Question

An 8.00 kg point mass and a 15.0 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point between the two masses 20.0 cm from the 8.00 kg mass along the line connecting the two fixed masses.
Find the magnitude and direction of the acceleration of the particle. Answer from book: 2.2 x 10^-9 m/s^2 - toward the 8.00 kg mass....

Answer 1

Answer:$a=2.23\times 10^{-9} m/s^2$

Explanation:

Given

Mass of first Point mass $m_1=8 kg$

Mass of second Point $m_2=15 kg$

distance between them $d=50 cm$

third point mass $m_3=m$

Distance between $m\ and\ m_1 is\ 20 cm$

Distance between  $m\ and\ m_2 is 30 cm$

Force Due to $m_1\ and\ m F_1=\frac{Gmm_1}{d_1^2}$

$F_1=\frac{8Gm}{(0.2)^2}$

$F_1=200 mG$

$F_2=m\frac{Gmm_2}{d_2^2}$

$F_1=m\frac{15Gm}{(0.3)^2}$

$F_2=166.67 mG$

Net Force

$F_{net}=F_1-F_2$

$=200 mG-166.67 mG$

$=33.33 mG$

$F_{net}=222.33\times 10^{-11} N$

$F_{net}=2.23m\times 10^{-9} N$

acceleration $a=\frac{2.23m\times 10^{-9}}{m}$

$a=2.23\times 10^{-9} m/s^2$

towards 8 kg mass