A girl of mass M stands on the rim of a frictionless merry-go-round of radius R and rotational inertia I that is not moving. She throws a rock of mass m horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is ν. Afterward, what are (a) the angular speed of the merry-go-round and (b) the linear speed of the girl?

Explanation:

As no external force is applied , therefore the momentum of the system will remain conserved .

The initial momentum = 0 , because the girl with stone in hand is at rest .

When stone is thrown horizontally , its momentum is = m v

Let the velocity of girl after this throwing of stone is v₀

Thus its momentum is M v₀

Therefore the sum of these two momentum should be zero

Hence m v + M v₀ = 0

or v₀ =   This will be the linear velocity of the girl

The angular velocity ω = =

here R is the radius of wheel

this is a good question! i'll do some research and come back when i have the answer!

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!