A famous relation in physics relates 'moving mass' m to the rest mass m0 of a particle in terms of its speed v and the speed of light c. (This relation first arose as a consequence of special relativity by Albert Einstein). A student recalls the relation almost correctly but forgets where to put the constant c . He should write: m=

Question

A famous relation in physics relates 'moving mass' m to the rest mass m0 of a particle in terms of its speed v and the speed of light c. (This relation first arose as a consequence of special relativity by Albert Einstein). A student recalls the relation almost correctly but forgets where to put the constant c . He should write: m= (A) (m0 c/(1-v2)(1)2) (B) (m0 c2/(1-v2)(1)2) (C) (m0/(1-(v2)c)(1)2 (D) (m0/(1-(v2)c2)(1)2

Tardigrade Admin · Accepted Answer

As [(v2/c2)]= dimensionless. So [(m0/(1-(v2)c2)(1)2]=[m] All others, do not have the dimensions of mass.

$\frac{m _{0} c}{( 1 - v ^{2} ) ^{\frac{1}{2}}}$

$\frac{m _{0} c ^{2}}{( 1 - v ^{2} ) ^{\frac{1}{2}}}$

$\frac{m _{0}}{( 1 - \frac{v ^{2}}{c} ) ^{\frac{1}{2}}}$

$\frac{m _{0}}{( 1 - \frac{v ^{2}}{c ^{2}} ) ^{\frac{1}{2}}}$

As $[\frac{v ^{2}}{c ^{2}}] =$ dimensionless.
So $[\frac{m _{0}}{( 1 - \frac{v ^{2}}{c ^{2}} ) ^{\frac{1}{2}}}] = [m]$
All others, do not have the dimensions of mass.

(1−v2)21​m0​c​

(1−v2)21​m0​c2​

(1−cv2​)21​m0​​

(1−c2v2​)21​m0​​

As [c2v2​]= dimensionless. So [(1−c2v2​)21​m0​​]=[m] All others, do not have the dimensions of mass.

$\frac{m _{0} c}{( 1 - v ^{2} ) ^{\frac{1}{2}}}$

$\frac{m _{0} c ^{2}}{( 1 - v ^{2} ) ^{\frac{1}{2}}}$

$\frac{m _{0}}{( 1 - \frac{v ^{2}}{c} ) ^{\frac{1}{2}}}$

$\frac{m _{0}}{( 1 - \frac{v ^{2}}{c ^{2}} ) ^{\frac{1}{2}}}$

As $[\frac{v ^{2}}{c ^{2}}] =$ dimensionless.
So $[\frac{m _{0}}{( 1 - \frac{v ^{2}}{c ^{2}} ) ^{\frac{1}{2}}}] = [m]$
All others, do not have the dimensions of mass.