Problem-11 | Reviving mechanics for JEE main

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  • Опубликовано: 21 апр 2021

Комментарии • 3

  • @bokiiibrother9995
    @bokiiibrother9995 3 года назад +1

    maximum height that A can go is 80m. Although that A ball reach the maximum height within 4seconds...that means when the A ball in the maximum height the B ball starts to project vertically. According to the principle of conservation of mechanic energy this two balls collide 40m above from the floor. It already mentioned the above collision is a elastic one...then the Energy of the entire system is a constant...It can be consider as the kinetic energy of that 2 balls. If we take the mass of one ball equals to "m" we can get the entire energy of the system is 1600m. When The collide is happening the potential and kinetic energy both balls equals to 1600m. And we know that collide is happen 40m above from the floor .so we can get the potential energy of that two balls as 800m ...then the kinetic energies of that two balls equals to 1600m-800m= 800m// ....if we take "v" as the velocity of that two balls when collide is happening...the entire kinetic energy = mv^2 and that equals to 800m then the answer should be 800^1/2=v
    2 DAYS AGO
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    • @10xphysics
      @10xphysics  3 года назад

      Nice explanation 😄thanks

  • @studywithraushan1116
    @studywithraushan1116 3 года назад +1

    Ha ab thik se dikhta hai questions