A glass marble is dropped down an elevator shaft

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Homework Help Overview

The problem involves a glass marble dropped in an elevator shaft, which hits a glass plate on a descending elevator. The marble's collision with the plate is elastic, and the task is to determine how high the marble will rise after the collision, relative to its initial drop point.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relevant equations and clarify the meanings of variables involved. There is an exploration of the marble's speed relative to both the elevator and the ground, as well as the implications of the elastic collision on the marble's subsequent motion.

Discussion Status

Participants are actively engaging with the problem, questioning assumptions about initial conditions and relative velocities. Some have provided calculations for the marble's speed before and after the collision, while others are seeking clarification on how to determine the height the marble will reach after bouncing.

Contextual Notes

There is an ongoing discussion about the definitions of variables and the setup of the problem, particularly regarding the initial conditions of the marble and the elevator's motion. Participants are also considering the implications of the elastic collision on the marble's velocity.

Salk13
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Homework Statement


[/B]
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?

Homework Equations


v 2 - u 2 = 2gh
v 2 = u 2 + 2gh
[/B]
H = v 2 / 2g
H-h= required height

The Attempt at a Solution

[/B]
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
not right answer though!
 
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Hello Salk, welcome to PF :smile: !

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.
 
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 !
 
Salk13 said:
v^2=(2.10 m/s)^2+ 2(9.81)(7.5m)
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.
 
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
 
Salk13 said:
I thought about finding the speed of the marble before impact by using conservation of energy.
That's good - what answer do you get?
Salk13 said:
I am having trouble with getting the velocity of the marble with respect to the "ground"
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?
 
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.
 
Salk13 said:
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.
Are the marble and the glass roof traveling in the same direction or opposite directions before bounce?
 
they are traveling in the same direction
 
  • #10
Salk13 said:
they are traveling in the same direction
If the marble is falling at 12.13m/s and the glass plate at 2.1 m/s, what's the relative velocity?
 
  • #11
14.23 m/s
 
  • #12
Salk13 said:
14.23 m/s
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
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
Salk13 said:
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.
Right, so do the same with the marble falling at 12.13 m/s instead of 2.1m/s.
 
  • #15
The relative velocity would be 10.03 m/s
 
  • #16
Salk13 said:
The relative velocity would be 10.03 m/s
Yes.

That's much better.
 
  • #17
Salk13 said:
The relative velocity would be 10.03 m/s
Right, so what would the relative velocity be immediately after the bounce?
 
  • #18
10.03 m/s because the collision is elastic and kinetic energy is conserved
based on that would height be 5.127m ?
 
  • #19
Salk13 said:
10.03 m/s because the collision is elastic and kinetic energy is conserved
based on that would height be 5.127m ?
Height is relative to what ?
 
  • #20
If you're talking about the height asked for in the problem statement, the answer is no.
 
  • #21
yes I was talking about the height as in answer to question
 
  • #22
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 I am not seeing how to get this answer?
 
  • #23
Check how you calculated the 5.13 m and think about what that means: starting from where is that 5.13 m ?
 
  • #24
Salk13 said:
10.03 m/s because the collision is elastic and kinetic energy is conserved
based on that would height be 5.127m ?
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
Bvu it should be from the ground which is the glass plate

haruspex -10.03 m/s
 
  • #26
i know 10.03 m/s is relative to Earth but i don't know how it is different after the bounce
 
  • #27
The velocity relative to the elevator was 10.03m/s before the collision, and it is the same after the collision. The elevator is moving, so the velocity relative to Earth will be different.
 
  • #28
it is 7.93 m/s since the marble is going up and the elevator is going down
 
  • #29
Salk13 said:
it is 7.93 m/s since the marble is going up and the elevator is going down
Good. So calculate the height to which it will rise from the position at which it bounced.
 

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