Question

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.)"

Answer 1

Answer:

The velocity of the ball after collision is  $v_{2x} = -4 m/s$

Explanation:

From the question we are told that

     The mass of the ball is  $m_b = 1.00\ kg$

      The velocity of the ball is  $v_{1x}= 12.0 \ m/s$

      The mass of the rectangular door is  $m_d = 30 \ kg$

        The width of the door is  $a = 1.00 \ m$

         The distance of impact from the hinge is  $L = 75 \ cm = \frac{75}{100} = 0.75 \ m$

         The angular speed of the door is  $w = 1.20 \ rad/s$

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

            $I = \frac{1}{3} M a^2$

substituting values

           $I = \frac{1}{3} * 30 * (1)$

           $I = 10 \ kg \cdot m^2$

According to the law of angular momentum conservation

           $L_i = L_f$

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

           $L_i = m_b * v_{1x} + Iw_i$

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

So    

                 $L_i = m_b * v_{1x}$

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

         $L_f = I w + m_b v_{2x} * L$

So  

      $m_b * v_{1x} = I w + m_b v_{2x} * L$

Substituting values

      $1 * 12 * 0.75 = 10* 1.2 * v_{2x} * 0.75$

         $v_{2x} = -4 m/s$

The negative sign show a reversal in the balls direction