Question

The theory of relativity predicts that an object whose mass is m0 when it is at rest will appear heavier when it is moving at speeds near the speed of light. When the object is moving at speed v, its mass m is given by

Answer 1

The theory of relativity predicts that an object whose mass is \[m _{0}\] when it is at rest will appear heavier when it is moving at speeds near the speed of light. When the object is moving at speed v, its mass m is given by \[m=\frac{ m _{0} }{ \sqrt{1-(v^2/c^2)} }\] where c is the speed of light. Find dm/dv and explain in terms of physics what this quantity tells you

Answer 2

am I just taking the derivative of this?

Answer 3

yeah you would take the derivative

Answer 4

m = m_0 - (1-(v^2/c^2))^1/2

Answer 5

\[m = m_0 - (1-(v^2/c^2))^-1/2\]

Answer 6

then \[m' = 1 +1/2(1-(v^2/c^2))^{-3/2}\]

Answer 7

Where did the minus sign come from?

Answer 8

because it was originally division

Answer 9

should I use the quotient rule?

Answer 10

\[\large \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} = m_0((1 - \frac{v^2}{c^2})^{-1/2})\]

proceed with the chain rule *or yes if you prefer, the quotient rule

Answer 11

You can always write division as multiplication....but there would be no minus sign added on

Answer 12

\[m_0((1-\frac{ v^2 }{ c^2 })^{-1/2}*\frac{ -1 }{ 2 }(1-\frac{ v^2 }{ c^2 }*(c^2)(2v)-v^2(2c))\]

Answer 13

how wrong am I?