A three-phase, 12 pole induction motor is excited by a 60 Hz three-phase source. At rated load and speed, the induction motor runs at 860 rpm and absorb 90kW power from the three-phase AC input. The kinetic energy of all revolving parts is 100kJ at rated load and speed.
a. Calculate the slip of the motor at rated speed.
b. If the voltage induced at the rotor bar is 10V when the rotor is locked, what is the voltage induced at the rotor bar when the motor is at rated speed?
c. What is the I^2R loss at rated load and speed?
d. Plugging is applied to stop the induction motor and then reverse the rotation to 860 rpm, how much power dissipated in the rotor circuit during this process?

    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