direction of acceleration problem

[[Tycho?]]

The qestion is: what is the direction of acceleration in these diagrams?

 

 

Here are the options to choose from.

 

 

I got:

1) A

2) A

3) B

4) B

5) A

 

Which is wrong. So I changed 3 to E and tried it again, and it was still wrong. I dont see where my error could lay, so any suggestions would be appreciated.

[5614]

I would say that 3 is E (0) because in that instant the car is at rest, only once the sponge expands again will there be an acceleration.

 

At the point in the picture where the sponge is completely squashed the net force acting on the car is 0 because gravity cancels out because the sponge is pushing back on the car with an 'equal but opposite force' and the sponge is not yet expanding itself, so the only force you need worry about it gravity, which we just dealt with.

 

But you say that answer is wrong...

 

Well, I'd agree with your answer for 1, 2, 4, 5... so 3 is the real dodgy one.

 

Try 3F

[timo]

4F

At the very point where the cart hits the sponge the answer would be A. At the point where the sponge is completely squished, the answer is B (with any scaling). At any sponge squish between 3 and 4, the anwer lies between A and B depending on how much the sponge is squished. Since this info isn´t given, you can´t know the force.

 

EDIT:

@5614: A zero velocity does not at all mean zero accelleration. Accelleration is the (time-) derivative of velocity. A function can very well have points with f(t)=0 and df/dx (x) != 0. The easiest example would be f(x)=x for x=0.

[[Tycho?]]

4F

At the very point where the cart hits the sponge the answer would be A. At the point where the sponge is completely squished' date=' the answer is B (with any scaling). At any sponge squish between 3 and 4, the anwer lies between A and B depending on how much the sponge is squished. Since this info isn´t given, you can´t know the force.

 

EDIT:

@5614: A zero velocity does not at all mean zero accelleration. Accelleration is the (time-) derivative of velocity. A function can very well have points with f(t)=0 and df/dx (x) != 0. The easiest example would be f(x)=x for x=0.[/quote']

 

While I agree that v=0 does not mean a=0, in this case the sponge is completely squished; it isn't being compressed anymore, and it isn't expanding yet. So I would say the acceleration for 3 is indeed zero. What do you think about this?

[timo]

I started to write a rather long text explaining what´s going on in detail but in the end I think it´s best to keep this as simple as possible and let you make your own picture of the physical background of it.

 

The first reason I why I assume the force in 3) is upwards was already menationed in the previous version of this post: If you´re at point x with velocity 0 and acceleration 0, then due to x(t) = x(t=0) +v*t + a/2 * t², you won´t get away from there (note: This statement is incorrect but it´ll hopefully help you understand why a must be upwards). Since you move upwards from 3) to 4) and have v=0 at 3), your only option left is an upwards acceleration.

 

Another reasoning would go like this: If you place the cart on the sponge it will squish a bit until the setback force of the sponge compensates gravity. But if you come in with some additional kinetic energy, the sponge must squish more to also compensate for this. Now assuming that more squishing leads to a bigger setback force, it should be clear that at 3), the setback force (which is upwards) exceeds the gravitational one ... because you squished bob beyond the point where it´s equal.

 

I hope either of those two reasonings are a help for you to understand what´s going on. A very similar physical standard problem is the ideal spring oscillator (with the solution x(t) = sin(t)), in case you´re interested in further readings.

[[Tycho?]]

Well. I tried

 

A

A

A

F

A

 

and

 

A

A

A

B

A

 

I'm pretty darn confused now, and I only have 3 tries left.

[ecoli]

What is this for?

[timo]

Yes, you are pretty confused. Thinking it´s 3E is a mistake I can understand up to some point. But why did you change your mind from (originally) 3B to 3A now?

 

Perhaps you should slowly go through 1 to 5 in order and give your anwer with an explanation why you think this is the answer.

[5614]

Atheist, in 3 the sponge is completely squashed, all of the car's KE has been turned into PE within the sponge... the car is at rest with no net forces acting on it until the sponge starts to expand.

 

Actually, maybe 3 is 'cant tell' because although the sponge is completely squashed you don't know if it has started to exert a force (start expanding) again or not.

 

Or maybe it is 3B... I can see why it could be, but not why it can't be 0 or can't tell.

 

I know v=0 [math]\neq[/math] a=0 but by saying v=0 I was representing the fact that the car was at rest with no net force on it ie. a=0.

 

As for 4F... why?

 

The cart is rebounding off the sponge, ie. the sponge is still expanding applying a force to the cart.... ahh, I got an idea. 4F because you can't tell what exerts more force on the cart, gravity acting one way, or the sponge acting in the other direction. Noo, that's wrong, because the cart has a velocity upwards, so there must be a bigger force in that direction (supplied by the expanding sponge). Or, it is 4F because the expanding sponge expands with less force as it expands more... so although initially it had more force than gravity (ie. it accelerated the car in direction B) now it is has partially expanded it may no longer be expanding with force greater than gravity so the net force on the car may already be in direction A (downwards).

 

So try:

 

A

A

B

F

A

[Tycho?]'s

 

or

 

A

A

E

F

A

Mine.

[timo]

Atheist, in 3 the sponge is completely squashed, all of the car's KE has been turned into PE within the sponge... the car is at rest with no net forces acting on it until the sponge starts to expand.

If you throw a ball into the air it will be at rest at the uppermost point of its trajectory. Surely, you are not going to tell me that there´s no force acting on it at this point.

 

 

The cart is rebounding off the sponge, ie. the sponge is still expanding applying a force to the cart.... ahh, I got an idea. 4F because you can't tell what exerts more force on the cart, gravity acting one way, or the sponge acting in the other direction.

Exaclty.

 

Noo, that's wrong, because the cart has a velocity upwards, so there must be a bigger force in that direction (supplied by the expanding sponge).

It isn´t wrong. Again: An upwards velocity doesn´t mean upwards acceleration. You´re giving the explanation yourself in the following sentences:

Or, it is 4F because the expanding sponge expands with less force as it expands more... so although initially it had more force than gravity (ie. it accelerated the car in direction B) now it is has partially expanded it may no longer be expanding with force greater than gravity so the net force on the car may already be in direction A (downwards).

[swansont]

'']While I agree that v=0 does not mean a=0, in this case the sponge is completely squished; it isn't being compressed anymore, and it isn't expanding yet. So I would say the acceleration for 3 is indeed zero. What do you think about this?

 

I disagree. If F=0 and v=0, by Newton's first law, the object must remain at rest (like the sponge was a crash pad that deflated). Because we know that the object rebounds, this can't be the case. If there is potential energy, there must be a force: F = -grad(U)

--

 

for #4, the diagram shows the sponge still compressed, so there should be no ambiguity that there is still an acceleration up the plane

[timo]

for #4' date=' the diagram shows the sponge still compressed, so there should be no ambiguity that there is still an acceleration up the plane[/quote']

Are you sure? lets assume a gravitational force G>0 and the force excerted by the sponge to be S = -c*x with c>0 and x>0 being the ammount of compression of the sponge.

How are you going to tell the sign of the net force F = S + G = G - c*x without knowing x?

[5614]

If you throw a ball into the air it will be at rest at the uppermost point of its trajectory. Surely, you are not going to tell me that there´s no force acting on it at this point.

Obviously there is a force on the ball caused by gravity... but this case is different because gravity is canceled out by the "equal and opposite" force of the sponge. At the instant in 3 the sponge is totaly deflated, it is exerting no force on the cart as it hasn't started to expand and gravity has no overall effect due the "equal/opposite reaction" on it due to the sponge.

 

for #4, the diagram shows the sponge still compressed, so there should be no ambiguity that there is still an acceleration up the plane

I agree with Atheist here, you cannot be sure because as the sponge expands the force it exerts becomes weaker, at that point in its expansion you do not know whether the force it exerts is greater than or weaker than gravity so you cannot know the net force on the cart.

[swansont]

Are you sure? lets assume a gravitational force G>0 and the force excerted by the sponge to be S = -c*x with c>0 and x>0 being the ammount of compression of the sponge.

How are you going to tell the sign of the net force F = S + G = G - c*x without knowing x?

 

Ok, I see your point. The net acceleration will change sign while there is still compression.

[swansont]

Obviously there is a force on the ball caused by gravity... but this case is different because gravity is canceled out by the "equal and opposite" force of the sponge. At the instant in 3 the sponge is totaly deflated' date=' it is exerting no force on the cart as it hasn't started to expand and gravity has no overall effect due the "equal/opposite reaction" on it due to the sponge.

[/quote']

 

I think use of "equal and opposite" here is either wrong or misleading. The reaction forces of Newton's third law are always of the same type, and that phrase is tied in with this law. The reaction force to gravity is also a gravitational force.

 

As I pointed out before, the net force on the object can't be zero, otherwise it would remain at rest. If the sponge has stored energy, it necessarily exerts a force. At the point where the net force is zero, there must be a velocity, other wise there would be no further motion - you would have a static system.

[5614]

What I meant by "equal and opposite" was that graivty was supplying a force in one direction and the sponge was supply a counteracting force of equal proportion in the opposite direction, the net result being f=0 and consequently a=0. As in the floor is exerting a force on me opposite/equal to gravity, hence I don't fall through the floor.

 

Anyway, yeah, I accept that 3 cannot be 0 and that the sponge which has PE will be exerting a force on the cart due to that PE so this combination should be the one:

 

A

A

B

F

A

 

2 other things:

 

Like ecoli said, what is this for?

 

And the other thing was that I like this thread, it isn't as obvious or easy as it might seem. Like you look through it first and it seems simple, but then when you think deeper it is quite complex, nice thread!

[swansont]

What I meant by "equal and opposite" was that graivty was supplying a force in one direction and the sponge was supply a counteracting force of equal proportion in the opposite direction' date=' the net result being f=0 and consequently a=0. As in the floor is exerting a force on me opposite/equal to gravity, hence I don't fall through the floor.

[/quote']

 

The objection I have is that "equal and opposite" is a phrase drawn from Newton's third law, and there's enough confusion around about that already, especially with first-semester physics students who might think that this is an application of that law. That's why I was saying it could be misleading - not to imply that it was done intentionally - and so it could lead to a wrong application. I think it's better to say that they cancel, or add to zero.