Consider a uniform solid sphere of radius r and mass m rolling without slipping. which form of its kinetic energy is larger, translational or rotational?

Translational kinetic energy is equal to one half the product of mass andthe square of the linear velocity. Rotational kinetic energy, on the other hand, is one half the product of the mass of the object in circular motion and the square of the angular velocity where angular velocity is equal linear velocity divided by the radius of the circular motion. To compare these energies, we do as follows:

KE(t) = 0.5mv^2
KE(r) = 0.5mw^2 = 0.5mv^2/r^2
KE(r) = KE(t)/r^2

From the relation, rotational kinetic energy is 1/r^2 times than the translational kinetic energy. Therefore, translational kinetic energy would be bigger than the rotational kinetic energy.

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    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