Question

A 30 kg boy lying on a surface of negligiblefriction throws horizontally a stone of mass 2kg with a speed of 24 m/s away from him. Asa result with what kinetic energy he moves back​

Answer 1

$kinetic \: energy \: = \frac{1}{2} {mv}^{2} \\ = \frac{1}{2} 2 \times {24}^{2} \\ = 1 \times {24}^{2} \\ = 576$

so the kinetic energy is 576 J

Answer 2

Answer:

960J

Explanation:

$\bf{\huge\mathcal{\boxed{\rm{\mathfrak\blue{Hey Mate:}}}}}$

Let m represent mass of stone,u be its speed while M & V denote mass of person & velocity of Man respectively .

mu = MV

2(24) = 30(V)

V = 8

So , energy with which he moves back

= ( 1/2 )MV²

= (0.5)(30)(8)²

= 15(8)²

= 960 J