Wham!

We first state the brief overview of how to solve this problem. Firstly, we note that this is an elastic collision. Therefore, both energy and momentum are conserved in this collision. Hence, we will form relationships linking angular momentum to kinetic energy, solve the resultant equations and then arrive at the answer

Now for the specifics. We take the ball and the door to be a system. In addition, throughout this solution, $v$ , will denote the velocity of the ball when it bounced back, $w$ , will denote the angular velocity of the door and, $I$ , will denote the moment of inertia of the door.

$I$ = $\frac {15 * 1^2}{3}$ = $5$

The initial kinetic energy of the system is

$0.5 \times 0.4 \times 35^2$ = 245

The final kinetic energy of the system is

$0.5 \times I \times w^2$ + $0.5 \times 0.4 \times v^2$ = $2.5 \cdot w^2$ + $0.2 \cdot v^2$

The kinetic energy relationship can thus be represented by

$2.5 \cdot w^2$ + $0.2 \cdot v^2$ = 245

Next, we look at the angular momentum of the system. Note that in this system, the hinge of the door acts like the center of the circle and the distance from the hinge to the point where the ball impacts the door is considered the radius of the circle. This distance is $\frac {1}{2}$ = 0.5 The initial angular momentum of the system is given by

$0.4 \times 35 \times 0.5$ = 7

as the ball hits the door it rebounds elastically. by using conservation of momentum at the time of impact we can see change in momentum of the ball will be same for that of the door. now if we assume the door to be a 1 meter wide square plate we can calculate its moment of inertia with axis of rotation along one side of the plate containing the hinge. let it be I . now just after impact its angular velocity be w . from conservation of angular momentum mvr = Iw where m is the mass of the ball, v it's velocity and r is the perpendicular distance of point of impact from the axis. with all the info given we can calculate w . now time taken to open is same as the time taken to rotate 90 degrees. time taken, t = \frac {pi} {2 w }

The moment of inertia of the door is $I = \frac{1}{3} M L^{2}$ where $M$ is its mass and $L$ its length. According to angular moment conservation, if $m$ is the mass of the ball and $v_{0}$ its initial velocity, we have

$m v_{0} \frac{L}{2} = - m v_{0} \frac{L}{2} + \frac{1}{3} M L^{2} \omega \Rightarrow \omega = \frac{3 m v_{0}}{M L}$

Where $\omega$ is the angular velocity the door gains when the ball hits. But, as we know, $\omega = \frac{2 \pi}{T}$ , where $T$ is the period of the movement. As we want to know how long the door takes to complete $1/4$ of its movement, we have

$t = \frac{T}{4} = \frac{\pi}{2 \omega} = \frac{\pi}{6} \frac{ML}{m v_{0}} \approx 0.572 s$

the energy of ball is kinetic energy which is 245 joules......this energy is transferred to the door ....... this energy is used by the door to travel 90 degrees.....(the energy is kinetic as the door moves). so we can find the distance and velocity and the time....