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

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
9 replies · 2K views
toboldlygo
Messages
32
Reaction score
2

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!
 
Physics news on Phys.org
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

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?
 
  • Like
Likes   Reactions: toboldlygo
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?