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## anonymous 4 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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1. anonymous

No sure but the answer should be either h or 0.

2. anonymous

|dw:1353298638078:dw|

3. anonymous

So, the ball was not dropped vertically

4. anonymous

Nope...and sorry i Forgot to Draw the Pic !

5. anonymous

|dw:1353299020369:dw| The motion of the ball would be somewhat like that. Do u mean to calculate that h' interms of h.

6. anonymous

Yup...

7. anonymous

But The answer shuld be 2h/3 ....No prob...Show the Way u Did This..)

8. anonymous

Velocity at Bottom is sqrt 2gh

9. anonymous

|dw:1353300955720:dw|

10. anonymous

So..According to u h =h1

11. anonymous

i mean vg' is not equal to 0...,

12. anonymous

@Vincent-Lyon.Fr

13. 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.

14. anonymous

Ball is on Translation..)

15. anonymous

Not need to Consider Moment of Inertia

16. anonymous

@Vincent-Lyon.Fr @experimentX @siddhantsharan @ghazi @ajprincess @CliffSedge @akash123 @Aperogalics

17. 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?

18. anonymous

The answer shuld be 2h/3

19. anonymous

@CliffSedge @CliffSedge @CliffSedge @CliffSedge

20. anonymous

If there is no friction...then I suppose the height should be halved...

21. anonymous

@Vaidehi09

22. anonymous

i also got the ans as h/2. wonder where i'm going wrong....is the ans really 2h/3 ?

23. anonymous

It's H/2. Recheck your solutions.

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