Question

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?

Short answer

a) The linear momentum of an object is $\left(-4.0\times {10}^4\mathrm{kg}.\frac{\mathrm{m}}{\mathrm{s}}\right)\hat{i}$.

b) The direction of motion of the object is along the negative x-axis.

c) The net force on the object is 0.

Step-by-step solution

Listing the given quantities

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

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

Understanding the concept of the law of conservation of momentum

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

Formula:

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

(a) Calculation of linear momentum

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, $\overrightarrow{P}=\left(-4.0\times {10}^4\mathrm{kg}.\frac{\mathrm{m}}{\mathrm{s}}\right)$.

(b) Direction of motion

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 .

(c) Calculation of net force

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,$\overrightarrow{F}=0$