Question

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?

Answer 1

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