# Question on linear momentum

1. Dec 3, 2005

### 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.

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?

2. Dec 3, 2005

### Staff: Mentor

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. Dec 3, 2005

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

4. Dec 6, 2005

### 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
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. Dec 6, 2005

### Staff: Mentor

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. Dec 7, 2005

### Lisa...

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

7. Dec 7, 2005

### Staff: Mentor

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