InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A plastice ball is dropped from a height of 1m and rebounds several times from the floor. IF1.3 s elaspes from from the moment it is dropped to the second impact with the floor, what is the coefficient of restitution? |
|
Answer» Here, `h=1m, u=0.` Velocity with which the ball strikes the floor for the first time, `v_(1)=sqrt(2gh)=sqrt(2xx9.8xx1)=4.427m//s` Time taken by the ball to strike the floor for the first time, `t_(1)=sqrt((2h)/(g))=sqrt((2xx1)/(9.8))=0.452s` Time taken by the ball between first and second impact with the floor is `t_(2)=1.3-t_(1)=1.3-0.452=0.848s. ` Time taken by the ball for upward journey after first imapct impact `=(t_(2))/(2)=(0.848)/(2)=0.424s` `:.` upward velocity after first impact `v_(2)=gxx(t_(2))/(2)=9.8xx0424=4.155m//s` `:` Coefficient of restitution. `e=(v_(2))/(v_(1))=(4.155)/(4.427)=0.939` |
|