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Q80P

Expert-verifiedFound in: Page 1

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**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 t in seconds. The radar station’s x axis points east, its y axis north, and its z axis 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?**

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.

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:

$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}\phantom{\rule{0ex}{0ex}}\overrightarrow{P}=m\overrightarrow{v}$

The equation for velocity is,

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

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

$\overrightarrow{v}=\frac{d\left(\left(3500-160t\right)\hat{i}+2700\hat{j}+300\hat{k}\right)}{dt}\phantom{\rule{0ex}{0ex}}=-160\hat{i}$

Linear momentum can be calculated as,

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

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

$\overrightarrow{P}=250\times \left(-160\hat{i}\right)\phantom{\rule{0ex}{0ex}}=\left(-4.0\times {10}^{4}\mathrm{kg}.\frac{\mathrm{m}}{\mathrm{s}}\right)$

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, $\overrightarrow{F}=0$

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