anonymous
  • anonymous
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 ?
Physics
chestercat
  • chestercat
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anonymous
  • anonymous
No sure but the answer should be either h or 0.
anonymous
  • anonymous
|dw:1353298638078:dw|
anonymous
  • anonymous
So, the ball was not dropped vertically

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anonymous
  • anonymous
Nope...and sorry i Forgot to Draw the Pic !
anonymous
  • anonymous
|dw:1353299020369:dw| The motion of the ball would be somewhat like that. Do u mean to calculate that h' interms of h.
anonymous
  • anonymous
Yup...
anonymous
  • anonymous
But The answer shuld be 2h/3 ....No prob...Show the Way u Did This..)
anonymous
  • anonymous
Velocity at Bottom is sqrt 2gh
anonymous
  • anonymous
|dw:1353300955720:dw|
anonymous
  • anonymous
So..According to u h =h1
anonymous
  • anonymous
i mean vg' is not equal to 0...,
anonymous
  • anonymous
@Vincent-Lyon.Fr
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
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.
anonymous
  • anonymous
Ball is on Translation..)
anonymous
  • anonymous
Not need to Consider Moment of Inertia
anonymous
  • anonymous
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?
anonymous
  • anonymous
The answer shuld be 2h/3
anonymous
  • anonymous
If there is no friction...then I suppose the height should be halved...
anonymous
  • anonymous
anonymous
  • anonymous
i also got the ans as h/2. wonder where i'm going wrong....is the ans really 2h/3 ?
anonymous
  • anonymous
It's H/2. Recheck your solutions.

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