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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 ?
 one year ago
 one year 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 ?
 one year ago
 one year ago

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sauravshakyaBest ResponseYou've already chosen the best response.0
No sure but the answer should be either h or 0.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
So, the ball was not dropped vertically
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
Nope...and sorry i Forgot to Draw the Pic !
 one year ago

sauravshakyaBest 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.
 one year ago

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

Yahoo!Best ResponseYou've already chosen the best response.0
Velocity at Bottom is sqrt 2gh
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.1
dw:1353300955720:dw
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
So..According to u h =h1
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.1
i mean vg' is not equal to 0...,
 one year ago

VincentLyon.FrBest 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.
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
Ball is on Translation..)
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
Not need to Consider Moment of Inertia
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
@VincentLyon.Fr @experimentX @siddhantsharan @ghazi @ajprincess @CliffSedge @akash123 @Aperogalics
 one year ago

ShadowysBest 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?
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
The answer shuld be 2h/3
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
@CliffSedge @CliffSedge @CliffSedge @CliffSedge
 one year ago

ShadowysBest ResponseYou've already chosen the best response.0
If there is no friction...then I suppose the height should be halved...
 one year ago

Vaidehi09Best 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 ?
 one year ago

siddhantsharanBest ResponseYou've already chosen the best response.0
It's H/2. Recheck your solutions.
 one year ago
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