A ball of mass m1 = 8.0 × 10−2 kg starts from rest and falls vertically downward from a height of 3.0 m. After colliding with the ground, it...

First we have to calculate the velocity of ball immediately before collision and immediately after it.

We can use the equation of motion: v^2 = u^2 + 2as

where, v is the final velocity, u is the initial velocity, a is acceleration (acceleration due to gravity in this case) and s is the distance traveled.

Before collision, s = 3 m, u = 0 m/s and a = g = 9.81 m/s^2

Thus, v^2 = 2 x 9.81 x 3 

solving this, we get, v = 7.67 m/s

Thus, momentum immediately before collision = mv = 8 x 10^-2 x 7.67 kg m/s

= 0.614 kg m/s 

Similarly, after the collision, ball rises to a height of 2 m.

Thus, s = 2 m, v = 0, a = -g = -9.81 m/s^2

Thus,  0 = u^2 + 2 x (-9.81) x 2

solving this, we get, u = 6.26 m/s

Thus, momentum immediately after collision = mu = 8 x 10^-2 x 6.26 kg m/s

= 0.501 kg m/s

b) Force exerted can be calculated as:

F = m(v-u)/t = 8 x 10^-2 x (7.67 - 6.26)/(5 x 10^-3) = 22.6 N

c) Impulse = Ft = 22.6 x 5 x 10^-3 = 0.113 kg m/s

Hope this helps.