If a 2 kg ball traveling at 10 m/s hits a wall and stops in 0.03 seconds, then how much force will the ball experience?

Newton's second law is:
F=m*a,

where a=dv/dt, so

F=m*(dv/dt)

Rearranging gives:
F*dt=m*dv.

Basic integration gives:
F*t=m(vf-v0),
where vf and v0 are the final and initial velocities of the object respectively.

In your case vf=0, because the ball stops completely, and v0=10m/s.

Rearranging the last expression gives F=(m(vf-vo))/t.

Plug in numbers to find F=(2*10)/0.03=666.6 N

1)

2)

Explanation:

Using the second equation of angular motion we have

Since the wheels start from rest we ahve

Applying the given values in the equation we have

Now by newton's second law of motion in angular motion we have

1) For Hoop We have

Thus  

2)For disc We have

Thus