1. Jan 25, 2007

### ReoFonzo

1. The problem statement, all variables and given/known data
Ok so my teacher used this example in class, he had this flat wheeled cart rolling on the floor. The object was to release a bucket of sand on top of the flat cart to see if the cart would slow down or accelerate. Easily, all of us the students predicted that the cart would slow down, which was the correct awnser because momentum is velocity times mass. And mass does matter, however when we inversed the experiment, starting the bucket of sand on top of the cart and removing it halfway of its pattern to see wether the cart would accelerate or stay the same ? Many of us said it would accelerate... however it stayed the same.... how ?

2. Relevant equations
velocity times mass = momentum

3. The attempt at a solution
Well I was thinking that maybe, mass can be canceled out since weight is a force... and I also though that velocity cannot be canceled out? I'm really not sure on this one... hopefully I can get some hints.

Last edited: Jan 25, 2007
2. Jan 25, 2007

uh, well correct me if I'm wrong but as far as I know, momentum isn't force x mass.

$$p=mv$$
momentum is mass by velocity, surely.
$$F = \frac {dp}{dv}$$
force is the rate of change of momentum... i.e the derivative of the momentum time graph.

As to why your cart didn't speed up when the sand is removed....

always always always but always remember, that wherever there is a force, there is acceleration (F=ma, follows on from F=dp/dv > F=m(dv/dt) >> f=ma when mass is constant). And the converse is ALWAYS but always true as well, wherever we see an acceleration, there must be a force. Now, by removing the bucket from the trolley, there was no resultant force acting on the trolley... and hence there could be no acceleration. ever.

As to why it slows down when mass is added, this isn't wholly a momentum consideration either: I don't know how much you know about friction, but this is part of the answer. The friction that acts to slow down a moving object ( I'm talking about sliding resistance here) is often modelled as proportional to the normal reaction force on the object from the ground. Now, this force is balancing the weight. By adding more mass, you're adding more weight, hence increasing the normal reaction force on the cart, and hence increasing the amount of friction force on the cart. I can't be sure this is the whole answer, because the wheels are rotating which adds a whole new degree of complexity, but you have always to think about what forces are acting, and if there are none, there won't be any acceleration

3. Jan 25, 2007

### ReoFonzo

Oh yeah, my bad... velocity times mass... I always mix up with impulse lol :S, but yeah I guess friction is part of the experiment ? However... when you take the bucket of sand off, the friction stays the same ?

4. Jan 25, 2007

### ranger

ReoFonzo, since your object has wheels. The situation is a little more complicated. You must now consider rolling friction as opposed to sliding friction.

5. Jan 25, 2007

### ReoFonzo

Rolling friction ? Never heard of ?

6. Jan 25, 2007

### ranger

When you just have the wheels as opposed to the entire bottom of the object in contact with the ground, its an entirely different story.

$$F_R = \mu_R \cdot W$$

$\mu_R$ is the coefficient of rolling friction for the two surfaces
W is the weight of the wheel plus the weight of the object it carries

You can determine $\mu_R$ experimentally:
$$\mu_R = v \cdot \frac{g}{t}$$

v is the initial velocity
g is acceleration dude to gravity
t is time it takes to stop

It depends on the surfaces that are in contact. A soft wheel is slowed by the depression of the wheel. However, if the wheel is hard, static friction slows it down. If the surface is now soft, I think you get the idea what will happen.

Last edited: Jan 25, 2007
7. Jan 25, 2007

### drpizza

ReoFonzo, what was the momentum of *the cart* before you removed the sand.

Or, to turn your problem 90 degrees, imagine I have two carts held together by a very small piece of tape, that barely holds them together. As they are moving along, a tree jumps in front of the cart on the right. It stops dead in its tracks. The tape brakes free from the carts, (slowing the cart on the left a negligible amount). If the carts started with the same mass, would you expect the one on the left to slow to half its speed?

8. Jan 25, 2007

### PhanthomJay

The cart is moving horizontally at a certain speed. Just before the bucket of sand hits the cart, it (the sand) has no horizontal speed. When it hits the cart, it must accelerate up to a certain speed, matching that of the cart. It is the friction force between the cart and sand that accelerates the sand forward in the direction of the cart. By newton's third law, the friction force of the sand on the cart acts in the opposite direction to slow it down. In avery short impulse of time, they move off together at a speed determined by conservation of momentum principle, and no forces longer act (I'm neglecting rolling friction, which slows down the cart regardless of whether the sand is added). When the sand is removed, there are no friction forces acting to begin with, and no forces acting when it is removed, therfore, no speed change. Note further that if you were running with the sand at the same speed of the cart before dropping it onto the cart, such that the sand's initial speed was the same as the cart's speed, then once you dropped the sand on the cart, there would be no change in the speed of the system. Of course, rolling friction complicates the problem, you might want to consider a cart without wheels sliding at constant speed on a frictionless surface, which will yield the same result as you witnessed.