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This is second of our many quizzes on Relativity. Click on this to access more such competitive quizzes on Modern Physics and Semiconductors.

Relativity Quiz 2

physics

The relativistic momentum of a proton is given by $1.56\times {{10}^{-21}}\text{ kgm}{{\text{s}}^{-1}}$. Calculate its speed. 

 Figure out the required expression for relativistic momentum, recall the mass of a proton and use trial and error methods to solve the derived equation. 

  $ \text{ We know that relativistic momentum is given by}\\ p=\gamma mu\\ \Rightarrow p=\dfrac{mu}{\sqrt{1-\dfrac{{{u}^{2}}}{{{c}^{2}}}}}\\ \text{ Also do we know that the rest mass of proton is}\,1.67\times {{10}^{-27}}\text{ kg}\\ \text{Clearly,}\\ p=\dfrac{mu}{\sqrt{1-\dfrac{{{u}^{2}}}{{{c}^{2}}}}} \\ \Rightarrow p{{u}^{2}}+mu{{c}^{2}}={{c}^{2}}p \\ \Rightarrow (1.56\times {{10}^{-21}}){{u}^{2}}+(1.67\times {{10}^{-27}})\times {{c}^{2}}u={{c}^{2}}\times (1.56\times {{10}^{-21}}) \\ \text{ Solving this quadratic equation will give the value for u as 0.985c.}\\ \text{ Another easy method which can be followed is trial and}\\ \text{ error method by putting the values given in options and checking for the equality of LHS and RHS.}\\ \text{ This method saves time and it can be seen that u=0.985 c satisfies the derived equation.}$

Evaluate the relativistic energy of $1\,g$ mass moving with a velocity equal to one-fourth the velocity of light. 

From the special theory of relativity, the expression for relativistic energy needs to be recalled. 

  $\text{ We know that relativistic energy is given by}\\ E=\dfrac{m{{c}^{2}}}{\sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}}}\\ \text{Given}\, m=1\text{ g    and  }v=\dfrac{c}{4}\\ E=\dfrac{1\times {{10}^{-3}}\times {{(3\times {{10}^{8}})}^{2}}}{\sqrt{1-\dfrac{1}{16}}} \\ \Rightarrow E=\dfrac{3\times {{10}^{13}}}{0.968} \\ \therefore E=3.1\times {{10}^{13}}\text{ J}$

The collision of two particles $A$ and $B$ in three-dimensional space produces another particle $C$, which relocates itself with a velocity equal to $0.95\,c$. When observed from C’s frame of reference, it exists with this velocity for a time duration of $1.52\,\mu \text{s}$ only. How long does this particle exist for Daniel, who stands on the surface of the Earth? 

Using special theory of relativity, we can directly arrive at the answer from the expression relating proper time duration $(\Delta {{t}_{0}})$ and the time duration $(\Delta t)$ as observed from the Earth. 

$\text{ Using special theory of relativity, we have}\\ \Delta t=\gamma \Delta {{t}_{0}} \\ \Rightarrow \gamma =\dfrac{1}{\sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}}} \\ \Rightarrow \gamma =\dfrac{1}{\sqrt{1-{{(0.95)}^{2}}}} \\ \Rightarrow \gamma =3.2 \\ \Rightarrow \Delta t=3.2\times 1.52\\ \therefore \Delta t=4.87\mu \text{s}$

 The distance covered by a particle in its frame of reference above the surface of the earth is $1.65\text{ km}$ and the distance it covers for Daniel who is observing the same through a powerful device is $2.5\text{ km}$. The inverse of ratio of its speed to the velocity of light is 

Using special theory of relativity and the concept of length contraction, we can easily arrive at the required ratio. 

 $\text{ According to the special theory of relativity and}\\ \text{ the concept of length contraction, we have}\\ L=\dfrac{{{L}_{0}}}{\gamma }={{L}_{0}}\sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}}\\ \text{Substituting the given values in the above expression, we have}\\ 1.65=2.5\sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}} \\ \Rightarrow 1-\dfrac{{{v}^{2}}}{{{c}^{2}}}=0.4356 \\\Rightarrow \dfrac{{{v}^{2}}}{{{c}^{2}}}=0.5644 \\ \Rightarrow \dfrac{v}{c}=0.751 \\ \therefore \dfrac{c}{v}=1.33$

 An artificial rocket is moving towards the earth with a velocity given by $0.5\,c$. At a particular instant, the rocket directly projects a cylindrical vessel of chemical gases towards the earth, with a speed given by $0.75\,c$. What will its velocity be for an observer standing on the control tower? On the other hand, if this vessel was directed away from the earth, what would its velocity be for the same observer?<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/a64576ad-2b30-444b-9efa-6b69b894922d78932558660249143.png"/> 

Since the projected vessel and the rocket move towards the earth at relativistic speeds, the speed of the velocity of the vessel for an observer on the earth can be determined using relativistic vector addition.

 $ \text{ From relativistic vector addition, we have}\\ u=\dfrac{v+u'}{1+\dfrac{vu'}{{{c}^{2}}}}\\ \text{For the first case, we have}\, v=0.500c\text{  and  }u'=0.750c\\ \text{Clearly, the velocity of the vessel for the observer standing on the control tower in this case is:}\\ u=\dfrac{0.500c+0.750c}{1+\dfrac{0.500c\times 0.750c}{{{c}^{2}}}} \\ \Rightarrow u=\dfrac{1.25c}{1.375} \\ \Rightarrow u=0.909c \\ \text{ For the second case, we have}\, v=0.500c\text{  and  }u'=-0.750c\\ \text{ Clearly, the velocity of the vessel for the observer standing on the control tower in this case is:}\\ u=\dfrac{0.500c-0.750c}{1-\dfrac{0.500c\times 0.750c}{{{c}^{2}}}} \\ \Rightarrow u=\dfrac{-0.25c}{1-0.375} \\ \therefore u=-0.400c$