A ball of mass 1.00 kg is thrown at a door with velocity v1x = 12.0 m/s. The rectangular door has a mass of 30.0 kg and is 1.00 m wide. The ball strikes the door perpendicular to it, a horizontal distance 75.0 cm away from the hinges, and causes the door to swing open with an angular speed of 1.20 rad/s. What is the velocity v2x of the ball after striking the door? (The moment of inertia of a thin rectangular plate with mass M and dimensions a × b and the axis along the edge of length b is I = 1/3Ma2.)"

The velocity of the ball after collision is  

Explanation:

From the question we are told that

     The mass of the ball is  

      The velocity of the ball is  

      The mass of the rectangular door is  

        The width of the door is  

         The distance of impact from the hinge is  

         The angular speed of the door is  

So the moment of inertia of the door is given from the question as

substituting values

According to the law of angular momentum conservation

Where   is the initial angular momentum of the system(the door and the ball) which is mathematically represented as

so   is the initial angular speed of the door which is zero

So    

is the final angular momentum of the system(the door and the ball) which is mathematically represented as

So  

Substituting values

The negative sign show a reversal in the balls direction

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer