Chapter 12: Q30P (page 537)

particle’s kinetic energy is ntimes its rest energy, what is its speed?

Short Answer

Expert verified

The speed is $u = \frac{\sqrt{n (n + 2)}}{n + 1} c$ .

Step by step solution

Expression for the particle’s kinetic energy:

Let the speed of a particle be u, and the rest mass be m.

It is given that the particle’s kinetic energy is n times its rest energy, write the equation based on the given condition $kinetic energy = n \times Rest energy E_{k in} = n \times E_{rest}$ …… (1)

Write the energy for the kinetic energy.

$E_{kin} = \frac{{mc}^{2}}{\sqrt{1 - \frac{u^{2}}{c^{2}}}} - {mc}^{2}$

Write the expression for the rest energy.

$E_{r e s t} = m c^{2}$

Determine the speed of a particle:

Substitute $\frac{m c^{2}}{\sqrt{1 - \frac{u^{2}}{c^{2}}}} - m c^{2} f o r E_{k i n} a n d m c^{2} f o r E_{r e s t} i n e q u a t i o n (1) . \frac{m c^{2}}{\sqrt{1 - \frac{u^{2}}{c^{2}}}} - m c^{2} = n \times m c^{2} m c^{2} (\frac{1}{\sqrt{1 - \frac{u^{2}}{c^{2}}}} - 1) = n \times m c^{2} \frac{1}{\sqrt{1 - \frac{u^{2}}{c^{2}}}} - 1 = n \frac{1}{\sqrt{1 - \frac{u^{2}}{c^{2}}}} = n + 1$

On further solving,

$\sqrt{1 - \frac{u^{2}}{c^{2}}} = \frac{1}{n + 1} 1 - \frac{u^{2}}{c^{2}} = {(\frac{1}{n + 1})}^{2} 1 - \frac{u^{2}}{c^{2}} = \frac{1}{{(n + 1)}^{2}} - \frac{u^{2}}{c^{2}} = \frac{1}{{(n + 1)}^{2}} - 1$

Solve the R.H.S:

$- \frac{u^{2}}{c^{2}} = \frac{1 - {(n + 1)}^{2}}{{(n + 1)}^{2}} - \frac{u^{2}}{c^{2}} = \frac{1 - (n^{2} + 1 + 2 n)}{{(n = 1)}^{2}} u = \frac{\sqrt{n (n + 2)}}{n + 1} c T h e r e f o r e, t h e s p e e d i s u = \frac{\sqrt{n (n + 2}}{n + 1} c .$