The Bouncing Ball: Understanding Energy Conservation

In summary, Time tries to crush a 0.85kg basketball, but is only able to compress it by 1.5cm when a force of 50N in applied. When he throws it against the floor, it is compressed by 3.4cm. How high will the ball bounce?
  • #1
EricPowell
26
0

Homework Statement


Time tries to crush a 0.85kg basketball, but is only able to compress it by 1.5cm when a force of 50N in applied
a) What is the spring constant of the ball?
b) When he throws it against the floor, it is compressed by 3.4cm. How high will the ball bounce?

Homework Equations


##F=kx## where k is the spring constant, and x is the length of the compression.

The Attempt at a Solution


The first part was easy enough.
a)##
F=3333\frac{N}{m}*0.034m
\\
=113.322N
##

b) I thought that maybe I could find two forces, the one due to gravity and the other due to the compression of the ball, and add those two forces together.
##
F_g=mg
\\
=0.58kg*(-9.81\frac{m}{s^2})
\\
=-5.6898N##

##
F_c=3333\frac{N}{m}*0.034m
\\
=113.322N
##
where Fc is the force due to compression.

##F_{Net}=113.322N-5.6898N
\\
=107.6322N
##

##
F=ma
\\
107.6322N=0.58kg*a
\\
a=185.57\frac{m}{s^2}
##

Could somebody push me in the right direction here?
 
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  • #2
As the ball decompresses, the force is not a constant.
That's why this approach is proving tricky.

One way forward would be to use conservation of energy.

When the ball is compressed by x, how much energy is stored?
How much energy is stored by the ball rising to a height h?
 
  • #3
from the initial conditions F=kx you get k=10000/3 then put .5kx^2=mgh
 
  • #4
Just noticed that that question should say "Tim", not "Time".

I should also note that my physics class is an introductory one. This question that I have posted was marked as "bonus" on the worksheet, so I would not be surprised if solving this question involves using something that we haven't learned in class yet.

Simon Bridge said:
As the ball decompresses, the force is not a constant.
That's why this approach is proving tricky.

One way forward would be to use conservation of energy.

When the ball is compressed by x, how much energy is stored?
How much energy is stored by the ball rising to a height h?

I'm not quite sure what the conservation of energy is. Could you show me how this works?
 
Last edited:
  • #5
EricPowell said:
I'm not quite sure what the conservation of energy is. Could you show me how this works?
Conservation of energy works like conservation of momentum - but for total energy from all sources i.e. kinetic and potential.

What sort of energy does the ball have when compressed?
What sort of energy does it have at the top of it's bounce?
Do you know the expressions for each of these?
How are they related to each other?
 

1. What factors affect the height of a bouncing ball?

The height of a bouncing ball is affected by several factors such as the material and elasticity of the ball, the surface it bounces on, and the force applied to it.

2. How does the height of a bouncing ball change over time?

The height of a bouncing ball decreases each time it bounces due to the loss of energy from impacting the surface and air resistance. Eventually, the ball will stop bouncing and come to rest.

3. What is the relationship between the initial height of a bouncing ball and its final height?

The final height of a bouncing ball is typically less than its initial height due to the loss of energy with each bounce. However, in a perfect system with no energy loss, the final height would be equal to the initial height.

4. How does the surface affect the height of a bouncing ball?

The surface a bouncing ball impacts on affects its height by either absorbing or reflecting energy. A softer surface, such as grass or carpet, will absorb more energy, resulting in a lower bounce compared to a harder surface, like concrete or wood.

5. Can the height of a bouncing ball be predicted or calculated?

Yes, the height of a bouncing ball can be predicted or calculated using mathematical equations that take into account factors like gravity, initial height, and energy loss. However, these calculations may not be accurate in real-world scenarios due to variables like air resistance and surface imperfections.

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