Disk A, with a mass of 2.0 kg and a radius of 70 cm , rotates clockwise about a frictionless vertical axle at 50 rev/s . Disk B, also 2.0 kg but with a radius of 50 cm , rotates counterclockwise about that same axle, but at a greater height than disk A, at 50 rev/s . Disk B slides down the axle until it lands on top of disk A, after which they rotate together. After the collision, what is their common angular speed (in rev/s) and in which direction do they rotate?

    w = - 197.5 rad / s

The negative sign indicates that the rotations are clockwise

Explanation:

To solve this exercise, let's use the concept of conservation of the angular number.

We create a system formed by the two discs, in this case the forces last the shock are internal

initial instant .. just before shock

           L₀ = I₀ w₀ + I₁ w₁

instnte final. Right after crash

           L_f = (I₀ + I1) w

angular momentum is conserved

           I₀ w₀ + I₁ w₁ = (I₀ + I₁) w

           w = I₀ w₀ + I₁ w₁ / Io + I1

The moment of inertia of a disk with an axis passing through its thermometric center

          I₀ = ½ m² r₀²

          I₁ = ½ m₁ r₁²

         we substitute

          I₀ = ½ 2.0 0.70²

          I₀ = 0.49 kg m

          I₁ = ½ 2.0 0.5²

          I₁ = 0.25

₁  

let's reduce the magnitudes the SI system

          w₀ = -50 rev / (2pi rad / 1rev) = -314.15 rad / s

          w₁ = 70 rev (2pi rad / 1rev) = 439.82 rad / s

we will assume that the counterclockwise turns are positive

           w = -0.49 314.15 + 0.25 439.82 / (0.49 + 0.25)

            w = (- 4.696 + 1.0995) 102) / 0.74

            w = -197.75 + 0.25

            w = - 197.5 rad / s

The negative sign indicates that the rotations are clockwise

momentum of soft ball three times less than baskey ball

there are no contact forces mentioned on the list of choices that you've provided.