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Abhisar
 one year ago
A 1 Kg ball falls from a height of 25 cm and rebounds upto a height of 9 cm. The coefficient of restitution is
Abhisar
 one year ago
A 1 Kg ball falls from a height of 25 cm and rebounds upto a height of 9 cm. The coefficient of restitution is

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Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1Umm.. I got it this. I'll solve it though if any one else needs it. Final velocity of the ball will be 2.21 m/s [using \(\sf V^2=U^2+2aS\)] Similarly, the ball's initial velocity after the collision will be 1.32 m/s [ [using \(\sf V^2=U^2+2aS\)] Coefficient of restitution = \(\sf \frac{relative~velocity~after~the~collision}{relative~velocity~before~the~collision}\) Since, ground is our frame of reference, relative velocities before and after collision will become 2.21 and 1.32 Substituting the values you get 0.6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got 0.6 doing a different way.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You should not have solved it without someone posting :(

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1@shalante ,Why don't you post your way too, we will all get to learn different ways c:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Abhisar I'll post in a while since I am in class right now.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It is just the kinetic and potential energy way It is an isolated system so \[\Delta KE +\Delta PE=0\] \[\frac{ 1 }{ 2 }mv _{f}^2\frac{ 1 }{ 2 }mv _{i}^2+mgh _{f}mgh _{i}=0\] \[\frac{ 1 }{ 2}mv _{i}^2+mgh _{i}=\frac{ 1 }{ 2 }mv _{f}^2+mgh _{f}\] Cancel all m \[\frac{ 1 }{ 2 }v _{i}^2+gh _{i}=\frac{ 1 }{ 2}v _{f}^2+gh _{f}\] L In a coefficient of restitution, when an object collides with a fixed object, the formula becomes similar to this \[coefficient of restiuition=\frac{ final velocity after collision }{ final velocity before collision }\] Convert 25 cm to 0.25 m and 9 cm to 0.09 m and since acceleration is in meters Time to plug and chug! To get final velocity before collision let \[v _{i}=0m/s, g=9.8 m/s ^{2} h _{i}=0.25m, h _{f}=0m\] Solve for final velocity So final velocity before collision is 2.21m/s To get final velocity after collision let \[v _{i}=0m/s, g=9.8m/s^2, h _{i}=0m, h _{f}=0.09m\] So final velocity before collision is 1.32m/s \[COR=\frac{ 1.32m/s }{ 2.21m/s }0.6\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oops meant to be \[COR=\frac{ 1.32m/s }{ 2.21m/s }=0.6\] Made a typo by putting a subtraction sign when it is supposed to be an equal sign. Well, this way is longer, but probably more useful in other problems that a bit similar to this where your formula does not work.
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