Question

a person of mass M is equal to 90 kgs standing on a smooth horizontal plane of ice throws a body of mass is equal to 10 kgs horizontally on the same surface if the distance between the person and body after 10 seconds 10 metres the the kinetic energy of the person in joules is how much

Answer 1

total distance between 90kg man and 10kg object = 10m
distance tracelled by man = (10/(10+90))*10m = (1/10)*10 = 1m
time = 10s

speed = 1/10 = 0.1 m/s

KE = 1/2 mv² = 1/2 * 90 * 0.1² = 0.45J

Answer 2

We know that linear momentum of the two bodies: person and 10 kg mass is conserved in the horizontal direction.  There is no external force acting on them in horizontal direction.  When the person throws the mass, he/she exerts a force(impulse) during a short duration and gives momentum to the mass.  Since an equal and opposite force acts on him/her the person travels also in backward direction.  These velocities remain constant as there is no friction on ice.

   mass of person = M = 90kg                  mass of the body = m = 10kg
   let the velocity of the person be - V.       velocity of the body = v

   relative velocity between the two = v - (-V) = v + V, as they separate.
          10 meters = (V + v) * 10 sec.
       =>   V + v = 1 m/sec    --- (1)

 conservation of linear momentum =>
            M V  + m v = 0           =>  - 90 V + 10 v = 0
             => v =  9 V        ---- (2)
 
   Solve the two equations.... V = 0.1 m/s  in the backward direction.
                   v = 0.9 m/s  in the forward direction.
 
   the kinetic energy of the person = 1/2 M V*V = 1/2 * 90 * 0.01  Joules
                     = 0.45 J

  KE of the body = 1/2 * 10 * 0.81 = 4.05 Joules.

  We may think that energy is conserved.  The total kinetic energy before the person threw the mass was 0.  So where did the energy come from...
 
  The person does work during the time he/she exerts force on the mass while throwing.  That work is converted into the KE of both.