(Part 1) A merry-go-round rotates at the rate of
0.4 rev/s with an 99 kg man standing at a
point 2.5 m from the axis of rotation.
What is the new angular speed when the
man walks to a point 1 m from the center?
Consider the merry-go-round is a solid 73 kg
cylinder of radius of 2.5 m.
Answer in units of rad/s.

(Part 2)

What is the change in kinetic energy due to
this movement?
Answer in units of J.

Earthplanet yplanet y5=earth     *many different equationsplanet x1.5=earthplanet x is two times planet yobject 5,2500 mass5,1,3,2,4object 1 is four times obect 4

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

That big glowy thing in the sky is called the sun