Q34E - University Physics with Modern Physics

Question

(a) How much work must be done on a particle with mass m to accelerate it (a) from rest to a speed of 0.090c and (b) from a speed of 0.900c to a speed of 0.990c? (Express the answers in terms of mc^2 .) (c) How do your answers in parts (a) and (b) compare?

Step-by-Step Solution

Answer

a) The work done on a particle to accelerate from rest to speed of 0.090c is $\mathbf{4}\mathbf{.}\mathbf{08}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{*}\mathit{m}{\mathit{c}}^{\mathbf{2}}$  

b) The work done on a particle to accelerate from 0.900c to 0.990c is $\mathbf{4}\mathbf{.}\mathbf{80}\mathit{m}{\mathit{c}}^{\mathbf{2}}$   

c) The greater amount of work is needed to accelerate a high speed particle than a low speed particle.

Step 1:Formula for Kinetic energy(K) and work done.

Kinetic energy,

$$\begin{gathered} \mathit{K}\mathbf{=}\frac{\mathbf{m}{\mathbf{c}}^{\mathbf{2}}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{v}}^{\mathbf{2}}\mathbf{/}{\mathbf{c}}^{\mathbf{2}}}}\mathbf{-}\mathit{m}{\mathit{c}}^{\mathbf{2}} \\ \mathit{K}\mathbf{=}\left(\gamma -1\right)\mathit{m}{\mathit{c}}^{\mathbf{2}} \end{gathered}$$ 

The work done is equal to change in kinetice energy.

  $\mathit{W}\mathbf{=}\mathit{\Delta }\mathit{K}$

Step 2:Calculating the work on a particle to accelerate from rest to speed of 0.090c.

$$\begin{gathered} \mathit{W}\mathbf{=}\mathbf{∆}\mathit{K}\mathbf{=}\mathbf{(}\mathbf{\gamma }\mathbf{-}\mathbf{1}\mathbf{)}\mathit{m}{\mathit{c}}^{\mathbf{2}}\mathbf{-}\mathbf{0} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{(}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}\mathbf{0}\mathbf{.}{\mathbf{090}}^{\mathbf{2}}}}\mathbf{-}\mathbf{1}\mathbf{)}\mathit{m}{\mathit{c}}^{\mathbf{2}}\mathbf{-}\mathbf{0} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{08}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{)}\mathit{m}{\mathit{c}}^{\mathbf{2}} \end{gathered}$$

Step 3:Calculating the work done on a particle to accelerate from 0.900c to 0.990c.

$$\begin{gathered} \mathit{W}\mathbf{=}\mathbf{∆}\mathit{K} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}{\mathit{K}}_{\mathbf{f}}\mathbf{-}{\mathit{K}}_{\mathbf{i}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\left({\gamma }_f-1\right)\mathit{m}{\mathit{c}}^{\mathbf{2}}\mathbf{-}\left({\gamma }_i-1\right)\mathit{m}{\mathit{c}}^{\mathbf{2}}\mathbf{ } \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\left({\gamma }_f-{\gamma }_i\right)\mathit{m}{\mathit{c}}^{\mathbf{2}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\left(\frac{1}{\sqrt{1-0.{990}^2}}-\frac{1}{\sqrt{1-0.{990}^2}}\right)\mathit{m}{\mathit{c}}^{\mathbf{2}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{80}\mathit{m}{\mathit{c}}^{\mathbf{2}} \end{gathered}$$