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

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A tennis ball with a mass of 0.20 kg is released from a height of 2.0 m and rebounds to 1.5 m. The kinetic energy just before impact is calculated to be 4 J, with a speed of approximately 6.3 m/s. After the bounce, the kinetic energy is determined to be 1.6 J, corresponding to a speed of about 5.48 m/s. The energy loss between the initial release and the maximum height after rebounding is found to be 2.4 J. Consistent unit usage is emphasized for accurate calculations.
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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.
 
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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.
 


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!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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