91Ó°ÊÓ

a)

An object at rest has some energy called rest mass energy and is expressed as

$E=m{c}^2$

Lets use 1,2, and 3 subscripts for three particles respectively. The rest mass energy for three particles are given below

$\begin{array}{l}{E}_1=A\\ E_2=A\\ E_3=3A\end{array}$width="66">E1=AE2=AE3=3A

As rest mass energy is only dependent on mass with $c^2$ term being constant. Therefore, the particle with greatest rest energy will have greatest mass, here it is the third particle. As 1^st and 2^nd particle has same rest energy will have same mass.

$m_3>m_1=m_2$

(b)

Kinetic energy can be expressed as total energy minus rest mass energy.

$k=T-E$

Here, in the problem the total energies and rest energies are given, therefore the kinetic energies of the three particles are

$\begin{array}{l}{K}_1=2A-A=A\\ K_2=3A-A=2A\\ K_4=4A-3A=A\end{array}$

Therefore,

$K_2>K_1=K_3$

The result of 2^nd postulate of the special theory of relativity is that the clocks run slower for a moving object when measured from a rest frame. The factor by which the clock is running differently is called the Lorentz factor.

The expression for Lorentz factor is

$Y=\frac{1}{\sqrt{1-{\mathrm{β}}^2}}$

Here $\mathrm{β}$is the speed parameter $v/c$.

The expression for total energy of an object is

$T=\mathrm{γ}m{c}^2$

The total energies of the three particles is

$\begin{array}{l}{T}_1=2A\\ T_2=3A\\ T_3=4A\end{array}$

Rankings will be

$$\begin{gathered} T_3>T_2>T_1 \\ {\mathrm{γ}}_3{m}_3{c}^2>{\mathrm{γ}}_2{m}_2{c}^2>{\mathrm{γ}}_1{m}_1{c}^2 \\ {\mathrm{γ}}_3{m}_3>{\mathrm{γ}}_2{m}_2>{\mathrm{γ}}_1{m}_1 \end{gathered}$$

Let’s consider particle 1 and 2

${\mathrm{γ}}_2{m}_2>{\mathrm{γ}}_1{m}_1$

As $m_1=m_2$, therefore, ${\mathrm{γ}}_2>{\mathrm{γ}}_1$.

And for particle 3, the mass and total energy, both are greatest, therefore will be greatest.

${\mathrm{γ}}_3>{\mathrm{γ}}_2>{\mathrm{γ}}_1$

(d)

And, as Lorentz factor is related to speed with relation

$$\begin{gathered} \mathrm{γ}=\frac{1}{\sqrt{1-{\mathrm{β}}^2}} \\ \mathrm{β}=\sqrt{1-\frac{1}{{\mathrm{γ}}^2}} \\ v=c\left(\sqrt{1-\frac{1}{{\mathrm{γ}}^2}}\right) \end{gathered}$$

Therefore,

$V_3>V_2>V_1$

Hence, greatest $\mathrm{γ}$will have greatest speed.