# Homework Help: A glass marble is dropped down an elevator shaft

1. Apr 16, 2015

### Salk13

1. The problem statement, all variables and given/known data

A glass marble is dropped down an elevator shaft and hits a thick glass plate on top of an elevator that is descending at a speed of 2.10 m/s. The marble hits the glass plate 7.5 m below the point from which it was dropped. If the collision is elastic, how high will the marble rise, relative to the point from which it was dropped?

2. Relevant equations
v 2 - u 2 = 2gh
v 2 = u 2 + 2gh

H = v 2 / 2g
H-h= required height
3. The attempt at a solution
v^2=(2.10 m/s)^2+ 2(9.81)(7.5m)
v= 12.31
12.31^2 / 2(9.81)
7.72-7.5= .22

2. Apr 16, 2015

### BvU

Hello Salk, welcome to PF !

There are a few things you must explain to me a bit further before I can help sensibly.

What is it the relevant equations are saying ?
Not $v_2 = u_2 + 2 gh$ because, at least if $v_2$ and $u_2$ are speeds, with the dimension of m/s, that doesn't fit dimensionally.
Perhaps you mean $v^2 = u^2 + 2 gh$ ? But what are $v$ and $u$ ?

From the next one I almost gather $u^2 = 0$ but I find it hard to believe that's what you mean.

So a bit more verbose, please, and we can get going.
Nice exercise, this is.

3. Apr 16, 2015

### BvU

Oh, and do you get a green bar above the tekst you are editing ? With an x2 for subscripts and an x2 for exponents ? Very nifty !

4. Apr 16, 2015

### haruspex

That would be right if the marble were descending at 2.1 m/s initially, and fell a further 7.5m before hitting the glass. But the marble starts from rest - it's the 'ground' that's moving.

5. Apr 16, 2015

### Salk13

BvU sorry for the confusion on equation I meant to say v[2] = u[2] + 2(a)x with x=7.5 u= speed of elevator and v= velocity of ball before impact

haruspex okay I thought about finding the speed of the marble before impact by using conservation of energy. I am having trouble with getting the velocity of the marble with respect to the "ground" so i can calculate height using v[2] equation

BvU i tried using the green bar for exponents but they arent working so those are v squared

6. Apr 16, 2015

### haruspex

That's good - what answer do you get?
From energy you can calculate the speed relative to the lift shaft. If the elevator is descending at 2.1 m/s, what's the marble's speed relative to that?

7. Apr 16, 2015

### Salk13

I got 12.13 m/s relative to the shaft. Then, using relative motion, the marble should be 14.23 m/s relative to ground.

8. Apr 16, 2015

### haruspex

Are the marble and the glass roof travelling in the same direction or opposite directions before bounce?

9. Apr 16, 2015

### Salk13

they are travelling in the same direction

10. Apr 16, 2015

### haruspex

If the marble is falling at 12.13m/s and the glass plate at 2.1 m/s, what's the relative velocity?

11. Apr 16, 2015

### Salk13

14.23 m/s

12. Apr 16, 2015

### haruspex

No.
Consider this, if they were both descending at 2.1m/s, what would the relative velocity be? Hint: it wouldn't be 4.2m/s.

13. Apr 16, 2015

### Salk13

it would be 0. I would use the equation Vac= Vab+Vbc with a= marble, b=elevator and c= ground. If they are both falling to the ground, then Vac=Vbc=2.1 m/s, leaving the relative velocity to 0.

14. Apr 17, 2015

### haruspex

Right, so do the same with the marble falling at 12.13 m/s instead of 2.1m/s.

15. Apr 17, 2015

### Salk13

The relative velocity would be 10.03 m/s

16. Apr 17, 2015

### SammyS

Staff Emeritus
Yes.

That's much better.

17. Apr 17, 2015

### haruspex

Right, so what would the relative velocity be immediately after the bounce?

18. Apr 17, 2015

### Salk13

10.03 m/s because the collision is elastic and kinetic energy is conserved
based on that would height be 5.127m ?

19. Apr 17, 2015

### SammyS

Staff Emeritus
Height is relative to what ?

20. Apr 17, 2015

### vela

Staff Emeritus

21. Apr 17, 2015

### Salk13

yes I was talking about the height as in answer to question

22. Apr 17, 2015

### Salk13

height is relative to the point where the marble first dropped but how do you find the height (the answer) is there another way cause im not seeing how to get this answer?

23. Apr 17, 2015

### BvU

Check how you calculated the 5.13 m and think about what that means: starting from where is that 5.13 m ?

24. Apr 17, 2015

### haruspex

One step at a time.... Yes, the relative velocity after the bounce is 10.03 m/s (in the other direction), so what now is the marble's velocity relative to the earth?

25. Apr 17, 2015

### Salk13

Bvu it should be from the ground which is the glass plate

haruspex -10.03 m/s