Question

A certain ball of mass M = 0.5 Kg is dropped from a height of 10 meters.
This ball will pop if a force greater than 20 Newtons is applied to it . When it hits the
ground, it stops completely in .1 seconds. Does the ball pop? What could you do to
prevent this?

Answer 1

The force on impact will equal the change in momentum over the change in time:\[F=\frac{\Delta \rho}{\Delta t}=\frac{mv_f-mv_i}{0.1s}\]You already know the mass...so you need to find the velocity at the time of impact. The simplest way to do this is to use:
\[V_f=\pm \sqrt{2gh}=\pm14m/s\]But you can also do it another way - start by calculating the time it takes for the ball to hit the ground from 10m:
\[d=V_it+\frac{1}{2}at^2\]\[10m=0m/s^2+\frac{1}{2}(9.8m/s^2)t^2\]\[t=1.429s\]
Now determine the ball's velocity at that instant:
\[V_f=V_i+at=0m/s+(-9.8m/s^2)(1.429s)=-14m/s\]
Note that the sign is only an indication of direction.

Now that we know the velocity on impact we can determine the momentum on impact:\[\rho = mv = (0.5kg)(-14m/s)=-7kgm/s\]Of course, the final momentum will be 0 so the change in momentum is 7kgm/s.

The force at this instance is:\[F=\frac{\Delta \rho}{\Delta t}=\frac{7kgm/s}{0.1s}=70N\]
Therefore, the ball will break. If you want to prevent this you need to increase the amount of time in the above equation so that the total force <= 20N:
\[F=20N=\frac{7kgm/s}{Xs}\]Solving for X you get 0.35s. So the impact must occur over a time interval of > 0.35s.

Answer 2

*That should have been a >= sign in the last sentence.