t what speed is the magnitude of the relativistic momentum of a particle three times the
nagnitude of the nonrelativistic momentum?

Question

t what speed is the magnitude of the relativistic momentum of a particle three times the nagnitude of the nonrelativistic momentum?

UpStudy AI · Accepted Answer

We know that the relativistic momentum is given by  
$$
p_{\text{rel}} = \gamma m v,
$$  
where  
$$
\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}.
$$  
The nonrelativistic momentum is  
$$
p_{\text{nr}} = m v.
$$

The problem states that  
$$
p_{\text{rel}} = 3p_{\text{nr}}.
$$

Substituting the expressions for the momenta gives:
$$
\gamma m v = 3(m v).
$$

Assuming $$m \neq 0$$ and $$v \neq 0$$, we can cancel $$m v$$ from both sides, which yields:
$$
\gamma = 3.
$$

Substitute the definition of $$\gamma$$:
$$
\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = 3.
$$

Solve for the square root:
$$
\sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{3}.
$$

Square both sides:
$$
1 - \frac{v^2}{c^2} = \frac{1}{9}.
$$

Rearrange to solve for $$\frac{v^2}{c^2}$$:
$$
\frac{v^2}{c^2} = 1 - \frac{1}{9} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9}.
$$

Take the square root on both sides:
$$
\frac{v}{c} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}.
$$

Thus, the speed is:
$$
v = \frac{2\sqrt{2}}{3} c.
$$
The speed is $$ \frac{2\sqrt{2}}{3} $$ times the speed of light.

t what speed is the magnitude of the relativistic momentum of a particle three times the
nagnitude of the nonrelativistic momentum?

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