# 2 masses of mass 10 kg and 20 kg are kept 1 meter apart and released. Assuming that only mutual...

## Question:

2 masses of mass 10 kg and 20 kg are kept 1 meter apart and released. Assuming that only mutual gravitational forces are acting, find the speed of particles at separation.

## Conservation of Energy:

Energy is defined as the ability to do work. A type of energy is the kinetic energy. Kinetic energy is the energy that is produced by the motion of an object. Another type of energy is the potential energy. This type of energy is dependent on the position of the object.

Using conservation of momentum, we write:

{eq}m_1 v_1 + m_2 v_2 = 0 \\ 10v_1 + 20v_2 = 0 v_1 = 2v_2 {/eq}

using conservation of energy, we write:

{eq}\frac{Gm_1m_2}{r} = \frac{Gm_1m_2}{x} + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 {/eq}

where:

x is the separation,

Inserting the values of the available parameters, we write:

{eq}\frac{(6.67\times10^{-11})(10)(20)}{1} = \frac{(6.67\times10^{-11})(10)(20)}{x} + \frac{1}{2}(10)v_1^2 + \frac{1}{2}(20)v_2^2 \\ 13.34\times10^{-9} = \frac{13.34\times10^{-9}}{x} + 30v_2^2 \\ v_2 = 13.34\times10^{-9} - \frac{13.34\times10^{-9}}{x} {/eq}

the other speed is given as:

{eq}v_1 = 2v_2 \\ v_1 = 2(13.34\times10^{-9} - \frac{13.34\times10^{-9}}{x}) {/eq}