Please show all work. A solid wood door, 90.0 cm wide by 2 m tall has a mass of 35kg. It is ajar and at rest. A ball with a mass of 500g is thrown perpendicular to the door with a speed of 20m/s and hits the door 60cm from the hinged side. The ball rebounds with a speed of 16 m/s along the same line. What is the angular speed of the door after the collision? Thanks
Change in momentum of the ball = m(v2 - v1) = 0.5(20+16) = 18 kg-m/s
By Newton's Second law of motion, the ...
This Solution contains calculations to aid you in understanding the Solution to this question.