Question

A 60kg boy lying on a surface of negligible friction throws horizontally a stone of mass 1kg with a speed of 12m/s away from him .As a result with what kinetic energy he moves back?

Answer 1

Answer:

1.2 J

Explanation:

Mv = constant

V =12/60= 0.2

Kinetic energy =M Vsquare /2

= 60*0.04/2 = 1.2 J

Answer 2

Answer:

the kinetic energy of the boy will be 1.2 J.

Explanation:

Given,

mass of boy,M = 60 kg

mass of stone,m = 1 kg

velocity of stone,v = 12 m/s

momentum of stone = mv

                                  = 12 x 1 kg-m/s

                                   = 12 kg-m/s

according to law of conservation of momentum,

momentum of stone = momentum of boy

=> 12 = M.V

=> 12 = 60 x V

$=>\ V\ =\ \dfrac{12}{60}\ m/s$

$=>\ V\ = 0.2\ m/s$

So, the kinetic energy of boy can be given by

$K\ =\ \dfrac{1}{2}.M.V^2$

   $=\ \dfrac{1}{2}\times 60\times (0.2)^2$

     = 1.2 J

So, the kinetic energy of the boy will be 1.2 J.