Physics Help!?

Question

A superball of mass 0.05 kg is dropped from a height of 10 cm above a table top. It bounces off the table and rises to the same height.
a. If the superball was in contact with the table for 30 ms, calculate the averge force exerted on the ball by the table.

Answer 1

impulse (I) = change in momentum (dp)
dp = m x dv
first we need velocity
v^2 = u^2 +2as
where u is initial velocity (u = 0 because its dropped)
v^2 = 0 +2g(0.1)
v = sqrt (0.2g)
= 1.4 m / s

Impulse = m x dv
dv = 1.4 - 0 = 1.4
Impulse = 0.05 x 1.4
= 0.07 Ns

b)
use Impulse = Ft
find time using
v = u+at
t = v/a where a =g
= 1.4/9.81 = 0.142s
Impulse = F x 0.142
0.07 = F x 0.142
F = 0.07 / 0.142
= 0.493 N
==0.49 N

Answer 2

Step 1: figure out how fast the ball was traveling when it hit the table. Use the formula: v = sqrt(2gh).

Step 2: Notice that the ball was traveling at the same speed (but in the opposite direction) immediately after the bounce. The clue for this is that the ball reached its original height after the bounce.

Step 3: Calculate how much the ball's momentum changed during the bounce. (Remember that momentum is a vector, so you have to take into account the change in direction as well as the change in amount (if any).)

Step 4: Use the formula: "Impulse = change in momentum". Remember that impulse = (average force)×(time of contact). Therefore:
(average force)×(time of contact) = (change in momentum).
They tell you the "time", and you calculated the change in momentum. So solve for "average force."