A ball of mass m= 0.08 kg, starting at rest, is dropped from a height h1= 3.0 meters above the ground and bounces back up to a height h2 = 2.0 meters. The collision with the ground occurs over a time of .005 seconds.

Solution:

The momentum of the ball before it hit the ground = m v₁

v₁² = u² + 2gs (Since the ball is dropped u = 0)

v₁² = 2 x 9.8 x 3 = 58.8

v₁ = 7.6 m/s

Momentum p₁ = 0.08 x 7.6

= - 0.614 kg m/ s

(The momentum is pointing downwards, so Negative)

b. what is the momentum of the ball immediately after it hit the ground?

Solution:

The momentum of the ball after hit the ground = m v₂

v₂² = u² + 2gs

0 = u² - 2gs    (since 'g' is negative)

u² = 2gs = 2 x 9.8 x 2

= 39.18

u = 6.26 m/s

Momentum p₂ = 0.08 x 6.26

= +0.501 kg m/s(The momentum is pointing upwards, so Positive)

c. what is the average force of the ground on the ball?

Solution:

Average Force F = Change in momentum/ ∆t

Since p₁ is acting downwards it posses negative sign.

F = p₂ - (-p₁) / ∆t

F = p₂ + p₁ / ∆t

F = (0.501 + 0.614) / 0.005

F = 1.115 / 0.005

F = 223 N

d. what impulse is imparted to the ball during its collision with the ground?

Solution:

Impulse = Change in linear momentum

I = p₂ -(-p₁)

I = p₂ + p₁

I = (0.501 + 0.614)

I = 1.115 kg m/s

Energy lost = Change in Kinetic energy before and after collision

= ½ m (V₂² – V₁²)

= ½ 0.08 (6.26 ² - 7.6² )

= - 0.786 Joules

Negative sign indicates lost in kinetic energy