Two particles A and B of masses 1kg and 2kg respectively are kept 1m apart and are released to move under mutual attraction. find the speed of A when that of B is 3.6cm/hr. what is the separation between the particles at this instant?........(in the solution to this question the linear momentum was taken as zero. if so why? becoz when the external force is zero the linear momentum must be conserved not taken as zero.plz explain that also)

initially velocity of particles is zero when they released ....

initial  linear momentam system = P_i = maU_a + m_bU_b = 0            (coz both U and U_{1 are 0 initially)}

finally momentam of system is P_{f = maVa +} m_bV_b

                                                =Va + 2V_b

                                        V_{a =} -2V_b      .........1

now applying conservation of energy

         (KE)_{i + (PE)f}  = (KE)_{f + (PE)f         .........2           ( KE , PE  is kinetic and potential energy)}

   initially KE = 0  &  PE = -Gm_a .m_b/r

                                PE =-2G                  ( r =1m initially)

  finally  KE  is  maVa² _{/2 +}  m_bV_{b² /2}

   finally PE is -Gm_am_b/r 

 putting these values in eq 2

we get

     0 + -2G  =  -Gm_am_b/r + maV_a² /2 + mbVb² _/2

          -2G = -2G/r + ( Va² + 2V_{b² )}/2 .......................3

       -2G = -2G/r + 3 V_{b²                                       ( putting Va = 2Vb)}

       on solving

                              r = 30.5cm  ans

        seperation bw particles is 30.5 cm at this instant...

Just put the formula and calculate the velocity..I am trying to describe the next part..Here the external force u r talking about is the mutal force gravitation..If it is not there then the bodies wouldnt move and so the linear momentum would have been 0 always..It becomes a case of absolute rest..which is impratical..because if motion is there force should be there..