StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q115P

Expert-verifiedFound in: Page 256

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**At time t = 0****, force $\overrightarrow{\mathbf{f}}{\mathbf{=}}{\mathbf{(}}{\mathbf{\_}}{\mathbf{4}}{\mathbf{.}}{\mathbf{00}}{\mathbf{i}}{\mathbf{^}}{\mathbf{+}}{\mathbf{5}}{\mathbf{.}}{\mathbf{00}}{\mathbf{j}}{\mathbf{^}}{\mathbf{)}}$**** acts on an initially stationary particle of mass**** ${\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{10}}{\mathbf{\times}}{\mathbf{3}}{\mathbf{}}{\mathbf{kg}}$and force $\overrightarrow{\mathbf{f}}{\mathbf{=}}{\mathbf{2}}{\mathbf{\_}}{\mathbf{(}}{\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{\mathbf{i}}{\mathbf{^}}{\mathbf{\_}}{\mathbf{4}}{\mathbf{.}}{\mathbf{00}}{\mathbf{j}}{\mathbf{^}}{\mathbf{)}}$**** acts on an initially stationary particle of mass ${\mathbf{4}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{10}}{\mathbf{\times}}{\mathbf{3}}{\mathbf{kg}}$****. From time ${{\mathbf{t}}}_{{\mathbf{0}}}$**** to t = 2.00 ms****, what are the (a) magnitude and (b) angle (relative to the positive direction of the x axis) of the displacement of the center of mass of the two particle system? (c) What is the kinetic energy of the center of mass at t = 2.00 ms**

- Magnitude of displacement of center of mass is 0.745 mm.
- Angle relative to the positive direction of x axis is $153\xb0$.
- Kinetic energy of center of mass at
*t*= 2.00 s is 0.00167 J.

**Calculate displacement of two particles individually. Use this displacement to calculate the displacement of the center of mass. Calculate velocity of center of mass of two particles. Using this velocity, find the kinetic energy of two particle system.**

Formulae are as follow:

$F=ma\phantom{\rule{0ex}{0ex}}x-{x}_{0}={v}_{0}t+0.5a{t}^{2}\phantom{\rule{0ex}{0ex}}v={v}_{0}+at$

Here, *m* is mass, *v* is velocity, *a* is an acceleration, *t* is time, *F* is force and *x* is displacement.

According to Newton’s second law, displacement is as follows:

For particle 1,

${d}_{1}={v}_{i}t+0.5{a}_{1}{t}^{2}$

As the initial velocity is zero,

${d}_{1}=0.5{a}_{1}{t}^{2}$ ….. (1)

From Newton’s second law of motion:

. ${F}_{1}={m}_{1}{a}_{1}\phantom{\rule{0ex}{0ex}}\therefore {a}_{1}=\frac{{F}_{1}}{{m}_{1}}$ …… (2)

Substituting equation (2) in equation (1) and solve as:

${d}_{1}=0.5\frac{{F}_{1}}{{m}_{1}}{t}^{2}\phantom{\rule{0ex}{0ex}}{d}_{1}{m}_{1}=0.5{F}_{1}{t}^{2}$ ….. (3)

For particle 2, write the equation as:

${d}_{2}={v}_{i}t+0.5{a}_{2}{t}^{2}\phantom{\rule{0ex}{0ex}}{d}_{2}=0.5{a}_{2}{t}^{2}$ ….. (5)

Now, write the equation as:

${F}_{2}={m}_{2}{a}_{2}\phantom{\rule{0ex}{0ex}}{a}_{2}=\frac{{F}_{2}}{{m}_{2}}$ …… (5)

Substituting equation (5) in equation (4) and solve as:

${d}_{2}=0.5\frac{{F}_{2}}{{m}_{2}}{t}^{2}\phantom{\rule{0ex}{0ex}}{d}_{2}{m}_{2}=0.5{F}_{2}{t}^{2}$ ….. (6)

Now, center of mass of two particles is as follows:

${d}_{cm}=\frac{{d}_{1}{m}_{1}+{d}_{2}{m}_{2}}{{m}_{1}+{m}_{2}}\phantom{\rule{0ex}{0ex}}{d}_{cm}=\frac{{d}_{1}{m}_{1}}{{m}_{1}+{m}_{2}}+\frac{{d}_{2}{m}_{2}}{{m}_{1}+{m}_{2}}$

Substitute the values from equation (4) and (6) and solve as:

role="math" localid="1661314168405" ${d}_{cm}=0.5\times \frac{(-4\hat{i}+5\hat{j})+(2\hat{i}-4\hat{j})}{(2\times {10}^{-3}+4\times {10}^{-3})}\times {(2\times {10}^{-3})}^{2}\phantom{\rule{0ex}{0ex}}{d}_{cm}=-6.67\times {10}^{-4}\hat{i}+3.33\times {10}^{-4}\hat{j}$

Solve for the magnitude as:

${d}_{cm}=\sqrt{{(-6.67\times {10}^{-4})}^{2}+{(3.33\times {10}^{-4})}^{2}}\phantom{\rule{0ex}{0ex}}{d}_{cm}=7.45\times {10}^{-4}\mathrm{m}$

Hence, magnitude of displacement of center of mass is 0.745mm.

Consider the equation

${d}_{cm}=-6.67\times {10}^{-4}\hat{i}+3.33\times {10}^{-4}\hat{j}$

So, the angle is,

$\mathrm{tan}\theta =\frac{3.33\times {10}^{-4}}{-6.67\times {10}^{-4}}\phantom{\rule{0ex}{0ex}}\theta =153\xb0$

Hence,** **angle relative to the positive direction of x axis is $153\xb0$.

Velocity of two masses is as follows:

${v}_{1}={a}_{1}\times t\phantom{\rule{0ex}{0ex}}{a}_{1}=\frac{{F}_{1}}{{m}_{1}}\phantom{\rule{0ex}{0ex}}{v}_{1}=\frac{{F}_{1}t}{{m}_{1}}\phantom{\rule{0ex}{0ex}}{v}_{1}{m}_{1}={F}_{1}t$ ……. (7)

Solve further as:

${v}_{2}={a}_{2}\times t\phantom{\rule{0ex}{0ex}}{a}_{2}=\frac{{F}_{2}}{{m}_{2}}\phantom{\rule{0ex}{0ex}}{v}_{2}=\frac{{F}_{2}t}{{m}_{2}}\phantom{\rule{0ex}{0ex}}{v}_{2}{m}_{2}={F}_{2}t$ ….. (8)

Now, velocity of center of mass is as follows:

${v}_{cm}=\frac{{m}_{1}{v}_{1}+{m}_{2}{v}_{2}}{{m}_{1}+{m}_{2}}$

Now, Substitute the values from equation (7) and (8) as follows:

${v}_{cm}=\frac{{F}_{1}+{F}_{2}}{{m}_{1}+{m}_{2}}\times t\phantom{\rule{0ex}{0ex}}{v}_{cm}=\frac{(-4\hat{i}+5\hat{j})+(2\hat{i}-4\hat{j})}{2\times {10}^{-3}+4\times {10}^{-3}}\times (2\times {10}^{-3}).\phantom{\rule{0ex}{0ex}}{v}_{cm}=-0.667\hat{i}+0.333\hat{j}$

Now, magnitude of velocity is,

${v}_{cm}=\sqrt{{(-0.667)}^{2}+{(0.333)}^{2}}\phantom{\rule{0ex}{0ex}}{v}_{cm}=0.745\mathrm{m}/\mathrm{s}$

Net mass is,

$m=6\times {10}^{-3}\mathrm{kg}$

Now, kinetic energy is calculated as:

role="math" localid="1661314904075" $KE=\frac{1}{2}m{v}_{cm}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 6\times {10}^{-3}\times {(0.745)}^{2}\phantom{\rule{0ex}{0ex}}=0.00167\mathrm{J}$

Hence, kinetic energy of center of mass at *t* = 2.00 s is 0.00167 J.

Determine the acceleration using Newton’s second law. Use this acceleration to find displacement and velocity after the given time. Using this, find the displacement of center of mass and velocity of center of mass.

94% of StudySmarter users get better grades.

Sign up for free