Question

The equation you found in Part D may be factored to give(E2−E1)(E2+E1)=p22c2−p21c2.Consider the term E2+E1. Since we are looking at the "everyday" limit of relativity, where speed is far lower than the speed of light, the kinetic energy will be negligible compared to the rest energy. Using this approximation, find the value of E2+

Answer 1

Complete Question

The equation you found in Part D may be factored to give(E2−E1)(E2+E1)=p22c2−p21c2.Consider the term E2+E1. Since we are looking at the "everyday" limit of relativity, where speed is far lower than the speed of light, the kinetic energy will be negligible compared to the rest energy. Using this approximation, find the value of E2+E1

Answer:

The  value  is $(E_2 + E_1) = 2mc^2$  

Explanation:

Generally in the question we are given this equation  

    $(E_2 - E_1 )(E_2 + E_1) = p_2^2 c^2 - p_1^2 c^2$

and the objective is to obtain  the value of  $(E_2 + E_1)$ in terms of  m and  c

Generally kinetic energy is mathematically represented as

      $K = \frac{1}{2} m v^2$

=>   $K = \frac{1}{2}* \frac{m^2 v^2}{m}$

recall that momentum is mathematically represented as $p = m * v$

Hence

=>   $K = \frac{1}{2}* \frac{p^2}{m}$

Hence  $E_1 = \frac{p_1^2}{ 2m}$     and   $E_1 = \frac{p_2^2}{ 2m}$

So

       $E_2 - E_1 = \frac{p_2^2}{ 2m} - \frac{p_1^2}{ 2m}$

=>  $E_2 - E_1 = \frac{p_2^2-p_1^2 }{ 2m}$

So from the given equation

     $( \frac{p_2^2-p_1^2 }{ 2m})(E_2 + E_1) = p_2^2 c^2 - p_1^2 c^2$

=>  $( \frac{p_2^2-p_1^2 }{ 2m})(E_2 + E_1) = (p^2_2 -p^2_1 )c^2$

=>  $(E_2 + E_1) = (p^2_2 -p^2_1 )c^2 * ( \frac{p_2^2-p_1^2 }{ 2m})$

=>  $(E_2 + E_1) = 2mc^2$