Kinematics

Why does it have a slower speed after leaving?

Why does it have a slower speed after leaving?

Wouldn't it decelerate inside the doughnut?


Posted from TSR Mobile

Gonna add another question. is it because of the donut causing a small gravitational force on the ball bearing:?

Yes, but if so why can it not also accelerate?

Aha right. So it will accelerate due to the gravitational attraction between them?


Posted from TSR Mobile

Assuming this is what the question is expecting (where has it come from?) then the ball bearing is attracted to the doughnut as it approaches and also as it moves away. So how do you think this force will affect its speed
A. As it approaches
B. as it moves away?

As it approaches, it will accelerate towards it increasing its velocity and the opposite happens when it leaves. So speed stays the same then?



Posted from TSR Mobile

Correct.
But what does the graph look like, then?

Correct.
But what does the graph look like, then?

Let me just try one more time.

So I did understand the the gravitational attraction between them will cause an increase in speed as it approaches and the opposite as it leaves so the speed will be the same as before entering. And the graph might look like this:


Now till this part it's okay with me.

Now if I want to think about it further.
As the all approaches the donut particles, why there won't be any force that oppose the motion like friction force when a car moves on a road with a certain "mu" value.
I don't mean the force is due to friction here but just shooting an example.
This is what i mean
^ The bold arrows shows the gravitational attraction and the other dotted arrow is the force when they hit (?)

Sorry for confusing it, but I hope you'll be able to clarify my problem.

There is only one force here - gravitation. This happens in space so no air resistance. No viscosity. There is no physical contact between the ball and the ring. I have to assume the question is talking about a ring doughnut and the ball goes through it without touching. The question is poorly written as it gives no background, or this was not the complete question. Or it is open ended and not an exam style question. If you post questions like this you will not get a straight answer in most cases.

The whole process is symmetrical, the ball could have gone from left to right or right to left. The attractive force on approach varies in an identical way to the force as the ball moves away.The accelerating force at any point on the left of the doughnut is exactly equal to the decelerating force at the equivalent point on the right.
The graph is symmetrical.

In reality, the gravitational force of a ball bearing on a doughnut is incredibly small, so I can't imagine how the writer of this question is expecting anyone to take it that seriously, because you would never be able to measure the change in the velocity as it goes through. However, taking the spirit of the question to be that it wants you to talk about this incredibly small change in velocity, then your red line needs to be
a) symmetrical, not lop sided
b) smooth and without a sharp discontinuity, rising, levelling out, then falling back to the original value.

You didn't say where you found this question.

Wow! That's beyond brilliant! Thanks a ton
You made everything clear.


oh sorry, I didn't know that you asked me.
I found it here: https://sites.google.com/site/oxbridgeinterviewquestions/why-is-water-so-important-to-life
Check the first question. I think this was asked in an interview or something :/

Yes, well interview questions require a different approach. They are usually starters for some sort of discussion and/or designed to see how you think. The last one is a good example.

"What happens if I drop an ant?"

If you asked this on here with no background you would get a number of different replies depending on how the reader has interpreted it. Your reply to this is more about the way you interact with the interviewer, and demonstrate your thinking process, than coming up with the "correct" answer.

There is only one force here - gravitation. This happens in space so no air resistance. No viscosity. There is no physical contact between the ball and the ring. I have to assume the question is talking about a ring doughnut and the ball goes through it without touching. The question is poorly written as it gives no background, or this was not the complete question. Or it is open ended and not an exam style question. If you post questions like this you will not get a straight answer in most cases.

The whole process is symmetrical, the ball could have gone from left to right or right to left. The attractive force on approach varies in an identical way to the force as the ball moves away.The accelerating force at any point on the left of the doughnut is exactly equal to the decelerating force at the equivalent point on the right.
The graph is symmetrical.

In reality, the gravitational force of a ball bearing on a doughnut is incredibly small, so I can't imagine how the writer of this question is expecting anyone to take it that seriously, because you would never be able to measure the change in the velocity as it goes through. However, taking the spirit of the question to be that it wants you to talk about this incredibly small change in velocity, then your red line needs to be
a) symmetrical, not lop sided
b) smooth and without a sharp discontinuity, rising, levelling out, then falling back to the original value.

You didn't say where you found this question.

Lol this was 9 years ago, But was wondering that surely if the ball has an initial velocity with respect to the donut - the time that the ball bearing would experience a force would be less as it leaves the donut because it was has accelerated untill it reached the centre of the donut then it decelerates once it has gone past the donut. Would there not suggest a disparty in the momentum lost as it leaves vs the momentum gained travelling in? If we take this we would surely find that the velocity of the ball bearing should be greater at very far distances

This would need to be in a system in which the donut has gained velocity - then it had previously, since if the ball bearning has gained velocity, the donut must have (in the negative direction) in order for momentum to be conserved.

An arguement to be constructed against this is, there would be an increase in energy of the system as both the ball bearing and the donut has increased in speed, and so increase in kinetic energy. But one could argue that the masses of the ball bearning and the donut decreases to account for the energy.

All this said, could you consider the interaction of the donut and the ball bearing as some sort of elastic momentum collision, where the donut imparts a force on the donut and so the momentum of the donut increases and the momentum of the ball bearing decreases. This is system could be consistent with both the conservation of energy and momentum. And so why would this model of whats going on be necessarily wrong.

Finally, a reason to potential support the first model as you have described could be that the time the force IS in fact experienced over as it travels in and out of the donut being equal because it is in the gravitational field of the donut and so the time it experiences relative to another person would be "Slower / faster" depending upon its motion through spacetime determined by the shape and mass of the donut. Supporting the idea that the momentum gained before reaching the centre of the donut would be equibalent to the momentum lost leaving the centre of the donut.

Just wondering if anyone would like to weigh in on what I have said. This is not really an answer per se, but Just considering other models ( gave two others) that may be consistent with at least my current knowledge of physics. It's a very interesting question!