0
$\begingroup$

I was wondering that a change in velocity can change the mass of an object slightly. I thought of this since p=mv so if we rearrange it to be m=p/v. Does this mean that a change in momentum or velocity can actually change the mass?

$\endgroup$
2
  • $\begingroup$ See here. $\endgroup$
    – Charlie
    Oct 19, 2020 at 14:40
  • 1
    $\begingroup$ What do you have in mind that can change momentum but not change velocity (or vice versa)? $\endgroup$ Oct 19, 2020 at 19:38

2 Answers 2

2
$\begingroup$

You have this equation, $p = mv$. If you increase the velocity and keep $p$ constant, yes you must have had an increase in mass. The question is, what does it mean for $p$ to be constant? It is defined by the product of mass and velocity. So you have done nothing to make predictions, you have just stated what you would call mass if you increased v and kept p the same. OR since mass and velocity are more readily known as measured quantities to the layman, you have said increasing v and lowering m keeps mv the same. See? you've said nothing of any value

$\endgroup$
0
$\begingroup$

Note, for a small change in $p$ (and $v$, as their correlation is 1, or $m$ depending on normalization):

$$ \delta m = \frac d {dp}({\frac p v})\delta p+ \frac d {dv}({\frac p v})\delta v $$

$$ \delta m = \frac{\delta p} v - \frac{p\delta v}{v^2}$$

Since:

$$ \delta p = m \delta v$$

we get:

$$ \delta m = \frac{\delta p} v - \frac{p\delta p/m}{v^2}$$

$$ \delta m = \delta p(\frac 1 v - \frac{v}{v^2})$$

$$ \delta m = \delta p(\frac 1 v - \frac{1}{v}) = 0$$

So, no.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.