An object is tracked by a radar station and determined to have a position vector given by$\overrightarrow{\mathbf{r}}\mathbf{=}\mathbf{(}\mathbf{3500}\mathbf{-}\mathbf{160}\mathbf{t}\mathbf{)}\hat{\mathbf{i}}\mathbf{+}\mathbf{2700}\hat{\mathbf{j}}\mathbf{+}\mathbf{300}\hat{\mathbf{k}}\mathbf{}\mathbf{,}\mathbf{with}\mathbf{}\overrightarrow{\mathbf{r}}$in meters and tin seconds. The radar station’s xaxis points east, its yaxis north, and its zaxis vertically up. If the object is a 250 kg meteorological missile, what are (a) its linear momentum, (b) its direction of motion, and (c) the net force on it?

The mass of an object is,$m=250\mathrm{kg}$.

The position vector of the object, $\overrightarrow{r}=\left(3500-160t\right)\hat{i}+2700\hat{j}+300\hat{k}$.

Here,we can apply the calculus knowledge along with Newton’s second law of motion.

Formula:

$$\begin{gathered} \overrightarrow{v}=\frac{d\overrightarrow{r}}{dt} \\ \overrightarrow{P}=m\overrightarrow{v} \end{gathered}$$

The equation for velocity is,

$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} \overrightarrow{v}=\frac{d\left(\left(3500-160t\right)\hat{i}+2700\hat{j}+300\hat{k}\right)}{dt} \\ =-160\hat{i} \end{gathered}$$

Linear momentum can be calculated as,

$\overrightarrow{P}=m\overrightarrow{v}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} \overrightarrow{P}=250\times \left(-160\hat{i}\right) \\ =\left(-4.0\times {10}^4\mathrm{kg}.\frac{\mathrm{m}}{\mathrm{s}}\right) \end{gathered}$$

Thus, the linear momentum of an object, $\overrightarrow{P}=\left(-4.0\times {10}^4\mathrm{kg}.\frac{\mathrm{m}}{\mathrm{s}}\right)$.

From the equation of velocity, we can say that the direction of motion of the object is (- x) direction.

Thus, the direction of motion is in the -x (west) direction .

The equation of force in terms of momentum is,

$\overrightarrow{F}=\frac{d\overrightarrow{P}}{dt}$

From part a, since the value of P does not change with time, we get,

$\overrightarrow{F}=0$

Thus, the net force on the object,$\overrightarrow{F}=0$