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Question Number 53325 by mr W last updated on 20/Jan/19

Commented by mr W last updated on 20/Jan/19

Two balls have equal size. Ball A has  mass m and rests. Ball B has mass  nm (n is integer greater than 1). Ball  B moves with velocity v towards ball  A and behind ball A there is a wall.  Assume there is no friction and all  collisions are elastic and the wall is  far away enough.  How many times will both balls   collide with each other?

Twoballshaveequalsize.BallAhasmassmandrests.BallBhasmassnm(nisintegergreaterthan1).BallBmoveswithvelocityvtowardsballAandbehindballAthereisawall.Assumethereisnofrictionandallcollisionsareelasticandthewallisfarawayenough.Howmanytimeswillbothballscollidewitheachother?

Commented by ajfour last updated on 20/Jan/19

too good a question Sir, let me two  days at least.

toogoodaquestionSir,letmetwodaysatleast.

Commented by mr W last updated on 20/Jan/19

thanks sir!  waiting for your result...

thankssir!waitingforyourresult...

Commented by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19

i have seen similar question in you tube...sharing  the link...

ihaveseensimilarquestioninyoutube...sharingthelink...

Commented by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19

Commented by mr W last updated on 20/Jan/19

thanks for this result but which i can not  follow yet. i′ll go to watch the video,  but at first i′ll try to solve in my  own way. my understanding is:  during the first x collisions the  energy will be transfered from ball B  to ball A. during  the next y collisions  the energy will be tranfered back to  ball B and ball B will be finally faster  than ball A and no further collision  is possible.

thanksforthisresultbutwhichicannotfollowyet.illgotowatchthevideo,butatfirstilltrytosolveinmyownway.myunderstandingis:duringthefirstxcollisionstheenergywillbetransferedfromballBtoballA.duringthenextycollisionstheenergywillbetranferedbacktoballBandballBwillbefinallyfasterthanballAandnofurthercollisionispossible.

Commented by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19

tanθ=(√(m_L /m_H ))   N_c θ=π  N_c =(π/θ)=(π/(tan^(−1) (√(m_L /m_H ))))  [N_c =number of collissiin]  1)example..(m_L /m_H )=(1/(100))  tan^(−1) ((√(1/(100))) )  tan^(−1) ((1/(10)))≈(1/(10))  N_c =(π/(1/(10)))=10π≈31 number of collision  2)example two  (m_L /m_H )=(1/(10000))  tan^(−1) ((√(m_L /m_H )) )  tan^(−1) ((1/(100)))  N_c =(π/(1/(100)))=100π=314 number of collission  3)(m_L /m_H )=(1/n)  tan^(−1) ((√(1/n)))≈(1/(√n))  N_c =(π/(1/(√n)))=π(√n) number of collission..  sir pls check the video in youtube...also  check my answer..  number of collission N_c =π(√n)

tanθ=mLmHNcθ=πNc=πθ=πtan1mLmH[Nc=numberofcollissiin]1)example..mLmH=1100tan1(1100)tan1(110)110Nc=π110=10π31numberofcollision2)exampletwomLmH=110000tan1(mLmH)tan1(1100)Nc=π1100=100π=314numberofcollission3)mLmH=1ntan1(1n)1nNc=π1n=πnnumberofcollission..sirplscheckthevideoinyoutube...alsocheckmyanswer..numberofcollissionNc=πn

Commented by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19

thank you sir...i have just tried to understand  the video...you pls go through it...am i really  understood it yet...you pls cross check...  hiwever your logic regarding collissin and  energy transfer really excellent...  thank you sir...

thankyousir...ihavejusttriedtounderstandthevideo...youplsgothroughit...amireallyunderstoodityet...youplscrosscheck...hiweveryourlogicregardingcollissinandenergytransferreallyexcellent...thankyousir...

Commented by mr W last updated on 22/Jan/19

very nice method in video!

verynicemethodinvideo!

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