Solving Linear Momentum: Glass Bead Stream Problem

[Lisa...]

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

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?

 

[Doc Al]

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.)

 

[11thgradephysics]

impulse- momentum
Ft = m(Vf - Vi)
solve for F

 

[Lisa...]

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
v_i= 0 m/s
v_f:

W= E_k
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= v_f

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= F_gravity/g= 0.04429/9.81= 4.52 g

Correct?

 

[Doc Al]

You made one error: in calculating Vf - Vi, Vi is the velocity of the bead just before hitting the pan. It's not zero.

 

[Lisa...]

Oops :) Okay so vi=vf= 4.43 m/s and m= 9 g right?

 

[Doc Al]

Right. (But I'd prefer that you say Vi = -4.43 m/s & Vf = +4.43 m/s.)