Homework Help: Solving for Kinetic Energy and Speed of a Rebounding Tennis Ball

In summary: Joules.In summary, the conversation is about solving a physics problem involving a tennis ball being released from rest at a height of 2.0m and rebounding to a height of 1.5m. The person has already solved part a) by calculating the kinetic energy and speed just before impact. They are struggling with parts b) and c) which involve calculating the kinetic energy and speed just after impact and the loss of energy between release and maximum height after rebounding. Another person explains that it is important to consider kinetic energy and potential energy in the problem and provides the equations and calculations for parts b) and c). The loss of energy is determined to be 1.5 Jou
  • #1
Turkish
34
0
Hi again.
I have 3 sets of questions which I spose are fairly easy for the ones here.
1) A tennis ball of mass 0.20kg is released from rest at a height of 2.0m above a concrete floor.
It rebounds to a height of 1.5m

a) Calculate its kinetic energy and speed just before impact
b) calculate its kinetic energy and speed just after impact
c) calculate its loss of energy between release and its maximum height after rebounding

I have managed to solve a) I did, V^2 = U^2 + 2as
So, 0^2 + 2*10*2.0 = 6.3 So the Velocity is 6.3m/s
Therefore, Kinetic energy = 1/2 mv^2
So, 0.20 * (6.3*6.3 = 40) / 2 = 4J

But I don't understand how I can do b and c, can someone please help me :) Thanks in advance.
 
Physics news on Phys.org
  • #2
Think about all of this in terms of kinetic energy. PE is all it has before it is released, and

PE = mgh = .2*10*2 = 4.

Just before impact, PE = 0, and so all energy is kinetic, i.e. KE = 4

.5*m*v^2 = 4, m*v^2 = 8, v^2 = 40, v = 6.32

So far so good.

Now since it came back up to 1.5 m, PE = mgh = .2*10*1.5 = 3. Did the ball lose any energy between right after hitting the floor and going back up to 1.5? Nope. It was in the air. So KE = 3. Do you see this? It's important. Because PE is 3 after the bounce, KE right after the bounce must equal this, because, once again, PE = 0 at that point.

Thus,

3 = .5*m*v^2, 6 = m*v^2, 30 = v^2, v = 5.48 or something.

And so what is the loss of energy? Why, it's simply the change in potential energy. Which is 1.
 
  • #3


For part b, you can use the same equation V^2 = U^2 + 2as, but this time the initial velocity (U) is the final velocity from part a (6.3 m/s). The final velocity (V) would be 0 m/s since the ball comes to a stop at the maximum height. The acceleration (a) would be -10 m/s^2 since the ball is decelerating due to gravity. Plugging these values into the equation, you should get a final kinetic energy of 1.6J.

For part c, you can find the difference between the kinetic energy just before impact (4J) and the kinetic energy just after impact (1.6J). This would give you the loss of energy between release and maximum height after rebounding, which is 2.4J.

I hope this helps! Remember to always pay attention to the units and make sure they are consistent throughout your calculations. Good luck!
 

What is motion?

Motion is defined as the change in position of an object over time.

What are the different types of motion?

The different types of motion include linear, circular, oscillatory, and rotational motion.

How is motion measured?

Motion is typically measured using units of distance, time, and speed, such as meters, seconds, and meters per second.

What are the laws of motion?

The laws of motion were formulated by Sir Isaac Newton and include the law of inertia, the law of acceleration, and the law of action and reaction.

How can you calculate motion?

Motion can be calculated using equations such as distance = speed x time and acceleration = change in velocity / time.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
728
Replies
1
Views
570
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
34
Views
694
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
32
Views
2K
  • Introductory Physics Homework Help
2
Replies
55
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
19
Views
672
  • Introductory Physics Homework Help
Replies
24
Views
1K
Back
Top