Question

A proton with the rest mass of 1.67 x 10 kg is initially moving at 0.90c in the +z-direction, relative to the reference frame of a physics laboratory. Then a constant force

F = (-3.0 x 10 12 N) i+ (5.0 × 10- N)j

is applied to the proton. What is the angle between the applied force and the resulting acceleration of the proton?

Answer 1

Answer

the angle between the applied force

$\tan \theta=\frac{F_y}{F_x}=\frac{5.0\times10^{12}}{ 3.0\times10^{12} }$

$\theta=\arctan{(1.67)}=60°$

calculating acceleration

$a=\frac{F}{m}$

For mass

$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{16.7}{0.44}=38Kg$

Force

$F=\sqrt{F^2_x+F^2_y}=5\times10^{6}N$

So acceleration

$a=\frac{5\times10^{6}}{38}=132 Km/sec^2$