Question

Particle A has half the mass and twice the speed of particle B. Is the momentum ${\mathrm{p}}_{\mathrm{A}}$ less than, greater than, or equal to ${\mathrm{p}}_{\mathrm{B}}$? Explain.

Short answer

Momentum A is greater than momentum of B.

Step-by-step solution

Given information

We have given,

Mass of A=$\frac{1}{2}$Mass of B

Speed of A= $2\times$Speed of B

We have to find the momentum of the A and B.

Simplify

Momentum of A, and B is relativistic here,

Then,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{A}}=\frac{{\mathrm{M}}_{\mathrm{A}}{\mathrm{V}}_{\mathrm{A}}}{\sqrt{1-{\displaystyle \frac{{\mathrm{V}}_{\mathrm{A}}^2}{{\mathrm{c}}^2}}}} \\ {\mathrm{P}}_B={\mathrm{M}}_{\mathrm{B}}{\mathrm{V}}_{\mathrm{B}}\times \frac{1}{\sqrt{1-{\displaystyle \frac{{\mathrm{V}}_{\mathrm{B}}^2}{{\mathrm{c}}^2}}}} \\ \frac{{\mathrm{P}}_{\mathrm{B}}}{{\mathrm{P}}_{\mathrm{A}}}=\frac{{\mathrm{M}}_{\mathrm{A}}{\mathrm{V}}_{\mathrm{A}}\sqrt{1-\frac{{\mathrm{V}}_{\mathrm{B}}^2}{{\mathrm{c}}^2}}}{{\mathrm{M}}_{\mathrm{B}}{\mathrm{V}}_{\mathrm{B}}\sqrt{1-{\displaystyle \frac{{\mathrm{V}}_{\mathrm{A}}^2}{{\mathrm{c}}^2}}}} \\ \frac{{\mathrm{P}}_{\mathrm{B}}}{{\mathrm{P}}_{\mathrm{A}}}=\frac{{\mathrm{M}}_{\mathrm{A}}\left(2{\mathrm{V}}_{\mathrm{B}}\right)\sqrt{1-\frac{{\mathrm{V}}_{\mathrm{B}}^2}{{\mathrm{c}}^2}}}{\left(2{\mathrm{M}}_{\mathrm{A}}\right){\mathrm{V}}_{\mathrm{B}}\sqrt{1-{\displaystyle \frac{4{\mathrm{V}}_{\mathrm{B}}^2}{{\mathrm{c}}^2}}}} \\ {\left(\frac{{\mathrm{P}}_{\mathrm{B}}}{{\mathrm{P}}_{\mathrm{A}}}\right)}^2=\frac{(1-\frac{{\mathrm{V}}_{\mathrm{B}}^2}{{\mathrm{c}}^2})}{(1-\frac{4{\mathrm{V}}_{\mathrm{B}}^2}{{\mathrm{c}}^2})} \\ {\left(\frac{{\mathrm{P}}_{\mathrm{B}}}{{\mathrm{P}}_{\mathrm{A}}}\right)}^2=\frac{({\mathrm{c}}^2-{\mathrm{V}}_{\mathrm{B}}^2)}{({\mathrm{c}}^2-4{\mathrm{V}}_{\mathrm{B}}^2)} \end{gathered}$$

Here we can see that the denominator will be greater than numerator then we can say that momentum of A is greater than momentum of B.