Question

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?

Short answer

The magnitude of rate of change of momentum is$448.5\mathrm{kg}-\mathrm{m}/{\mathrm{s}}^2$, and $\mathrm{magnitude}\mathrm{of}\mathrm{rate}\mathrm{of}\mathrm{the}\mathrm{net}\mathrm{force}\mathrm{is}448.5\mathrm{kg}-\mathrm{m}/{\mathrm{s}}^2$

Step-by-step solution

Given data

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 $\stackrel{\rightharpoonup }{p}=78\mathrm{kg}-\mathrm{m}/\mathrm{s}$.

The concept and the formula used

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

Calculate the value using the formula

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 .