A community for students.
Here's the question you clicked on:
 0 viewing
Yahoo!
 2 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 ?
Yahoo!
 2 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 ?

This Question is Closed

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

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

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

sauravshakya
 2 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.

Yahoo!
 2 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..)

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

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

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

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

VincentLyon.Fr
 2 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.

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

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

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

Shadowys
 2 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?

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

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

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

Vaidehi09
 2 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 ?

siddhantsharan
 2 years ago
Best ResponseYou've already chosen the best response.0It's H/2. Recheck your solutions.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.