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(a) Use numerical and graphical evidence to guess…

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Problem 54 Medium Difficulty

In the theory of relativity, the mass of a particle with velocity $ v $ is
$$ m = \frac{m_0}{\sqrt{1 - v^2/c^2}} $$
where $ m_0 $ is the mass of the particle at rest and $ c $ is the speed of light. What happens as $ v \to c^- $?

Answer

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Problem 55

Video Transcript

Okay. So M is the mass of an object and M sub not. Is the initial mass How much it way before he started moving around? Okay. V is the velocity of the mass and C. Is the speed of light? So the question is, what happens to this mass? This mass as your as its velocity approaches the speed of light? I put a little minus sign there because I mean I'm approaching it from numbers less than it. Because we have to slowly get up. We have to come from a slower speed to get to see. Okay, so as we get close to see then what happens to V squared over C squared? Well B and C are almost the same then V squared over C squared is almost one. Okay, so we have um sub zero over the square to one minus one nipple. Little minus sign here and buy this minus. I mean the same thing. It's a little bit less than one. Okay, so the bottom is one minus something. A tiny little bit less than it. So that's zero. Okay, but it's positive zero. It's a little bit bigger than zero because this is a little bit smaller than one. So it's square root is a little bit bigger than zero. So what happens if you have a number and you divide it by a tiny tiny tiny little thing. That's almost zero. Well it goes to infinity infinite mass. As its velocity approaches the speed of light, its mass becomes infinite

Linda H.

Oklahoma State University

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Problem

(a) Use numerical and graphical evidence to guess…

Problem 54 Medium Difficulty

In the theory of relativity, the mass of a particle with velocity $ v $ is
$$ m = \frac{m_0}{\sqrt{1 - v^2/c^2}} $$
where $ m_0 $ is the mass of the particle at rest and $ c $ is the speed of light. What happens as $ v \to c^- $?

Answer

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