After a 0.300-kg rubber ball is dropped from a height of 1.75 m, it bounces off a concrete floor

and rebounds to a height of 1.50 m. (a) Determine the magnitude and direction of the impulse
delivered to the ball by the floor. (b) Suppose the ball is in contact with the floor for 0.05 s.
Calculate the average force the floor exerts on the ball.

(a) Well, the magnitude of the impulse can be calculated using the equation J = Δp, where J is impulse and Δp is the change in momentum. The change in momentum is simply the final momentum minus the initial momentum.

Now, we know that impulse is equal to the average force multiplied by the time of impact. Since the time of impact is very short, we can assume that the average force delivered by the floor is constant.

So, we can use the equation impulse = average force * time to determine the magnitude of the impulse.

However, to find the direction of the impulse, we need to consider the direction of the velocity change.

In this case, since the ball is dropping from a height and then bouncing back up, the initial velocity vector points downwards (negative direction) and the final velocity vector points upwards (positive direction).

Therefore, the change in velocity is (-v_f) - (-v_i) = v_f + v_i.

Since the magnitude of the velocities is the same, the net velocity change is 2v_i.

Now, the magnitude of the impulse is given by |impulse| = m * |2v_i|, where m is the mass of the ball.

Given that the mass (m) is 0.300 kg, we can calculate the magnitude of the impulse.

(b) To calculate the average force, we can use the equation average force = impulse / time.

Given the impulse from part (a) and the time of impact, which is 0.05 s, we can compute the average force.

To solve this problem, we will use the principle of conservation of mechanical energy and the impulse-momentum theorem.

(a) Determining the magnitude and direction of the impulse:
1. Calculate the initial velocity of the ball just before it hits the floor. We can use the following equation of motion to find the velocity:

v^2 = u^2 + 2as

where
v = final velocity of the ball (-1.50 m/s as it moves upwards after the bounce)
u = initial velocity of the ball (unknown)
a = acceleration due to gravity (-9.8 m/s^2)
s = displacement (1.75 m)

Rearranging the equation, we get:

u^2 = v^2 - 2as
u^2 = (-1.50 m/s)^2 - 2(-9.8 m/s^2)(-1.75 m)
u^2 = 2.25 m^2/s^2 - 34.3 m^2/s^2
u^2 = 36.55 m^2/s^2
u = 6.04 m/s (rounded to two decimal places)

2. Next, we calculate the initial momentum of the ball just before it hits the floor. The momentum of an object is given by the equation:

p = mv

where
p = momentum (unknown)
m = mass of the ball (0.300 kg)
v = initial velocity of the ball (6.04 m/s)

p = (0.300 kg)(6.04 m/s)
p = 1.81 kg·m/s (rounded to two decimal places)

3. Since the ball changes direction after bouncing off the floor, the magnitude of the impulse delivered to the ball is equal to the change in momentum. We can calculate the change in momentum using the equation:

impulse = change in momentum

Given that the initial momentum is 1.81 kg·m/s and the final momentum is unknown, the change in momentum is:

impulse = final momentum - initial momentum

Since the final momentum is opposite in direction to the initial momentum, the impulse will have a negative sign. Therefore:

impulse = (-1.81 kg·m/s) - (1.81 kg·m/s)
impulse = -3.62 kg·m/s (rounded to two decimal places)

The magnitude of the impulse delivered to the ball by the floor is 3.62 kg·m/s, and its direction is downward.

(b) Calculating the average force exerted by the floor on the ball:
1. The impulse applied to an object is also equal to the average force applied multiplied by the time interval over which it acts. We can rearrange this equation to solve for average force:

impulse = force x time

Given:
impulse = -3.62 kg·m/s (from part a)
time = 0.05 s

Substituting these values into the equation, we get:

-3.62 N·s = force x 0.05 s

Solving for force:

force = (-3.62 N·s) / (0.05 s)
force = -72.4 N (negative sign indicates the force is acting in the opposite direction of motion)

The average force exerted by the floor on the ball is approximately 72.4 N (rounded to two decimal places) and it acts in the upward direction.

To solve this problem, we'll need to use the principles of impulse and momentum. Let's break it down step by step.

(a) To determine the magnitude and direction of the impulse delivered to the ball by the floor, we can use the concept of conservation of momentum. The momentum before the collision is equal to the momentum after the collision.

1. Calculate the initial velocity of the ball just before it hits the floor:
Using the equation v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is acceleration, and s is the distance traveled, we can plug in the values:
v = 0 m/s (as the ball momentarily comes to rest at the highest point)
u = ?
a = -9.8 m/s² (acceleration due to gravity)
s = 1.75 m (distance traveled before hitting the floor)

Rearranging the equation, we get:
u = √(v² - 2as)
u = √(0 - 2 * (-9.8) * 1.75)
u = √(0 + 34.3)
u = √34.3
u ≈ 5.85 m/s

2. Calculate the final velocity of the ball just after it bounces off the floor:
Using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is acceleration, and t is time taken, we can plug in the values:
v = ?
u = 5.85 m/s (the initial velocity after bouncing off the floor)
a = -9.8 m/s² (acceleration due to gravity)
t = ? (unknown)

At the highest point after bouncing, the final velocity will be momentarily zero, so we can solve for t:
v = u + at
0 = 5.85 - 9.8t
9.8t = 5.85
t = 5.85 / 9.8
t ≈ 0.597 s

3. Calculate the impulse delivered to the ball by the floor:
Impulse = change in momentum
Momentum before collision = mass * initial velocity
Momentum after collision = mass * final velocity

Mass of the ball = 0.300 kg
Initial momentum = 0.300 kg * 5.85 m/s = 1.755 kg·m/s
Final momentum = 0.300 kg * (-5.85) m/s = -1.755 kg·m/s (opposite direction)

Impulse = Final momentum - Initial momentum
Impulse = (-1.755) kg·m/s - 1.755 kg·m/s
Impulse = -3.51 kg·m/s

Therefore, the magnitude of the impulse delivered to the ball by the floor is 3.51 kg·m/s. Since the impulse is in the opposite direction of the initial velocity, the direction of the impulse is downward.

(b) To calculate the average force the floor exerts on the ball, we can use the equation:
Impulse = average force * time

Given:
Impulse = 3.51 kg·m/s (from part a)
Time = 0.05 s

Rearranging the equation, we get:
Average force = Impulse / Time
Average force = 3.51 kg·m/s / 0.05 s
Average force ≈ 70.2 N

Therefore, the average force the floor exerts on the ball is approximately 70.2 Newtons.

The impulse is the change of momentum of the ball

coming down:
(1/2) m v^2 = m g h
h = 1.75 meters
v down = -sqrt(2 g h) = -sqrt(2*9.81*1.75)
now v up
v up = +sqrt (2*9.81*1.50)
change in momentum = + m sqrt [ 2 * 9.81 (1.50 +1.75) ]
= impulse
then F * time = impulse
F = impulse / 0.05