W=\frac{3}{0{,}6}W=5\,\text{МэВ}
This is the total energy of a moving particle.
- K=W-E_0K=5 - 3K=2\,\text{МэВ}
Subtract the rest energy.
Answer2\,\text{МэВ}
Problem #139
Relativistic effectsParticle rest energy 6 MeV, the speed is 0{,}6 from the speed of light. Find the kinetic energy.
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- E
- 6\,\text{МэВ}
- b
- 0{,}6\,\text{доля c}
K - ?
- k=\sqrt{1-\beta^2}k=\sqrt{1 - {0{,}6}^{2}}k=0{,}8\,\text{безразм.}
Find the factor in the denominator.
- W=\frac{E_0}{k}W=\frac{6}{0{,}8}W=7{,}5\,\text{МэВ}
This is the total energy of a moving particle.
- K=W-E_0K=7{,}5 - 6K=1{,}5\,\text{МэВ}
Subtract the rest energy.
Answer1{,}5\,\text{МэВ}
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