1. Do we have to take the angular velocity of the ball into consideration for writing the impulse-momentum equation?
Nope, if you meant it was
linear momentum. Actually, the angular velocity is always taken into consideration, and it has to be so. BUT: it is finally canceled out. Here is the reason:
Consider a body (a system of mass points will do, not necessarily a rigid/ continuous body), whose COM is at point A. The total linear momentum of the body is the sum of linear momentums of all infinitesimal elements whose locations are conventionally points M
i:
[tex]\vec{p}_{total} = \sum m_i\vec{v}_i = \sum m_i(\vec{v}_{COM} + \vec{\omega} \times \vec{AM_i}) = \sum m_i\vec{v}_{COM} + \sum m_i(\vec{\omega} \times \vec{AM_i})[/tex]
But: [tex]\sum m_i(\vec{\omega} \times \vec{AM_i}) = \vec{\omega} \times \sum m_i\vec{AM_i} = \vec{0}[/tex]
as by definition of COM: [tex]\sum m_i\vec{AM_i} = \vec{0}[/tex]
Therefore: [tex]\vec{p}_{total} = \sum m_i\vec{v}_{COM} = m\vec{v}_{COM}[/tex]
where [tex]m = \sum m_i[/tex] is the total mass of the system.
So the conclusion is: for any system of mass points, the linear momentum of the whole system = "linear momentum" of COM; angular velocity doesn't explicitly appear, as it's already canceled out.
P.S.: The conservation of linear momentum applies to everything involved in the collision: sphere + mass A + mass B + spring + Earth, not just mass A + the point mass on sphere that touches mass A! However, spring is massless and thus has no momentum, mass B stays at rest during the collision, the Earth is not impacted horizontally (the 3rd point on the Earth needs a good reasoning behind it; find it yourself
), and so, for horizontal linear momentum, we are left with the sphere and mass A. The sphere = all the point masses contained in the sphere, not just the point of contact
2. Do we have to take the angular velocity of the ball into consideration for writing Newton's Law of Restitution equation?
I don't know
Newton's law of restitution is not a universial law, and doesn't appear very frequently in scientific calculations (I'm not sure about engineering, e.g. mechanical/ material engineering). I doubt that there is a unified and official definition of the coefficient of restitution; or perhaps it's my shortcoming.
Check your textbook which is widely used in your country to see how they define it. The exams should apply the same or similar definition, so stick to that definition.
3. How many unknowns to be determined for the collision?
4 actually. I see you listed R as an unknown. Those quantities like mass, length, initial speed, those that specify the intrinsic characteristics of the system & initial conditions, should be provided. (Classical) Mechanics problems are always deterministic provided that the system is well-specified and there are enough intial conditions.
If the problem doesn't provide the numerical value of R, then just leave R as R in the final result, after substituting all the values in
4. And how many equations to be written?
4 of course
Those are:
1. The conservation of horizontal component of linear momentum of the sphere + mass A.
2. The equation for the rotation of the sphere. The change of linear momentum of the sphere is resulted from the frictional force F, and this F also exerts a change in angular momentum. Relate those two and eliminate F, you will arrive at one equation containing some of the 4 unknowns only, not F.
3. The equation for the vertical component of linear momentum of the sphere. This change is due to the normal force N. The frictional force F is related to N by the frictional coefficient between the sphere and mass A. Again, you will arrive at another equation containing some of the 4 unknowns only, not N or F.
4. The restitution equation.
