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Yahoo! Group Title

A mass m is released from height h on a block of mass m which rests on a smooth floor after elastic collision with the surface the mass will rise to a height ?

  • 2 years ago
  • 2 years ago

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  1. sauravshakya Group Title
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    No sure but the answer should be either h or 0.

    • 2 years ago
  2. Yahoo! Group Title
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    |dw:1353298638078:dw|

    • 2 years ago
  3. sauravshakya Group Title
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    So, the ball was not dropped vertically

    • 2 years ago
  4. Yahoo! Group Title
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    Nope...and sorry i Forgot to Draw the Pic !

    • 2 years ago
  5. sauravshakya Group Title
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    |dw:1353299020369:dw| The motion of the ball would be somewhat like that. Do u mean to calculate that h' interms of h.

    • 2 years ago
  6. Yahoo! Group Title
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    Yup...

    • 2 years ago
  7. Yahoo! Group Title
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    But The answer shuld be 2h/3 ....No prob...Show the Way u Did This..)

    • 2 years ago
  8. Yahoo! Group Title
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    Velocity at Bottom is sqrt 2gh

    • 2 years ago
  9. gerryliyana Group Title
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    |dw:1353300955720:dw|

    • 2 years ago
  10. Yahoo! Group Title
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    So..According to u h =h1

    • 2 years ago
  11. gerryliyana Group Title
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    i mean vg' is not equal to 0...,

    • 2 years ago
  12. Yahoo! Group Title
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    @Vincent-Lyon.Fr

    • 2 years ago
  13. Vincent-Lyon.Fr Group Title
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    Is the ball rolling on the wedge? Even if it has pure translational KE, max height will be less than h because on top of the parabola, KE is not zero, due to horizontal component of motion.

    • 2 years ago
  14. Yahoo! Group Title
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    Ball is on Translation..)

    • 2 years ago
  15. Yahoo! Group Title
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    Not need to Consider Moment of Inertia

    • 2 years ago
  16. Yahoo! Group Title
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    @Vincent-Lyon.Fr @experimentX @siddhantsharan @ghazi @ajprincess @CliffSedge @akash123 @Aperogalics

    • 2 years ago
  17. Shadowys Group Title
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    Since it is taken as a particle, I will assume that after collision with the floor, it will jump at an angle of 45 degrees. To the maximum height on the other side, H, Using this, \(v^2_y-u^2_y = 2aH\), where \(v^2_y=0, u^2_y =2gh \sin 45, a =g\) Now sub it all in and we get, \(H=\frac{h}{2}\) Did I miss something?

    • 2 years ago
  18. Yahoo! Group Title
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    The answer shuld be 2h/3

    • 2 years ago
  19. Yahoo! Group Title
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    @CliffSedge @CliffSedge @CliffSedge @CliffSedge

    • 2 years ago
  20. Shadowys Group Title
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    If there is no friction...then I suppose the height should be halved...

    • 2 years ago
  21. Yahoo! Group Title
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    @Vaidehi09

    • 2 years ago
  22. Vaidehi09 Group Title
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    i also got the ans as h/2. wonder where i'm going wrong....is the ans really 2h/3 ?

    • 2 years ago
  23. siddhantsharan Group Title
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    It's H/2. Recheck your solutions.

    • 2 years ago
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