How does linear velocity of a spinning object affect its rebound

[WilkinzMicawber]

Consider the case where a basketball rotating about its center of mass strikes a wall head-on. The velocity given it parallel to the wall due to the spin is caused by the friction between the wall and the ball as the ball is sliding against the wall during the collision. Friction is proportional to the normal force acting on a surface. The faster the ball hits the wall, the greater is the normal force, but the time the ball touches the wall is also smaller. Does the linear velocity of the ball have any effect on the total amount of velocity given to the ball parallel to the wall?

 

[Cutter Ketch]

I don't think the time of contact is smaller. Think of the basketball as a spring. The kinetic energy of the ball 1/2 m v^2 gets converted to elastic potential energy 1/2 k x^2. So the distance the ball compresses is

x = sqrt(m/k) v0

The force from the first contact to maximum compression is -k x, so the acceleration is
a = - kx/m.

dx^2/dt^2 = -kx/m

Implies

x = C sin(sqrt(k/m) t)

Which knowing the max x from above must be

x = sqrt(m/k) v0 sin( sqrt(k/m) t)

where t=0 gives x=0 as the moment of initial contact. Maximum compression is reached when

sqrt(k/m) t = pi / 2

So tmax = sqrt(k/m) pi/2

So higher initial velocity gives greater compression but the time of contact is the same.

Regarding the effect that has on the transfer of the angular momentum for some of the time the ball slips and there is one torque then it sticks. Since the force is higher the torque is higher but once it's enough to stop the ball I suppose there isn't much difference if you go to yet higher velocity. I have no idea at what velocity the spin finally stops.

 

[WilkinzMicawber]

Thank you for your very in-depth reply. The main concern is whether the ball will travel further in the direction of the wall after rebounding, due to the spin. Your answer helped me realize the answer. The spin can only result in velocity along the wall as long as there is rotational energy left to convert to lateral velocity, through friction. Assuming the duration of contact is sufficient to transfer all of the rotational energy to kinetic energy in the cases where the striking linear velocity differs, the amount of spin transferred to lateral velocity should be similar.