a .140 kg baseball is dropped from rest from a height of 2.2m above the ground. it rebounds to a height of 1.6 m. what change in the balls momentum occurs when the ball hits the ground

To calculate the change in momentum when the baseball hits the ground, we can use the principle of conservation of momentum.

The equation for the change in momentum is given by:

Δp = m * Δv

where Δp is the change in momentum, m is the mass of the baseball, and Δv is the change in velocity.

First, we need to determine the initial velocity of the baseball just before it hits the ground.

Using the equation of motion, we can find the final velocity of the ball just before it rebounds:

vf^2 = vi^2 + 2 * a * d

where vf is the final velocity, vi is the initial velocity, a is the acceleration (in this case, due to gravity -9.8 m/s^2), and d is the displacement.

Plugging in the values:

0 = vi^2 + 2 * (-9.8 m/s^2) * (-2.2 m)
0 = vi^2 + 43.12 m^2/s^2
vi^2 = -43.12 m^2/s^2
vi = sqrt(-43.12 m^2/s^2)
vi = -6.57 m/s (taking the positive root as the speed is positive before it hits the ground)

The change in velocity when the ball rebounds is the difference between the initial and final velocities. Since the initial velocity was downward, and the final velocity is upward:

Δv = (-6.57 m/s) - (+6.57 m/s)
Δv = -13.14 m/s

Next, we can substitute the mass and change in velocity into the equation for the change in momentum:

Δp = (0.140 kg) * (-13.14 m/s)
Δp = -1.84 kg·m/s

Therefore, the change in the ball's momentum when it hits the ground is -1.84 kg·m/s.

To find the change in momentum of the ball when it hits the ground, we need to calculate the initial momentum (before the ball hits the ground) and the final momentum (after the ball hits the ground and rebounds).

Step 1: Calculate the initial momentum (before the ball hits the ground):
The initial momentum (p_initial) of an object is given by the formula: p_initial = m * v, where m is the mass of the object and v is its initial velocity.
In this case, the baseball is dropped from rest, so its initial velocity is zero. Therefore, the initial momentum is zero.

Step 2: Calculate the final momentum (after the ball hits the ground and rebounds):
The final momentum (p_final) of an object is given by the formula: p_final = m * v, where m is the mass of the object and v is its final velocity.
To calculate the final velocity (v), we need to use the equation of conservation of energy, assuming negligible energy losses due to air resistance.

Potential energy at height 1 (initial position): PE_initial = m * g * h1
Potential energy at height 2 (final position): PE_final = m * g * h2

The potential energy at height 1 equals the kinetic energy at height 2 (assuming negligible energy losses due to air resistance), so:
PE_initial = KE_final

m * g * h1 = (1/2) * m * v^2

Simplifying the equation and solving for v, the final velocity:
v = sqrt(2 * g * (h1 - h2))

Substituting the given values into the equation:
m = 0.140 kg
g = 9.8 m/s^2
h1 = 2.2 m
h2 = 1.6 m

v = sqrt(2 * 9.8 * (2.2 - 1.6)) = sqrt(2 * 9.8 * 0.6) = sqrt(11.76) ≈ 3.43 m/s

Now we can calculate the final momentum:
p_final = m * v = 0.140 kg * 3.43 m/s ≈ 0.48 kg·m/s

Step 3: Calculate the change in momentum:
The change in momentum (Δp) is given by the formula: Δp = p_final - p_initial.
In this case, as we found earlier, the initial momentum is zero.

Δp = p_final - p_initial = p_final - 0 = p_final = 0.48 kg·m/s

Therefore, the change in momentum of the baseball when it hits the ground is approximately 0.48 kg·m/s.

The momentum change is

M [V1 -(-V2)] = M(V1 + V2)
For the two speeds, use
V1 = sqrt[g*2.2/(2M)] = 8.77 m/s
V2 = sqrt[g*1.6/(2M)] = 7.48 m/s