Question

. A 60 kg boy lying on a surface of negligible
friction throws horizontally a stone of mass 1
kg with a speed of 12 m/s away from him. As
a result with what kinetic energy he moves
back?
Pease help me with this question
I know the answer I want to know the methord of this sum

Answer 1

dear ! This question is based on law of conservation of linear momentum.

as external force acts on system of man and stone is zero.

initial momentum = final momentum

or, 60kg × 0 + 1kg × 0 = 60kg × v + 1kg × 12m/s

or, 0 = 60v + 12

or, v = 12/60 = 0.2 m/s

so, man moves with speed 0.2m/s just opposite direction.

now, he moves back with kinetic energy , K.E = 1/2 Mv²

= 1/2 × 60 × (0.2)²

= 30 × 0.04 = 1.2 J

hence, the man moves with kinetic energy = 1.2 J