91Ó°ÊÓ

We are given the speed of the object v=c where c is the speed of light and equals $3×{10}^8\mathrm{m}/\mathrm{s}$ and the radius of the path is R=8m and the magnitude of the perpendicular force is${\mathrm{F}}_{âŠ\yen }=2×{10}^{-10}\mathrm{N}$

In Newtonian mechanics, linear momentum, translational momentum, or simply momentum is the product of the mass and velocity of an object. It is a vector quantity, possessing a magnitude and a direction.

We are asked to determine the magnitude of the momentum P. The net force on an object can be expressed as the sum of two parts and is equal the of the momentum. The two parts that we are taken about are the parallel rate of change of the momentum $\frac{d\stackrel{→}{t}}{dt}$and the perpendicular rate of change of the momentum $\frac{d\stackrel{→}{p}}{dt}$. So, the change of momentum and the net force are given by

$F_{net}=\frac{d\stackrel{→}{p}}{dt}=\frac{d\stackrel{→}{p}}{dt}|{|}_{\left|\right|}+\frac{d\stackrel{→}{p}}{d{t}_{âŠ\yen }}$

The parallel rate of change of the momentum changes the speed of the object and as we are given the speed is constant, therefore, the parallel is

$\frac{d\stackrel{→}{p}}{dt}|{|}_{\left|\right|}=0$

Hence, the rate change here is the change in direction due to the perpendicular rate of change. The magnitude of the perpendicular rate change at speeds much less than the speed of light is given by

$$\begin{gathered} F_{net}=\frac{d\stackrel{→}{p}}{dt} \\ =\frac{m{v}^2}{R} \\ {F}_{net}=\left(mv\right)\frac{v}{R} \\ {F}_{net}=p\frac{v}{R} \end{gathered}$$

Now we can plug our values for $F_{net}$,v and R into equation (2) to get p

$$\begin{gathered} p=\frac{F_{net}}{vIR} \\ =\frac{\left(2×{10}^{-10}\mathrm{N}\right)}{\left(3×{10}^8\mathrm{m}/\mathrm{s}\right)/\left(8\mathrm{m}\right)} \\ =5.34×{10}^{-18}\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

The magnitude of the momentum is $p=5.34×{10}^{-18}\mathrm{kg}.\mathrm{m}/\mathrm{s}$