After a 0.200-kg rubber ball is dropped from a height of 1.95 m, it bounces off a concrete floor and rebounds to a height of 1.45 m.

(a) Determine the magnitude and direction of the impulse delivered to the ball by the floor.

(b) Estimate the time the ball is in contact with the floor to be 0.08 seconds. Calculate the average force the floor exerts on the ball.

To solve this problem, we need to use the principle of conservation of energy and consider the bounce as an elastic collision.

(a) To determine the magnitude and direction of the impulse delivered to the ball by the floor, we need to calculate the change in momentum of the ball during the bounce.

The impulse is given by the equation:

Impulse = Change in momentum

The momentum of an object is given by the equation:

Momentum = Mass of the object * Velocity

Since the ball is dropped from a height and then rebounds, we can calculate its initial and final velocities using the principle of conservation of energy.

Initially:
Potential Energy = Mass * Gravity * Height
Potential Energy = (0.200 kg) * (9.8 m/s^2) * (1.95 m)

Finally:
Potential Energy = (0.200 kg) * (9.8 m/s^2) * (1.45 m)

Since the ball bounces elastically, its kinetic energy before and after the bounce will be the same:

Kinetic Energy = (1/2) * Mass * Velocity^2

Using the conservation of energy, we can equate the potential energy from before and after the bounce to the kinetic energy after the bounce:

(0.200 kg) * (9.8 m/s^2) * (1.95 m) = (1/2) * (0.200 kg) * Velocity^2

Solving for the final velocity:

Velocity = sqrt(2 * (9.8 m/s^2) * (1.95 m))

Now, we can calculate the initial momentum and final momentum using these velocities:

Initial Momentum = (0.200 kg) * (0 m/s)
Final Momentum = (0.200 kg) * Velocity

The change in momentum is the difference between the initial momentum and the final momentum:

Change in Momentum = Final Momentum - Initial Momentum

Substituting the values:

Change in Momentum = (0.200 kg) * Velocity - (0.200 kg) * (0 m/s)

(b) To estimate the time the ball is in contact with the floor, we are given a value of 0.08 seconds.

Finally, to calculate the average force the floor exerts on the ball:

Average Force = Change in Momentum / Time

Substituting the values:

Average Force = (Change in Momentum) / 0.08 seconds

By plugging in the numbers and solving these equations, you can determine the answers for part (a) and (b).

To solve this problem, we can use the principle of conservation of momentum and the equation for impulse.

(a) Let's first calculate the initial velocity of the ball just before hitting the ground. We can use the equation for gravitational potential energy:

mgh = (1/2)mv^2

where m is the mass of the ball (0.200 kg), g is the acceleration due to gravity (9.8 m/s^2), h is the height (1.95 m), and v is the initial velocity.

Rearranging the equation, we have:

v = sqrt(2gh)

v = sqrt(2 * 9.8 * 1.95)

v ≈ 6.09 m/s

Next, we can calculate the final velocity of the ball just after bouncing off the floor. Since the height is given, we can use the equation for potential energy again:

mgh' = (1/2)mv'^2

where h' is the rebound height (1.45 m) and v' is the final velocity.

Rearranging the equation, we have:

v' = sqrt(2gh')

v' = sqrt(2 * 9.8 * 1.45)

v' ≈ 5.05 m/s

Now, we can calculate the change in velocity (or delta-v) of the ball:

delta-v = v' - v

delta-v ≈ 5.05 m/s - 6.09 m/s

delta-v ≈ -1.04 m/s

Since the direction of the impulse is opposite to the direction of motion, the magnitude of the impulse is the mass multiplied by the absolute value of delta-v:

Impulse = m * |delta-v|

Impulse = 0.200 kg * |-1.04 m/s|

Impulse ≈ 0.208 kg·m/s

Therefore, the magnitude of the impulse delivered to the ball by the floor is approximately 0.208 kg·m/s. The direction of the impulse is downward.

(b) To find the average force exerted by the floor on the ball, we can use the equation for impulse:

Impulse = F * Δt

where F is the average force and Δt is the contact time.

Rearranging the equation, we have:

F = Impulse / Δt

F = 0.208 kg·m/s / 0.08 s

F ≈ 2.6 N

Therefore, the average force exerted by the floor on the ball is approximately 2.6 Newtons.

a. V^2 = Vo^2 + 2g*h = 0 @ max ht.

Vo^2 - 19.6*1.45 = 0
Vo^2 = 19.6*1.45 = 28.42
Vo = 5.33 m/s = Initial velocity after
striking the floor.

Impulse = m*V = 0.20*5.33 = 1.067 kg-m/s

b. V^2 = Vo^2 + 2g*h = 0 + 19.6*1.95 =
38.22
V = 6.18 m/s. = Velocity just before
striking the floor.

a = (Vo-V)/t =
(-5.33-6.18)/0.08 = -143.9 m/s^2

F = m*a = 0.20 * (-143.9) = -28.78 N.