Question on linear momentum

  • Thread starter Lisa...
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  • #1
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Could anybody help me out with the following problem?

A stream of glass beads, each with a mass of 0.1 g, comes out of a horizontal tube at a rate of 100 per second.

http://img4.imageshack.us/img4/11/beads1xr.th.gif [Broken]

The beads fall a distance of 1 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?
 
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  • #2
Doc Al
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Consider the average impulse that the beads exert on the pan--and vice versa. (Relate the impulse to the change in momentum of the beads.)
 
  • #3
impulse- momentum
Ft = m(Vf - Vi)
solve for F
 
  • #4
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What I've done is the following:

Ft = m(Vf - Vi)
F = (m(Vf - Vi) )/t

t= 1/f= 1/100= 0.01 s
m= 0.1 g = 0.1 E -3 kg
vi= 0 m/s
vf:

W= Ek
Fs= 1/2 m v2

where s= 1 m, m= 0.1 g = 0.1 E -3 kg and F= the gravity working on the beads= m* g

mgs= 1/2 m v2
gs= 1/2 v2
v2= 9.81/0.5
v= sqrt(9.81/0.5)= 4.43 m/s= vf

Ok so now filling this in the formula F = (m(Vf - Vi) )/t gives F= 0.04429 N

This is the force each bead exerts on the scale, therefore it needs to equal the force the mass m on the other side exerts on the scale which is it's gravity equalling m g.

m= Fgravity/g= 0.04429/9.81= 4.52 g

Correct?
 
  • #5
Doc Al
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You made one error: in calculating Vf - Vi, Vi is the velocity of the bead just before hitting the pan. It's not zero.
 
  • #6
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Oops :) Okay so vi=vf= 4.43 m/s and m= 9 g right?
 
  • #7
Doc Al
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Right. (But I'd prefer that you say Vi = -4.43 m/s & Vf = +4.43 m/s.)
 

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