A father racing his son has 13 the kinetic energy of the son

A father racing his son has 1/3 the kinetic energy of the son, who has 1/3 the mass of the father. The father speeds up by 1.3 m/s and then has the same kinetic energy as the son. What are the original speeds of (a) the father and (b) the son?

Solution

Kinetic energy of the father K = (1/3) kinetic energy of the son

              (1/2) MV ² = (1/3) (1/2) mv ²      --------------( 1)

Mass of son m = (1/3) mass of father

                   m = (1/3) M

Substitute this in equation(1) you get ,

(1/2) MV ² = (1/3) (1/2)(1/3) Mv ²  

           V ² = (1/9)v ²

             V = v / 3               ------------( 2)

When Speed of the father V \' = V+1.3 then ,

Kinetic energy of the father = kinetic energy of son

(1/2) MV \' ² = (1/2) mv ²

        MV \' ² = mv ²

        MV \' ² = (M/3)v ²

           V \' ² = v ² / 3

(V+1.3) ² = (1/3) v ²

= (1/3) (3V) ²           Since from equation( 2) ,V = v/3 =>3V = v

               = 3V ²

V + 1.3 = 1.732 V

0.732 V = 1.3

         V = 1.3 / 0.732

             = 1.775 m/s

Original speed of father V = 1.775 m/s

Original speed of son v = 3V

                                  = 3(1.775)

                                  = 5.327 m/s