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aquamarine08
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[SOLVED] Problem about the Conservation of Energy
When a certain rubber ball is dropped from a height of 1.25m onto a hard surface, it loses 18% of its mechanical energy on each bounce. (a) How high will the ball bounce on the first bounce? (b) How high will it bounce on the second bounce?(c) With what speed would the ball have to be thrown downward to make it reach its original height on the first bounce?
Eq.1 ~ [tex]V_{1}^{2}[/tex]= [tex]V_{0}^{2}[/tex]+2a([tex]d_{1}[/tex]-[tex]d_{0}[/tex])
Eq.2 ~ E= U+K (E is mechanical energy, U is potential energy, and K is kinetic energy)
Eq.3 ~ U=mgh
Eq.4 ~ K= [tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex]
I first solved for velocity using Eq.1 ...
[tex]V_{1}^{2}[/tex]= [tex]V_{0}^{2}[/tex]+2a([tex]d_{1}[/tex]-[tex]d_{0}[/tex])
[tex]0^{2}[/tex]= [tex]V_{0}^{2}[/tex]+ 2(-9.8)(1.25-0)
0=[tex]V_{0}^{2}[/tex]+24.5
-24.5 = [tex]V_{0}^{2}[/tex]
4.95 m/s = [tex]V_{0}[/tex]
and then I thought I would use this value to solve Eq.2...
E= U+K
E=mgh+[tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex]
then I saw a problem...I don't have the mass of the ball!
What I was planning on doing before seeing this mass situation...was to solve for the mechanical energy and then take off 18% of its energy and use the same Eq.2 to solve for the height and repeat this again for the second bounce.
Now I am lost...please help! Thank you!
Homework Statement
When a certain rubber ball is dropped from a height of 1.25m onto a hard surface, it loses 18% of its mechanical energy on each bounce. (a) How high will the ball bounce on the first bounce? (b) How high will it bounce on the second bounce?(c) With what speed would the ball have to be thrown downward to make it reach its original height on the first bounce?
Homework Equations
Eq.1 ~ [tex]V_{1}^{2}[/tex]= [tex]V_{0}^{2}[/tex]+2a([tex]d_{1}[/tex]-[tex]d_{0}[/tex])
Eq.2 ~ E= U+K (E is mechanical energy, U is potential energy, and K is kinetic energy)
Eq.3 ~ U=mgh
Eq.4 ~ K= [tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex]
The Attempt at a Solution
I first solved for velocity using Eq.1 ...
[tex]V_{1}^{2}[/tex]= [tex]V_{0}^{2}[/tex]+2a([tex]d_{1}[/tex]-[tex]d_{0}[/tex])
[tex]0^{2}[/tex]= [tex]V_{0}^{2}[/tex]+ 2(-9.8)(1.25-0)
0=[tex]V_{0}^{2}[/tex]+24.5
-24.5 = [tex]V_{0}^{2}[/tex]
4.95 m/s = [tex]V_{0}[/tex]
and then I thought I would use this value to solve Eq.2...
E= U+K
E=mgh+[tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex]
then I saw a problem...I don't have the mass of the ball!
What I was planning on doing before seeing this mass situation...was to solve for the mechanical energy and then take off 18% of its energy and use the same Eq.2 to solve for the height and repeat this again for the second bounce.
Now I am lost...please help! Thank you!
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