In Figure (a), particleAis fixed in place at$\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{m}$ on thexaxis and particleB, with a mass of 1.0 kg, is fixed in place at the origin. ParticleC(not shown) can be moved along thexaxis, between particleBand$x=\infty$.Figure (b)shows thexcomponent${\mathbf{F}}_{\mathbf{net},\mathbf{x}}$of the net gravitational force on particleBdue to particlesAandC, as a function of positionxof particleC. 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) particleAand (b) particleC?

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

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

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

d) Distance of particle C from origin,$x_C=0.4\text{m(fromfigure)}$

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:

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

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:

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

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