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A-ball-is-bouncing-down-a-flight-of-stairs-The-coefficiate-of-restitution-is-e-The-height-of-each-step-d-and-the-ball-descends-one-step-at-each-bounce-After-each-bounce-it-rebounds-to-heigt-h-above




Question Number 51107 by peter frank last updated on 24/Dec/18
A ball is bouncing down  a flight of stairs.The   coefficiate of restitution  is e.The height of each step  d and the ball descends  one step at each bounce.  After each bounce it rebounds  to heigt  h above the next  lower step.The height h is/  large enough compare with  width of a step that  the empacts are effectively   head on.show that  h=(d/(1−e^2 ))
Aballisbouncingdownaflightofstairs.Thecoefficiateofrestitutionise.Theheightofeachstepdandtheballdescendsonestepateachbounce.Aftereachbounceitreboundstoheigthabovethenextlowerstep.Theheighthis/largeenoughcomparewithwidthofastepthattheempactsareeffectivelyheadon.showthath=d1e2
Commented by peter frank last updated on 24/Dec/18
sir we are given that the ball descend  one step at each bounce.so   ball falls a distance h from  its heighest  position and   rebounds to a distance  (h−d) i think.
sirwearegiventhattheballdescendonestepateachbounce.soballfallsadistancehfromitsheighestpositionandreboundstoadistance(hd)ithink.
Answered by mr W last updated on 24/Dec/18
height of the ball before rebound: h  height of the ball after rebound: h−d  velocity of the ball before rebound: v_(before)   velocity of the ball after rebound: v_(after)   E_(before) =mgh=(1/2)mv_(before) ^2   E_(after) =mg(h−d)=(1/2)mv_(after) ^2   ⇒((h−d)/h)=((v_(after) /v_(before) ))^2   since v_(after) =ev_(before)   ⇒((h−d)/h)=e^2   ⇒h−d=e^2 h  ⇒h=(d/(1−e^2 ))
heightoftheballbeforerebound:hheightoftheballafterrebound:hdvelocityoftheballbeforerebound:vbeforevelocityoftheballafterrebound:vafterEbefore=mgh=12mvbefore2Eafter=mg(hd)=12mvafter2hdh=(vaftervbefore)2sincevafter=evbeforehdh=e2hd=e2hh=d1e2
Answered by peter frank last updated on 24/Dec/18
e=((velocity of separation)/(velocity of approach))  e=((√(2g(h−d)))/( (√(2gh))))  e(√(2gh)) =(√(2g(h−d)))   e^2 .2gh=2g(h−d)  e^2 .2gh=2gh−2gd  e^2 =1−(d/h)  1−e^2 =(d/h)  h=(d/(1−e^2 ))  hence shown
e=velocityofseparationvelocityofapproache=2g(hd)2ghe2gh=2g(hd)e2.2gh=2g(hd)e2.2gh=2gh2gde2=1dh1e2=dhh=d1e2henceshown
Commented by mr W last updated on 24/Dec/18
sorry, i misread the information about h.  now it′s fixed.
sorry,imisreadtheinformationabouth.nowitsfixed.
Commented by peter frank last updated on 24/Dec/18
okay sir.
okaysir.

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