1. Nov 29, 2009

### Boogeyman

1. The problem statement, all variables and given/known data
Sand is poured at a steady rate of 5g/s onto the pan of a direct reading balance calibrated in grams. If the sand falls from a height of 0.20m onto the pan and it does not bounce off the pan then, neglecting any motion of the pan, calculate the reading on the balance 10s after than sand first hits the pan.

3. The attempt at a solution
I thought the solution was to simply find the mass of sand that would fall after 10s i.e 50g. However this is not the answer.

Evidently, the height from which it falls must therefore affect the reading on the balance. Does the acceleration that occurs in the fall affect the reading? If this is the case, I am tempted to find Potential Energy.

PE=mgh
PE=(0.05)(10)(0.2)
PE=0.1J

Since energy is conserved then the total energy remains the same:
PE of sand before being poured=KE of sand before it hits the pan
KE=0.1J

KE=1/2 mv2

Now if any of this was correct, how do I find v?

2. Nov 29, 2009

### cepheid

Staff Emeritus
The net force on an object is equal to its rate of change of momentum. In this case, because the sand does not bounce, it seems continuously transferred from the sand to the pan.

3. Nov 29, 2009

### Boogeyman

You are saying f=ma right? How do I use this? And what does the second part of what you said mean?

4. Nov 29, 2009

### cepheid

Staff Emeritus
Not exactly. That's only a special case of Newton's Second Law for situations in which the mass is constant. The most general statement of Newton's Second law is that the net force on an object is equal to its rate of change of momentum. In this case, the pan's mass is changing at a certain rate, and therefore its momentum is changing at a certain rate, which is equal to the force it is experiencing.

Sorry, the second part of what I said had a really bad typo and should have read:

In this case, because the sand does not bounce, it seems momentum is continuously transferred from the sand to the pan.

5. Nov 29, 2009

### Boogeyman

Ok but then we need to find the velocity the sand is moving with. Did I do that correctly or do you not have to find it?

6. Nov 29, 2009

### cepheid

Staff Emeritus
Yes, you can find the velocity of the sand upon impact from conservation of energy as you have described:

(1/2)mv2 = mgh

v2 = 2gh ​

Or you could use the kinematics formula which says that:

vf2 = 2ah ​

You'll notice that these are in fact exactly the same equation.

7. Nov 29, 2009

### Boogeyman

p=mv
p=(5)(2)
p=10
p/t=F
F=10/10
F=1?
m=1/10
=WRONG!!

Dude I am not getting the answer from the back of the book. I used this value of v, and found change in momentum. Then I found the force after ten second.

8. Nov 29, 2009

### cepheid

Staff Emeritus
This is true

But this is not true. The 5g/s is NOT the mass. It is the *rate* at which mass is flowing onto the pan. Hence, grams PER second. In other words, it is the change in mass divided by the change in time:

$$\frac{\Delta m}{\Delta t} = 5\, \textrm{g/s}$$

Therefore, what IS true is that the rate of change of momentum is given by:

$$\frac{\Delta p}{\Delta t} = \left(\frac{\Delta m}{\Delta t}\right)v$$

By Newton's second law, this is equal to F. So there are two forces on the pan after 10 s. One is due to the *weight* of the total amount of sand accumulated on the pan (which is just the weight of 50 g). The second force on the pan is what we just calculated...the force that manifests itself because sand is flowing onto the pan and is therefore transferring vertical momentum to it at a certain rate. If add these together to get the total force on the pan, you'll see that the weigh scale is going to give you a measurement that is slightly higher than 50 g because it "feels" a higher force than just the weight of the sand, and therefore it assumes/concludes that it is being loaded by a slighter larger mass.

9. Nov 30, 2009

### Boogeyman

I UNDERSTAND!! Thanks so much. I have never done a question quite like this, and these rates questions are proving to be tricky.

So Fw + F$$\Delta$$p = Total downward force
Therefore, Fw=0.5N

F$$\Delta$$p=0.005 x 2

=0.01

Total downward force=0.5 + 0.01

=0.51N

Mass=51g