Chapter 26, Problem 18P

(a)

To determine

The error percentage of a particle travels at given speed.

Answer to Problem 18P

  $\text{-0}\text{.50\%}$ .

Explanation of Solution

Given:

Speed is $0.10c$

Formula used:

It is known that the momentum of proton or particle:

  $p=\frac{m_{0}v}{\sqrt{1-\left(\frac{v^2}{c^2}\right)}}$

Where,

  $m_0$ is rest mass

  $v$ is velocity

  $c$ is speed of light, $3.00\times {10}^{8}m/s$

Calculation:

The momentum of the particle is $p=\frac{m_{0}v}{\sqrt{1-\left(\frac{v^2}{c^2}\right)}}$

  $p=\frac{m_0\left(0.10c\right)}{\sqrt{1-{\left(0.10\right)}^2}}$

  $p=0.1005c{m}_0$

The classical formula gives $p=m_{0}v=0.10c{m}_0$

So, the error is

  $\frac{0.10c{m}_0-0.1005c{m}_0}{0.1005c{m}_0}=-0.005=-0.50\%$

So, the error is $-0.50\%$

(b)

To determine

The error percentage of a particle travels at given speed.

Answer to Problem 18P

  $\text{-13\%}$ .

Explanation of Solution

Given:

Speed is $0.50c$

Formula used:

It is known that the momentum of proton or particle:

  $p=\frac{m_{0}v}{{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}}$

Where,

  $m_0$ is rest mass

  $v$ is velocity

  $c$ is speed of light, $3.00\times {10}^{8}m/s$

Calculation:

The momentum of the particle is $p=\frac{m_{0}v}{{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}}$

  $p=\frac{m_0\left(0.50c\right)}{\sqrt{1-{\left(0.50\right)}^2}}$

  $p=0.57735c{m}_0$

The classical formula gives $p=m_{0}v=0.50c{m}_0$

So, the error is

  $\frac{0.50c{m}_0-0.57735c{m}_0}{0.57735c{m}_0}=-0.13=-13\%$

So, the error is $-13\%$