*Originally posted by uzless*

**You drop a 300 pound ball to the ground from a height of 1cm.
**

You then drop a 100 pound ball to the ground from a height of 10cm.

If you had a device to measure the impact on the ground, which would have a higher impact, the 100 pound ball or the 300 pound ball? Show your calcs.

Drawing a free body diagram around a ball, we see that the only force acting on the ball is the force of gravity acting in the downward direction:

F = ma = -mg

Dividing through by the mass "m", we see that:

a = -g

And therefore:

y" = -g

Integrating both sides, and assuming the ball is initially at rest, we get:

y' = -gt + 0 = -gt

Integrating both sides again, and assuming a height of "h", we get:

y = -(1/2)gt^2 + h

The ball will increase in velocity as it is dropped until it reaches a maximum just before impact. Letting y = 0 and solving for t, we have:

(1/2)gt^2 = h

t = SQRT(2h/g)

Subbing this value into the expression for the velocity, we have:

y' = -g*SQRT(2h/g) = -SQRT(2gh)

The maximum momentum of the ball is the product of this and its mass:

p = mv = -m*SQRT(2gh)

Subbing in the values provided in the question, we find the following:

m1 = 300 lbs = 300/2.2 kg =~ 136.36 kg

h1 = 1 cm = 0.01 m

g = 9.81 m/s^2

**p1 = -(136.36)*SQRT(2*9.81*0.01) = 60.4 kg*m/s**
m2 = 100 lbs = 100/2.2 kg =~ 45.45 kg

h2 = 10 cm = 0.1 m

g = 9.81 m/s^2

**p2 = -(45.45)*SQRT(2*9.81*0.1) = 63.7 kg*m/s**
Therefore the momentum of the second ball will be higher at impact. Now, the force of impact depends on a number of factors including the material properties of the ball, the material properties of the impact surface, the momentum of the ball, the angle of impact, etc... but we can make the simplifying assumption that both balls have the same material properties (including density). This assumption simplifies things because if they both have the same density, the second ball will be smaller. This means that the force of the second ball's impact will be spread over less area, resulting in greater pressure at impact. If the definition of "greater impact" is greater penetrating power, then the second ball clearly has greater impact.