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anonymous
 3 years ago
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 ?
anonymous
 3 years ago
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 ?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No sure but the answer should be either h or 0.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1353298638078:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So, the ball was not dropped vertically

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Nope...and sorry i Forgot to Draw the Pic !

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1353299020369:dw The motion of the ball would be somewhat like that. Do u mean to calculate that h' interms of h.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But The answer shuld be 2h/3 ....No prob...Show the Way u Did This..)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Velocity at Bottom is sqrt 2gh

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1353300955720:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So..According to u h =h1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i mean vg' is not equal to 0...,

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Is 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
 3 years ago
Best ResponseYou've already chosen the best response.0Ball is on Translation..)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Not need to Consider Moment of Inertia

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@VincentLyon.Fr @experimentX @siddhantsharan @ghazi @ajprincess @CliffSedge @akash123 @Aperogalics

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Since 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, \(v^2_yu^2_y = 2aH\), 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
 3 years ago
Best ResponseYou've already chosen the best response.0The answer shuld be 2h/3

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@CliffSedge @CliffSedge @CliffSedge @CliffSedge

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If there is no friction...then I suppose the height should be halved...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i also got the ans as h/2. wonder where i'm going wrong....is the ans really 2h/3 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It's H/2. Recheck your solutions.
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