Answer
The mass of particle A is $~~0.25~kg$
Work Step by Step
$F_{net,x}$ approaches $-4.17\times 10^{-10}~N$ as $x \to \infty$
Since the gravitational force due to particle C approaches 0 as $x \to \infty$, we can assume that the gravitational force due to particle A is $-4.17\times 10^{-10}~N$
We can find the mass of particle A:
$F = \frac{G~m_A~m_B}{(0.20~m)^2} = 4.17\times 10^{-10}~N$
$m_A = \frac{(4.17\times 10^{-10}~N)(0.20~m)^2}{G~m_B}$
$m_A = \frac{(4.17\times 10^{-10}~N)(0.20~m)^2}{(6.67\times 10^{-11}~N~m^2/kg^2)(1.0~kg)}$
$m_A = 0.25~kg$
The mass of particle A is $~~0.25~kg$.
