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Fundamentals Of Physics

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$\stackrel{\mathbf{\to }}{\mathbf{r}}{\mathbf{=}}\mathbf{\left(}\mathbf{3500}\mathbf{-}\mathbf{160}\mathbf{t}\mathbf{\right)}\stackrel{\mathbf{^}}{\mathbf{i}}{\mathbf{+}}{\mathbf{2700}}\stackrel{\mathbf{^}}{\mathbf{j}}{\mathbf{+}}{\mathbf{300}}\stackrel{\mathbf{^}}{\mathbf{k}}{\mathbf{}}{\mathbf{,}}{\mathbf{with}}{\mathbf{}}\stackrel{\mathbf{\to }}{\mathbf{r}}$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 $\left(-4.0×{10}^{4}\mathrm{kg}.\frac{\mathrm{m}}{\mathrm{s}}\right)\stackrel{^}{i}$.

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

c) The net force on the object is 0.

See the step by step solution

Step 1: Listing the given quantities

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

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

Step 2: 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:

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

Step 3: (a) Calculation of linear momentum

The equation for velocity is,

$\stackrel{\to }{v}=\frac{d\stackrel{\to }{r}}{dt}$

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

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

Linear momentum can be calculated as,

$\stackrel{\to }{P}=m\stackrel{\to }{v}$

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

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

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

Step 4: (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 .

Step 5: (c) Calculation of net force

The equation of force in terms of momentum is,

$\stackrel{\to }{F}=\frac{d\stackrel{\to }{P}}{dt}$

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

$\stackrel{\to }{F}=0$

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