The superball rebounds back in the opposite direction, with the same momentum. But during the rebound, the momentum acts in an opposite direction of the initial momentum during the hitting with the door. There is a negligible loss in the speed of the superball mass.
An impulse of the system is equal to the change in the momentum of the system. The impulse can be calculated as
\({I_2} = m\left( {{v_{{f_2}}} - {v_{{i_2}}}} \right)\).
Here,\({v_{{f_2}}}\)is the final velocity of the superball is equal to\(\left( { - {v_{{i_2}}}} \right)\);\({v_{{i_2}}}\)is the initial velocity of the superball; m is the mass of the superball.
Substitute the values in the above equation.
\(\begin{aligned}{c}{I_2} = m\left( { - {v_{{i_2}}} - {v_{{i_2}}}} \right)\\ = - 2m{v_{{i_2}}}\end{aligned}\)
Here, the negative sign indicates the opposite direction of velocity.
An impulse force exerted by the door in the opposite direction is equal to two times the momentum of the superball.
The momentum of the superball is equal to twice the value of the initial momentum; in the case of the superball, the magnitude of impulse increases. Thus, the superball will be more effective to throw at the door.