Momentum, Force, and Impulse: help with balance pan problem

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Homework Help Overview

The problem involves a stream of elastic glass beads falling onto a balance pan and bouncing back to their original height. Participants are discussing the relationship between momentum, force, and impulse in the context of this scenario.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the calculation of velocity before and after the beads hit the pan, using energy conservation principles and impulse equations. There are questions about the change in velocity and how it relates to the beads' behavior after impact.

Discussion Status

Some participants are sharing their calculations and questioning the values used for velocity. There is a productive exchange regarding the implications of the beads bouncing back and how that affects the overall analysis. Clarifications about the direction of velocity and its impact on momentum are being discussed.

Contextual Notes

Participants are navigating potential misunderstandings about the velocity of the beads after impact and the assumptions regarding their motion. The discussion is framed within the constraints of a homework problem, emphasizing the need for careful consideration of the physics involved.

toboldlygo
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Homework Statement


[/B]
A stream of elastic glass beads, each with a mass of 0.53 g, comes out of a horizontal tube at a rate of 99 per second. The beads fall a distance of 0.49 m to a balance pan and bounce back to their original height. How much mass must be placed in the other pan of the balance to keep the pointer at zero?

Homework Equations


impulse = force * Δp ==> F = mΔv/Δt
conservation of energy: potential energy = kinetic energy ==> mgh = 0.5mv2

The Attempt at a Solution



I think I'm supposed to find the velocity of the initial and final moments (when the beads exit the tube, and as they hit the pan, respectively), then use that to find the force (using the impulse equation), and then find mass from that.

So, I set mgh = 0.5mv2, and using the velocity that I obtained, I solved for F (I used 1/99 as Δt), and then I divided that number by 9.81 m/s/s to get my answer. I don't know where I'm going wrong, though. Any help would be much appreciated!
 
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toboldlygo said:
So, I set mgh = 0.5mv2, and using the velocity that I obtained, I solved for F (I used 1/99 as Δt), and then I divided that number by 9.81 m/s/s to get my answer.
That sounds ok, so please post your detailed working.
(I suspect you used the wrong value for Δv.)
 
haruspex said:
That sounds ok, so please post your detailed working.
(I suspect you used the wrong value for Δv.)

Here's my work. I've attached it to the reply.
 

Attachments

toboldlygo said:
Here's my work. I've attached it to the reply.
My suspicion is confirmed.
What happens to the bead after it hits the pan?
 
After it hits the pan, the beads bounce back up to their original height. But how does that factor into the velocity? The velocity of the bead is 3.1006 m/s before it hits the pan, and then zero when it does. Isn't that a change of 3.1006 m/s?
 
toboldlygo said:
After it hits the pan, the beads bounce back up to their original height. But how does that factor into the velocity? The velocity of the bead is 3.1006 m/s before it hits the pan, and then zero when it does. Isn't that a change of 3.1006 m/s?
If its velocity is zero after hitting the pan, how does it regain its original height?
 
haruspex said:
If its velocity is zero after hitting the pan, how does it regain its original height?

It couldn't, if its velocity were zero. Does that mean it hits the pan with twice the velocity? There can't be zero change in velocity, could there? I know what you're saying now, but I don't know how I would factor that into a solution.
 
toboldlygo said:
It couldn't, if its velocity were zero. Does that mean it hits the pan with twice the velocity? There can't be zero change in velocity, could there? I know what you're saying now, but I don't know how I would factor that into a solution.
Velocity and momentum are vectors; signs matter. If the velocity before impact is -x (up positive) what will it be after bouncing? What is the difference of the two?
 
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haruspex said:
Velocity and momentum are vectors; signs matter. If the velocity before impact is -x (up positive) what will it be after bouncing? What is the difference of the two?

Oh, that makes sense! So if I were to subtract initial from final, I'd get x-(-x), which is just 2x, right? Or did I completely miss the mark?
 
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toboldlygo said:
Oh, that makes sense! So if I were to subtract initial from final, I'd get x-(-x), which is just 2x, right? Or did I completely miss the mark?
That's it.
 
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