An object moving at a constant speed of $\mathbf{23}\mathbf{m}\mathbf{/}\mathbf{s}$is making a turn with a radius of curvature of $\mathbf{4}\mathbf{m}$(this is the radius of the kissing circle). The object's momentum has a magnitude of . What is the magnitude of the rate of change of the momentum? What is the magnitude of the net force?

An object moving at a constant speed of$v=23\mathrm{m}/\mathrm{s}$ is making a turn with a radius of curvature of $r=4\mathrm{m}$ (this is the radius of the kissing circle). The object's momentum has a magnitude of $\stackrel{\rightharpoonup }{p}=78\mathrm{kg}-\mathrm{m}/\mathrm{s}$.

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$\left|\stackrel{\rightharpoonup }{p}\right|\mathbf{=}\mathit{m}\mathit{v}$

From the definition of momentum $\mathrm{P}$.

$$\begin{gathered} m=\frac{\left|\stackrel{\rightharpoonup }{p}\right|}{v} \\ =\frac{78\mathrm{kg}-\mathrm{m}/\mathrm{s}}{23\mathrm{m}/\mathrm{s}} \\ =3.39\mathrm{kg} \\ \left|\frac{d\stackrel{\rightharpoonup }{p}}{dt}\right|=\left|\stackrel{\rightharpoonup }{\mathrm{F}}\right| \\ =m\frac{v^2}{r} \\ =3.38kg\times \frac{{23}^2{m}^2/s^2}{4m} \\ =448.5\mathrm{Kg}.\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

From Newton’s second law, the rate of change of momentum equals the net force. The radial force equals times the radial acceleration which is given by .

Therefore, The magnitude of rate of change of momentum is$448.5\mathrm{Kg}.\mathrm{m}/{\mathrm{s}}^2$, and $448.5\mathrm{Kg}.\mathrm{m}/{\mathrm{s}}^2$is .