Question

An 8.50 kg point mass and a 14.5 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 12.0 cm from the 8.50 kg mass along the line connecting the two fixed masses.Find the magnitude of the acceleration of the particle.

Answer 1

`a=2.8*10^(-9)m/s`

Step-by-step explanation:

From the question we are told that:

First Mass
`m=8.50kg`

2nd Mass
`m=14.5kg`

Distance

`d_1=50=>0.50m\\\\d_2=>12cm=>0.12m`

Generally the Newtons equation for Gravitational force is mathematically given by

`F_n=(Gm_nm)/((r_n)^2)`

Therefore

Initial force on m

`F_1=(Gm_1m)/((r_1)^2)`

Final force on m

`F_2=(Gm_2m)/((r_2)^2)\\\\F=(Gm_1m)/((r_1)^2)-(Gm_2m)/((r_2)^2)`

Acceleration of m

`a=(F)/(m)\\\\a=(Gm_1)/(r_1^2)-(Gm_2)/(r_2^2)`

`a=6,67*10^(-11){(8.5)/(0.12)}-(14.5)/(0.50)`

`a=2.8*10^(-9)m/s`