Question

Two particles A and B of masses 1 kg and 2 kg respectively are kept 1 m apart and are released to move under mutual attraction. Find The speed A when that of B is 3.6 cm/hr. What is the separation between the particles at this instant?

Answer 1

The linear momentum of the pair A+B is zero initially. As only mutual attractiois taken into account, which is interN/Al when A+B is taken as the system the linear momentum wil remain zero. The particles move in opposite directions. If the speed of A is v when the speed of B is 3.6 cm/hour `=10^-5ms^-2`.
(1kg)v=(2kg)(10^-5ms^2-1)`
v=2xx10^-5ms^-1`
The potential energy of the pair is `-(Gm_Am_B)/R` with usual symbols. Initial potential energy
`=-(6.67xx10^-1N-m^2/kg^2xx2kgxx1kg)/1m`
`=13.34xx10^-11J`
if the separstion at the given instant is d using conservation of energy
`-13.34xx10^-11J+0`
`=-(13.34xx10^-11J-m)/d+1/2(2kg)(10^-5ms^-1)^2`
`+1/2(1kg)(2xx10^-5ms^-1)^2`
Solving this d=0.31m