Question

at what velocity will the mass of a body is 2.25 times its rest mass

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Answer 1

Answer:

First I want to point out that relativistic mass is an outdated concept that is rarely used anymore. It is misleading as it implies that the object is changing in some sort of physical way, but its not.

That being said, to answer your question you need to find the speed at which γ =2.25.

Gamma ( γ ) is the lorentz factor which can be found using

γ=11−v2c2√

So if we set c to equal 1 and put 2.25 as γ then:

2.25=11−v21√

So after solving for v we find that v≈.8958c

2.252=11−v21

5.0625=11−v2

5.0625(1−v2)=1

5.0625−5.0625v2=1

−5.0625v2=−4.0625

v2=.80246913

v=.80246913−−−−−−−−√

v≈.8958

Answer 2

Answer:

At velocity equal to 0.9c, the mass will be 2,25 times its rest mass.

Explanation:

The relativistic mass is given as follows:

$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}} }$, where m is the relativistic mass, m₀ is the rest mass, v is the velocity, and c is the speed of light.

We have to find the velocity at which the relativistic mass becomes 2.25 times of its rest mass, i.e., m = 2.25m₀.

Substituting the value of m in the formula, we get

$2.25m_0 = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}} }$

$2.25 = \frac{1}{\sqrt{1-\frac{v^2}{c^2}} }$

$\sqrt{1-\frac{v^2}{c^2}} = \frac{1}{2.25}$

$\sqrt{1-\frac{v^2}{c^2}} = \frac{4}{9}$

Squaring both the sides,

$1 - \frac{v^2}{c^2} = \frac{16}{81}$

$1 - \frac{16}{81} = \frac{v^2}{c^2}$

$\frac{65}{81} = \frac{v^2}{c^2}$

$v^2 = \frac{65}{81} c^2$

Taking positive square roots on both sides, we get

$v = \frac{\sqrt{65}}{9} c$

v = 0.896c

v ≈ 0.9c

Therefore, the mass of the body will become 2.25 times the rest mass when its velocity will be 0.9 times the speed of light.

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