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Q12P

Expert-verifiedFound in: Page 379

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**In Figure (***a***), particle ****A ****is fixed in place at ${\mathbf{x}}{\mathbf{=}}{\mathbf{-}}{\mathbf{}}{\mathbf{0}}{\mathbf{.}}{\mathbf{20}}{\mathbf{}}{\mathbf{m}}$**** on the ****x ****axis and particle ****B****, with a mass of 1.0 kg, is fixed in place at the origin. Particle ****C ****(not shown) can be moved along the ****x ****axis, between particle ****B ****and ${x}{=}{\infty}$****.Figure (****b) ****shows the ****x ****component ${{\mathbf{F}}}_{\mathbf{net},\mathbf{x}}$****of the net gravitational force on particle ****B ****due to particles ****A ****and ****C****, as a function of position ****x ****of particle ****C****. The plot actually extends to the right, approaching an asymptote of $-4.17\times {10}^{10}\mathbf{N}$****as $\to \infty $****. What are the masses of (a) particle ****A ****and (b) particle ****C****?**

a) The mass of particle A is $0.25\text{\hspace{0.33em}kg}$

b) The mass of particle B is $1\text{\hspace{0.33em}kg}$

a) Mass of particle B,${m}_{B}=1\text{\hspace{0.33em}kg}$

b) Distance of particle A from origin,${x}_{A}=0.2\text{\hspace{0.33em}m (from figure)}$

c) Distance of particle B from origin,${x}_{B}=0\text{\hspace{0.33em}m}$

d) Distance of particle C from origin, ${x}_{C}=0.4\text{m (from figure)}$

e) Gravitational constant, $G=6.67\times {10}^{-11}\text{\hspace{0.33em}N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

**Force between two masses can be calculated by using Newton’s law of gravitation. According to Newton’s law of gravitation, the force of two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.**

Formula:

Gravitational force, $F=\frac{GMm}{{r}^{2}}$ .......(i)

Using equation (i),Force between mass A & B can be written as:

$\begin{array}{c}{F}_{AB}=\frac{G{m}_{A}{m}_{B}}{{{x}_{A}}^{2}}\\ 4.17\times {10}^{-10}\text{\hspace{0.17em}\hspace{0.17em}N}=\frac{6.67\times {10}^{-11}\text{\hspace{0.17em}\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\text{\hspace{0.17em}\hspace{0.17em}}\times \text{\hspace{0.17em}\hspace{0.17em}}{m}_{A}\text{\hspace{0.17em}\hspace{0.17em}}\times \text{\hspace{0.17em}\hspace{0.17em}}1\text{\hspace{0.17em}\hspace{0.17em}kg}}{{\left(0.2\text{\hspace{0.17em}\hspace{0.17em}m}\right)}^{2}}\text{\hspace{0.17em}\hspace{0.17em}}\\ {m}_{A}=0.25\text{\hspace{0.33em}kg}\end{array}$

Hence, mass of the particle A is $0.25\text{\hspace{0.33em}kg}$

Using equation (i), Force between mass B & C can be written as:

$\begin{array}{c}{F}_{BC}=\frac{G{m}_{B}{m}_{C}}{{x}^{2}}\\ 4.17\times {10}^{-10}\text{\hspace{0.17em}\hspace{0.17em}N}=\frac{6.67\times {10}^{-11}\text{\hspace{0.17em}\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\times 1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}kg\hspace{0.17em}}\times \text{\hspace{0.17em}\hspace{0.17em}}{m}_{C}}{{\left(0.4\text{\hspace{0.17em}\hspace{0.17em}m}\right)}^{2}}\\ {m}_{C}=1\text{\hspace{0.33em}kg}\end{array}$

Hence, mass of particle C is .$1\text{\hspace{0.33em}kg}$

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