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Corrected typo in eq. (2)
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Qmechanic
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Confused what equation I should use,
$$m=m_0+\frac12\frac{m_0v^2}{c^2}\tag{1}$$ or $$m=m_0\sqrt{1-\frac{v^2}{c^2}}\tag{2}$$$$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\tag{2}$$
when solving for relativistic mass. When I plugged in $0.833c$ for velocity and $5\ \mathrm{kg}$ for rest mass, the equations gave two different answers about $3\ \mathrm{kg}$ apart. I can find $(2)$ all over the internet, but finding $(1)$ is harder. Is $(1)$ a true relationship?

Confused what equation I should use,
$$m=m_0+\frac12\frac{m_0v^2}{c^2}\tag{1}$$ or $$m=m_0\sqrt{1-\frac{v^2}{c^2}}\tag{2}$$
when solving for relativistic mass. When I plugged in $0.833c$ for velocity and $5\ \mathrm{kg}$ for rest mass, the equations gave two different answers about $3\ \mathrm{kg}$ apart. I can find $(2)$ all over the internet, but finding $(1)$ is harder. Is $(1)$ a true relationship?

Confused what equation I should use,
$$m=m_0+\frac12\frac{m_0v^2}{c^2}\tag{1}$$ or $$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\tag{2}$$
when solving for relativistic mass. When I plugged in $0.833c$ for velocity and $5\ \mathrm{kg}$ for rest mass, the equations gave two different answers about $3\ \mathrm{kg}$ apart. I can find $(2)$ all over the internet, but finding $(1)$ is harder. Is $(1)$ a true relationship?

deleted 145 characters in body; edited title
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user59991
user59991

$m = m_0 + $m=m_0+(½1/2) m_0v^2m_0v^2/c^2$ vs $ m=m_0$m=m_0/\sqrt{1-v^2/c^2}$

Confused what equation I should use,
$\displaystyle{m = m_o + \frac{1}{2} \frac{m_ov^2}{c^2}}$ or $\displaystyle{m=m_o\sqrt{1-\frac{v^2}{c^2}}}$,$$m=m_0+\frac12\frac{m_0v^2}{c^2}\tag{1}$$ or $$m=m_0\sqrt{1-\frac{v^2}{c^2}}\tag{2}$$
when solving for relativistic mass. When I plugged in $0.833c$ for velocity and $5kg$$5\ \mathrm{kg}$ for rest mass, the equations gave two different answers about $3 kgs$$3\ \mathrm{kg}$ apart. I can find $\displaystyle{m=m_o\sqrt{1-\frac{v^2}{c^2}}}$$(2)$ all over the internet, but finding $\displaystyle{m = m_o + \frac{1}{2} \frac{m_ov^2}{c^2}}$$(1)$ is harder. Is $\displaystyle{m = m_o + \frac{1}{2} \frac{m_ov^2}{c^2}}$$(1)$ a true relationship?

$m = m_0 + (½) m_0v^2/c^2$ vs $ m=m_0/\sqrt{1-v^2/c^2}$

Confused what equation I should use,
$\displaystyle{m = m_o + \frac{1}{2} \frac{m_ov^2}{c^2}}$ or $\displaystyle{m=m_o\sqrt{1-\frac{v^2}{c^2}}}$,
when solving for relativistic mass. When I plugged in $0.833c$ for velocity and $5kg$ for rest mass, the equations gave two different answers about $3 kgs$ apart. I can find $\displaystyle{m=m_o\sqrt{1-\frac{v^2}{c^2}}}$ all over the internet, but finding $\displaystyle{m = m_o + \frac{1}{2} \frac{m_ov^2}{c^2}}$ is harder. Is $\displaystyle{m = m_o + \frac{1}{2} \frac{m_ov^2}{c^2}}$ a true relationship?

$m=m_0+(1/2)m_0v^2/c^2$ vs $m=m_0/\sqrt{1-v^2/c^2}$

Confused what equation I should use,
$$m=m_0+\frac12\frac{m_0v^2}{c^2}\tag{1}$$ or $$m=m_0\sqrt{1-\frac{v^2}{c^2}}\tag{2}$$
when solving for relativistic mass. When I plugged in $0.833c$ for velocity and $5\ \mathrm{kg}$ for rest mass, the equations gave two different answers about $3\ \mathrm{kg}$ apart. I can find $(2)$ all over the internet, but finding $(1)$ is harder. Is $(1)$ a true relationship?

improved formatting, corrected spelling
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UKH
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Confused what equation I should use, m = mo + ½ mov2/c2
$\displaystyle{m = m_o + \frac{1}{2} \frac{m_ov^2}{c^2}}$ or m=mo/sqrt(1-v2/c2)$\displaystyle{m=m_o\sqrt{1-\frac{v^2}{c^2}}}$, when
when solving for relativistic mass. When I plugged in .833c$0.833c$ for velocity and 5kg$5kg$ for rest mass, the equations gave two different answers about 3 kgs$3 kgs$ apart. I can find m=mo/sqrt(1-v2/c2)$\displaystyle{m=m_o\sqrt{1-\frac{v^2}{c^2}}}$ all over the internet, but finding m = mo + ½ mov2/c2$\displaystyle{m = m_o + \frac{1}{2} \frac{m_ov^2}{c^2}}$ is harder. Is m = mo + ½ mov2/c2$\displaystyle{m = m_o + \frac{1}{2} \frac{m_ov^2}{c^2}}$ a true relantionshiprelationship?

Confused what equation I should use, m = mo + ½ mov2/c2 or m=mo/sqrt(1-v2/c2), when solving for relativistic mass. When I plugged in .833c for velocity and 5kg for rest mass, the equations gave two different answers about 3 kgs apart. I can find m=mo/sqrt(1-v2/c2) all over the internet, but finding m = mo + ½ mov2/c2 is harder. Is m = mo + ½ mov2/c2 a true relantionship?

Confused what equation I should use,
$\displaystyle{m = m_o + \frac{1}{2} \frac{m_ov^2}{c^2}}$ or $\displaystyle{m=m_o\sqrt{1-\frac{v^2}{c^2}}}$,
when solving for relativistic mass. When I plugged in $0.833c$ for velocity and $5kg$ for rest mass, the equations gave two different answers about $3 kgs$ apart. I can find $\displaystyle{m=m_o\sqrt{1-\frac{v^2}{c^2}}}$ all over the internet, but finding $\displaystyle{m = m_o + \frac{1}{2} \frac{m_ov^2}{c^2}}$ is harder. Is $\displaystyle{m = m_o + \frac{1}{2} \frac{m_ov^2}{c^2}}$ a true relationship?

edited tags; edited title
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