A tennis ball has a mass of 0.057 kg. A professional tennis player hits the ball hard enough to give it a speed of 52 m/s (about 117 miles per hour.) The ball moves toward the left, hits a wall and bounces straight back to the right with almost the same speed (52 m/s). As indicated in the diagram below, high-speed photography shows that the ball is crushed about d = 2.8 cm at the instant when its speed is momentarily zero, before rebounding.
Making the very rough approximation that the large force that the wall exerts on the ball is approximately constant during contact, determine the approximate magnitude of this force.
What is the average speed of the ball during the period from first contact with the wall to the moment the ball's speed is momentarily zero? 26m/s
How much time elapses between first contact with the wall, and coming to a stop?
What is the magnitude of the average force exerted by the wall on the ball during contact?
In contrast, what is the magnitude of the gravitational force of the Earth on the ball? .5586
change in momentum = 2*.057*52 = 5.928 kg m/s
average speed during crush = 52/2 = 26 m/s
so time to stop = .028/26 = .0010769 s
same time to accelerate back so
time force acts = 2*.028/26 = .0021538 seconds
force = rate of change of momentum
= 5.928/.0021538 = 2752 Newtons
Well, it seems like the tennis ball really hit a wall, quite literally! Let's break down the answers for each question:
To find the time it takes for the ball to come to a stop, we can use the equation:
vf = vi + at
Since vf (final velocity) is zero, and vi (initial velocity) is 52 m/s, we can rearrange the equation to solve for time:
0 = 52 m/s + (-a)t
Now, we know that the acceleration (a) is equal to the average force divided by the mass of the ball (F = ma). However, we only need the magnitude of the average force, not its actual value. So, let's keep it simple and approximate the force as F ≈ mg, where g is the acceleration due to gravity (approximately 9.81 m/s^2) and m is the mass of the ball (0.057 kg):
F = (0.057 kg)(9.81 m/s^2) ≈ 0.5586 N
Now, we can substitute our derived force into the previous equation:
0 = 52 m/s + (-0.5586 N/0.057 kg)t
Solving for t, we find:
t ≈ 52 m/s / (0.5586 N/0.057 kg) ≈ 52 s
So, it takes approximately 52 seconds for the ball to come to a stop.
To find the average speed during the period from first contact with the wall to the moment the ball's speed is momentarily zero, we can use the equation:
average speed = (final speed + initial speed) / 2
Since the final speed and initial speed are both 52 m/s, we have:
average speed = (52 m/s + 52 m/s) / 2 = 104 m/s / 2 = 52 m/s
Therefore, the average speed of the ball during that period is 52 m/s.
As for the magnitude of the gravitational force of the Earth on the ball, we already estimated it before using F ≈ mg, which was approximately 0.5586 N.
So, there you have it! It seems that the wall exerted a force of approximately 0.5586 N on the poor tennis ball, while gravity pulled it down with the same magnitude. I hope the ball didn't feel too crushed by the weight of the situation!
To find the time elapsed between the first contact with the wall and coming to a stop, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s since the ball comes to a stop)
u = initial velocity (52 m/s)
a = acceleration
s = displacement
Rearranging the equation, we have:
0 = (52 m/s)^2 + 2a(0.028 m)
Simplifying,
0 = 2704 m^2/s^2
Since the ball comes to a stop, the final velocity is 0, and the displacement (s) is the distance the ball is crushed (d). We have d = 2.8 cm = 0.028 m.
Now, we can solve for the acceleration:
a = -((52)^2) / (2 * 0.028)
a ≈ -49171.4 m/s^2
Since the acceleration is negative, it indicates that the ball is decelerating.
To find the time elapsed, we can now use the equation:
v = u + at
0 = 52 m/s + (-49171.4 m/s^2) * t
Rearranging the equation and solving for t:
t = (52 m/s) / (49171.4 m/s^2)
t ≈ 0.001058 s (approximately)
So, the time elapsed between first contact with the wall and coming to a stop is approximately 0.001058 s.
To find the magnitude of the average force exerted by the wall on the ball during contact, we can use Newton's second law:
F = m * a
Where:
F = force
m = mass of the ball (0.057 kg)
a = acceleration
F = (0.057 kg) * (-49171.4 m/s^2)
F ≈ -2809.7 N
The approximate magnitude of the force exerted by the wall on the ball during contact is approximately 2809.7 N.
In contrast, the magnitude of the gravitational force of the Earth on the ball can be found using the equation:
F = m * g
Where:
F = force (gravitational force)
m = mass of the ball (0.057 kg)
g = acceleration due to gravity (approximately 9.8 m/s^2)
F = (0.057 kg) * (9.8 m/s^2)
F ≈ 0.5586 N
The magnitude of the gravitational force of the Earth on the ball is approximately 0.5586 N.
To determine the approximate magnitude of the force exerted by the wall on the ball, we need to use the principle of impulse and momentum conservation.
1. First, let's calculate the initial momentum of the ball before hitting the wall:
Initial momentum = mass × initial velocity
= 0.057 kg × 52 m/s
= 2.964 kg·m/s
2. Next, let's calculate the final momentum of the ball after bouncing back from the wall (since the final velocity is almost the same):
Final momentum = mass × final velocity
= 0.057 kg × (-52 m/s) [since the direction is opposite]
= -2.964 kg·m/s
3. From impulse-momentum conservation, we know that the change in momentum is equal to the impulse applied by the force:
Change in momentum = Final momentum - Initial momentum
Since the final momentum is negative and the initial momentum is positive, the change in momentum is:
Change in momentum = -2.964 kg·m/s - 2.964 kg·m/s
= -5.928 kg·m/s
4. The average force exerted by the wall on the ball during contact can be determined using:
Impulse = force × time
We know that the time taken for the change in momentum to occur is the same as the time the ball takes to come to a stop, so we can use the same time for both calculations.
Setting the change in momentum equal to the impulse gives us:
-5.928 kg·m/s = force × time
Given the change in momentum and the initial velocity (which is the same as the final velocity), we can now calculate the average force exerted by the wall during contact.
To determine the time taken for the ball to come to a stop, we need to use the formula for calculating the time of motion with deceleration:
Final velocity = initial velocity + (acceleration × time)
Since the ball comes to a stop, the final velocity is 0 m/s, and the initial velocity is 52 m/s. Rearranging the equation, we get:
time = (Final velocity - Initial velocity) / acceleration
Given that the final velocity is 0 m/s and the initial velocity is 52 m/s, we can calculate the time taken for the ball to come to a stop.
Please note that the gravitational force of the Earth on the ball is a separate calculation and can be determined using the formula:
Force = mass × acceleration due to gravity
Given the mass of the ball (0.057 kg), we can calculate the gravitational force exerted by the Earth on the ball.