Particle $A$ has mass $m$ and speed $|\vec{v}|$. Particle $B$ has mass $\frac{1}{3} m$ and speed $3|\vec{v}|$. How do the magnitudes of their relativistic momenta compare? A) $\left|\overrightarrow{p_{A}}\right|\left|\overrightarrow{p_{B}}\right|$ B) $\left|\overrightarrow{p_{A}}\right|

Question

Particle $A$ has mass $m$ and speed $|\vec{v}|$. Particle $B$ has mass $\frac{1}{3} m$ and speed $3|\vec{v}|$. How do the magnitudes of their relativistic momenta compare?
A) $\left|\overrightarrow{p_{A}}\right|>\left|\overrightarrow{p_{B}}\right|$
B) $\left|\overrightarrow{p_{A}}\right|<\left|\overrightarrow{p_{B}}\right|$
C) $\left|\overrightarrow{p_{A}}\right|=\left|\overrightarrow{p_{B}}\right|$

Quizplus · Accepted Answer

The relativistic momentum of a particle is given by $p = \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}$, where $m$ is the mass, $v$ is the speed, and $c$ is the speed of light. For particle $A$, the momentum is $p_A = \frac{m|\vec{v}|}{\sqrt{1-\frac{|\vec{v}|^2}{c^2}}}$. For particle $B$, the momentum is $p_B = \frac{\frac{1}{3}m \cdot 3|\vec{v}|}{\sqrt{1-\frac{(3|\vec{v}|)^2}{c^2}}}$ = $\frac{m|\vec{v}|}{\sqrt{1-\frac{9|\vec{v}|^2}{c^2}}}$. Because the denominator for $p_B$ is smaller due to the $9|\vec{v}|^2$ term (making the square root larger than in the denominator for $p_A$), the magnitude of $p_B$ is greater than $p_A$. 
Basically, the square root factor will enhance the momentum of objects traveling closer to the speed of light.

Particle AAA Has Mass mmm And Speed ∣v⃗∣|\vec{v}|∣v∣ Particle BBB Has Mass 13m\frac{1}{3} m31​m

Particle $A$ Has Mass $m$ And Speed $|\vec{v}|$ Particle $B$ Has Mass $\frac{1}{3} m$