Post-bounce:
A = A1
T = T1
N = + N1 (upwards)

N: In a perfect bounce, all the energy put into compressing the ping-pong ball would be recovered, so N0 = N1. In practice, there is likely to be some loss due to heat dissipation, so that N1 < N0.

T: due to the spin of the ball, there could be a friction-based impact that changes the tangential speed. For overspin, it would increase the speed; for underspin, decrease it. Let’s call the impact "p": then
T1 = T + p/m where m = mass of the ping-pong ball
and p is > 0 for overspin and < 0 for underspin.

A: when there is an impact due to spin, it will also affect the angular momentum of the ball, and it will always act to counter the existing spin. So for overspin:
A1 = A0 – r*p/I where r = radius of ball and I is the moment of inertia
and for underspin:
A1 = A0 – r*p/I (but notice now that a1 and a0 are negative, and p is also negative – so it still acts to reduce the magnitude of A)

So the equations are:
a) N1 < N0
b) T1 = T + p/m (where p is > 0 for overspin, < 0 for underspin)
c) A1 = A0 – r*p/I

I = integral ((x^2 + y^2) dm
= (M/(4pi*r^2) integral (x^2 + y^2) (2pi*r^2*sin(b))db
= (M/(4pi*r^2) integral (2pi)(r^4) sin(b)^3 db
= (M*r^2)/2 * integral (sin^3(b) ) db

integral (sin^2(b) db = – integral (sin^2(b) dcos(b)
= – integral (1 – cos^2(b)) dcos(b)
= integral (1 – x^2) dx {-1, 1}
= x – x^3/3 {-1, 1}
= 2 – (2/3)
= 4/3
So I = 2Mr^2/3
and c) becomes:
c) A1 = A0 – p/(2Mr/3)
= A0 – 3p/(2Mr)

So the complete equations are:
a) N1 < N0
b) T1 = T + p/m (where p is > 0 for overspin, < 0 for underspin)
c) A1 = A0 – 3p/(2Mr)

You have some options:
- For simplicity, you could set N1 = N0. It’s not realistic. Or you could arbitrarily reduce set N1/N0 to a factor less than 1.
- p is undetermined here. However, we are pretty sure that p cannot be so large as to change the sign of T. So you could set (p/M)/T0 to a factor less than 1.
- Once you do that, p/(Mr) is fixed. So you have to take whatever comes out of c).

Hope this helps!