Question

The Large Hadron Collider accelerates protons to energies of 14 TeV.

Part A

Determine the ratio of the proton energy to the kinetic energy of a 25-mg bug crawling at 2.0 mm/s.

K bug

K bug

Determine the ratio of the magnitude of the proton momentum to that of the same bug.

$\frac{p}{p_{bug}} =$

Answer 1

$K_{bug}=\frac{1}{2}m_{bug}v_{bug}^2\\ \implies K_{bug}=\frac{1}{2}\times 25\times 10^{-3}\times (2\times 10^{-3})^2\\ \implies K_{bug}=5\times 10^{-8}J$

$E=14TeV\\ \implies E=14\times 10^{12}\times 1.6\times 10^{-19}\\ \implies E=2.24\times 10^{-6}J$

So $\frac{E}{K_{bug}}=\frac{2.24\times 10^{-6}}{5\times 10^{-8}}\\ \implies \frac{E}{K_{bug}}=44.8$

For the bug, momentum is

$p_{bug}=mv\\ \implies p_{bug}=25\times 10^{-3}\times 2\times 10^{-3}\\ \implies p_{bug}=5\times 10^{-5}kg m/s$

For the proton, momentum has to be calculated using relativistic formula

$p=\beta \frac{E}{c}$

Now

$\beta=\sqrt{1-\frac{1}{\gamma^2}}\\ \implies \beta=\sqrt{1-\frac{(E_{rest})^2}{(E_{total})^2}}\\ \implies \beta=\sqrt{1-\frac{(938MeV)^2}{(14\times 10^6)^2}}\\ \implies \beta = 0.99$

So $p=0.99\times \frac{14\times 10^{9}}{3\times 10^8}\\ \implies p=46.2kg m/s$

The ratio is

$\frac{p}{p_{bug}}=\frac{46.2}{5\times 10^{-5}}\\ \implies \frac{p}{p_{bug}}=9.24\times 10^5$