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A ball is dropped from a height h on a floor if the coefficient of restitution ise. Find the (a). Speed of ball after the first second … n^(th) collision. (b). Maximum height attained by the ball, after the first, second … n^(th) collision. (c). Time taken by the ball to reach the highest point after the first, second, ...n^(th) collision. (d). total distance covered by the ball. (e). totol time of journey. |
Answer» Solution : O to A: Speed of the ball when it hits the ground `u=sqrt(2gh)` Time take by the ball to hit the floor, `h=(1)/(2)g t_(0^(2))impliest_(0)=sqrt((2h)/(g))` `v_(1)-v_(2)=-e(u_(1)-u_(2))` 1 ball, 2 floor `u_(2)=v_(2)=0` `v_(1)=-eu_(1)=-EU` Speed of the ball after the FIRST collision `=eu` A to B `0=v_(1)^(2)-2gh_(1)impliesh_(1)=(v_(1)^(2))/(2g)=(e^(2)u^(2))/(2g)=(e^(2)xx2gh)/(2g)=e^(2)h` HEIGHT attained by the ball after the first collision `=e^(2)h` `0=v_(1)-g t_(1)impliest_(1)=(v_(1))/(g)=(eu)/(g)=(e)/(g)sqrt(2gh)=esqrt((2h)/(g))=et_(0)` After every collision the speed becomes e times. Maximum height becomes `e^(2)` times. time to reach at the heighest point becomes e times. `{:(,1^(st)"collision","After 2nd collision",n^(th)" collsion"),("speed of ball",eu,e^(2)u,e^(n)u),("Maximum height",e^(2)h,e^(4)h,e^(2n)h),("Time to reach at highest point",et_(0),et_(0),e^(n)t_(0)):}` when `u=sqrt(2gh),t_(0)=sqrt(2h//g)` TOTAL distance covered by the ball `S=h+2h_(1)+2h_(2)+. . . infty` `=h+2e^(2)h+2e^(4)h+. . . infty` `=h+2e^(2)h{1+e^(2)+. . .infty}` `=h+2e^(2)h{(1)/(1-e^(2))}=((1+e^(2))/(1-e^(2)))h` Total time of journey `T=t_(0)+2t_(1)+2t_(2)+... infty` `=t_(0)+2et_(0)+2e^(2)t_(0)+. . . infty` `=t_(0)+2et_(0)(1+e+ . . .infty)` `=t_(0)+2et_(0)((1)/(1--e))=((1+e)/(1-e))t_(0)=((1+e)/(1-e))sqrt((2h)/(g))` |
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