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Q. 79- Excercises And Problems

Expert-verified
Found in: Page 746

### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

# Bead A has a mass of $15\mathrm{g}$ and a charge of $-5.0\mathrm{nC}$. Bead $\mathrm{B}$ has a mass of $25\mathrm{g}$ and a charge of$-10.0\mathrm{nC}$. The beads are held$12\mathrm{cm}$ apart (measured between their centers) and released. What maximum speed is achieved by each bead?

- Maximum speed achieved by band A is $1.77\mathrm{cm}/\mathrm{s}.$

- Maximum speed achieved by band B is$1.06\mathrm{cm}/\mathrm{s}.$

See the step by step solution

## Step 1: Given information

${m}_{a}=15\mathrm{gm}\phantom{\rule{0ex}{0ex}}{m}_{u}=25\mathrm{gm}\phantom{\rule{0ex}{0ex}}{q}_{a}=-5×{10}^{-7}\mathrm{C}\phantom{\rule{0ex}{0ex}}{4}_{a}=-10×{10}^{-9}\mathrm{C}\phantom{\rule{0ex}{0ex}}r=12\mathrm{cm}$

We know that initial energy of system is

$\mathrm{PE}-\frac{1}{4\mathrm{sin}\theta }\frac{{\mathrm{sin}}_{t}}{r}$

Where${E}_{i}$-Initial energy of the system

$qa$=charge on bead a de

${q}_{t}$= charge on bead b

$\mathrm{K}$= distance l in meters

Final energy of system is kinetic energy

Here,

${m}_{a}$= mass of bead a

${m}_{b}$= mass of bead b

${\mathrm{V}}_{\mathrm{a}}$$=$velocity of bead a

${v}_{b}$ =velocity of bead b

## Step 2: Calculation

According to energy conservation, Initial energy final energy ${E}_{i}={E}_{f}$

Plugging the values in the above equation,$\frac{1}{4\pi {ϵ}_{0}}\frac{4a{q}_{u}}{r}=\frac{1}{2}{m}_{a}{v}_{a}^{2}+\frac{1}{2}{m}_{b}{v}_{b}^{2}\dots \dots .\left(1\right)$

According to law of conservation of momentum

${m}_{u}{v}_{u}={m}_{b}{v}_{b}$

${v}_{b}=\frac{{m}_{1},{v}_{0}}{-\cdots \dots +1}\left(2\right)$

Putting (2) in (1)

$\frac{1}{2}\left(1+\frac{{m}_{a}}{{m}_{b}}\right){m}_{a}{v}_{a}^{2}=\frac{1}{4\pi {\in }_{0}}\frac{{q}_{b}{q}_{b}}{r}$

$va=\sqrt{\frac{{q}_{a}{q}_{b}}{r}\frac{1}{4\pi {\in }_{0}}\frac{2}{{m}_{a}}\frac{{m}_{b}}{{m}_{a}+{m}_{b}}}$

Putting the given values,

$va=\sqrt{\frac{2×25}{15}9×{10}^{9}\frac{2}{\left(.12\right)}\frac{\left(-5×{10}^{-9}\right)\left(-10×{10}^{-9}\right)}{\left(.025+.015\right)}}\phantom{\rule{0ex}{0ex}}$

localid="1650117105638" role="math" $=1.77$ $m/s$

$\text{Using Eq.}\left(2\right)$

$=\frac{15}{25}×1.77$

$vb=1.06$