Understanding Free Fall: Newton's Laws and Gravity

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Homework Statement

From a height of 50 meters two heavy masses are dropped at 1s interval.

Relevant Equations

They ask how high the second masses is when the first hits the ground

I don't even know where to begin. I assume that the masses are equal.

 

[BvU]

Hello dummy,
!

Please read https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

You have to post at least some attempt !

I don't even know where to begin

Start posting the equations you learned that may be useful here

I assume that the masses are equal.

Ass u me ?

designate them $m_1$ and $m_2$

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This it what i found

 

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View attachment 295885

This it what i found

Is it correct?

 

[BvU]

If I could read it I might be able to answer that


https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/ :

5 Do not simply post images of the problem statement or your work.​

Please make the effort to type up the problem statement and your work. Ask yourself "If I can't be bothered to spend my time typing it, why should they be bothered to spend their time reading it, much less responding to it?" Use images for supporting figures. You may, of course, attach an electronic copy of the problem statement in addition to the typed version. Indeed, if it's a complicated or long problem, you probably should, but you should always provide a typed version as well.​

6 Don't post poor images.​

When you do use an image in your post, make sure it's in focus, oriented the right way, well lit, etc. It seems like this should be obvious, but experience has shown that people frequently post incredibly poor images. Add images as attachments to the post. Don't host it externally. That way it will remain on PF indefinitely, and your thread will remain useful to future visitors.​

​

​

 

[PeroK]

Is it correct?

Is that $26.4m$?

 

[BvU]

But it seems you do know where to begin and you do know to find your way through this exercise, so not all is lost !

!

Teaser: what if $m_2 = 2 m_1$ ?

$\$

 

[A dummy progression]

If I could read it I might be able to answer that


https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/ :

5 Do not simply post images of the problem statement or your work.​

Please make the effort to type up the problem statement and your work. Ask yourself "If I can't be bothered to spend my time typing it, why should they be bothered to spend their time reading it, much less responding to it?" Use images for supporting figures. You may, of course, attach an electronic copy of the problem statement in addition to the typed version. Indeed, if it's a complicated or long problem, you probably should, but you should always provide a typed version as well.​

6 Don't post poor images.​

When you do use an image in your post, make sure it's in focus, oriented the right way, well lit, etc. It seems like this should be obvious, but experience has shown that people frequently post incredibly poor images. Add images as attachments to the post. Don't host it externally. That way it will remain on PF indefinitely, and your thread will remain useful to future visitors.​

​

​

 

[PeroK]

View attachment 295886

You have got the right answer, but I don't understand your method. If the initial height is $h_0 = 50m$, then, for mass $m_1$:$$h = h_0 -\frac 1 2 gt^2$$

 

[A dummy progression]

But it seems you do know where to begin and you do know to find your way through this exercise, so not all is lost !

!

Teaser: what if $m_2 = 2 m_1$ ?

$\$

You have got the right answer, but I don't understand your method. If the initial height is $h_0 = 50m$, then, for mass $m_1$:$$h = h_0 -\frac 1 2 gt^2$$

Thanks for the method. It is way simpler.

 

[BvU]

View attachment 295886

Oh well, at least it's legible now. Much better than what I achieved with my expensive image enhancement software .

So $h_t$ is not the height at time $t$, but the distance over which $m$ has fallen in time $t$ . Fine with me, but easily confusing, which may be disastrous later on.
So it's good you approve of Perok's approach.

And you didn't answer my teaser !

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Oh well, at least it's legible now. Much better than what I achieved with my expensive image enhancement software .

So $h_t$ is not the height at time $t$, but the distance over which $m$ has fallen in time $t$ . Fine with me, but easily confusing, which may be disastrous later on.
So it's good you approve of Perok's approach.

And you didn't answer my teaser !

$\$

Give me some time for the teaser. I am putting the homework on paper.

 

[BvU]

Let me save you some time: do you see an $m$ anywhere in the calculations to get the answer ?



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Let me save you some time: do you see an $m$ anywhere in the calculations to get the answer ?



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No, and i think that the m is of no use

 

[PeroK]

No, and i think that the m is of no use

Newton's law of gravity says that the force between two masses is proportion to the product of their masses. If $M$ is the mass of the Earth, and $R$ is the radius of the Earth, then the gravitational force on an object of mass $m$ on the surface of the Earth has magnitude:$$F_g = \frac{GMm}{R^2}$$Where $G$ is the universal gravitational constant.

Also, Newton's second law says that the acceleration of a body of mass $m$ subject to a force of magnitude $F$ is given by:$$a = \frac F m$$Putting these two laws together we see that the acceleration of a body at the surface of the Earth is independent of its mass:$$g = \frac{F_g}{m} =\frac{GM}{R^2} \approx 9.8m/s^2 \ \ (1)$$This is why in free fall problems we can omit the masses and forces and skip directly to the equation $$y = y_0 + u_yt - \frac 1 2 gt^2$$Where I've taken upwards to be positive, hence $a = -g$.

If we were dealing with an electromangentic force on a particle of change $q$ and mass $m$ then we would have to take both the charge and mass of the particle into account (as well as the EM field strength) in order to calculate the acceleration.

Note that equation (1) holds approximately for anywhere near the surface of the Earth. It doesn't hold for problems involving satellites or the Moon or any motion that is sufficiently far from the Earth's surface. For any problem where the heights involved are no more than a few kilometres, you can assume that $g$ is approximately a constant of $9.8m/s^2$.

 

[A dummy progression]

Newton's law of gravity says that the force between two masses is proportion to the product of their masses. If $M$ is the mass of the Earth, and $R$ is the radius of the Earth, then the gravitational force on an object of mass $m$ on the surface of the Earth has magnitude:$$F_g = \frac{GMm}{R^2}$$Where $G$ is the universal gravitational constant.

Also, Newton's second law says that the acceleration of a body of mass $m$ subject to a force of magnitude $F$ is given by:$$a = \frac F m$$Putting these two laws together we see that the acceleration of a body at the surface of the Earth is independent of its mass:$$g = \frac{F_g}{m} =\frac{GM}{R^2} \approx 9.8m/s^2 \ \ (1)$$This is why in free fall problems we can omit the masses and forces and skip directly to the equation $$y = y_0 + u_yt - \frac 1 2 gt^2$$Where I've taken upwards to be positive, hence $a = -g$.

If we were dealing with an electromangentic force on a particle of change $q$ and mass $m$ then we would have to take both the charge and mass of the particle into account (as well as the EM field strength) in order to calculate the acceleration.

Note that equation (1) holds approximately for anywhere near the surface of the Earth. It doesn't hold for problems involving satellites or the Moon or any motion that is sufficiently far from the Earth's surface. For any problem where the heights involved are no more than a few kilometres, you can assume that $g$ is approximately a constant of $9.8m/s^2$.

Golden explanation