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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
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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sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
No sure but the answer should be either h or 0.
 2 years ago

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dw:1353298638078:dw
 2 years ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
So, the ball was not dropped vertically
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
Nope...and sorry i Forgot to Draw the Pic !
 2 years ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
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

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

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
Velocity at Bottom is sqrt 2gh
 2 years ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
dw:1353300955720:dw
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
So..According to u h =h1
 2 years ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
i mean vg' is not equal to 0...,
 2 years ago

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@VincentLyon.Fr
 2 years ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.0
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

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
Ball is on Translation..)
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
Not need to Consider Moment of Inertia
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
@VincentLyon.Fr @experimentX @siddhantsharan @ghazi @ajprincess @CliffSedge @akash123 @Aperogalics
 2 years ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
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, \(v^2_yu^2_y = 2aH\), 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?
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
The answer shuld be 2h/3
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
@CliffSedge @CliffSedge @CliffSedge @CliffSedge
 2 years ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
If there is no friction...then I suppose the height should be halved...
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.0
i also got the ans as h/2. wonder where i'm going wrong....is the ans really 2h/3 ?
 2 years ago

siddhantsharan Group TitleBest ResponseYou've already chosen the best response.0
It's H/2. Recheck your solutions.
 2 years ago
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