Free fall of solid bodies

In summary: No, and i think that the m is of no useNewton's law of gravity says that the force between two masses is proportion to the product of their masses. If...The force between two masses is proportional to the product of their masses.
  • #1
New poster has been reminded to always show their work when starting schoolwork threads
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.
 
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  • #3
New poster has been reminded not to post illegible images of their work
IMG_20220123_131108_595.jpg


This it what i found
 
  • #5
If I could read it I might be able to answer that :rolleyes:


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.​
 
  • #6
A dummy progression said:
Is it correct?
Is that ##26.4m##?
 
  • #7
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 !

:welcome: !

Teaser: what if ##m_2 = 2 m_1 ## ?

##\ ##
 
  • #8
BvU said:
If I could read it I might be able to answer that :rolleyes:


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.​
IMG_20220123_132817_852.jpg
 
  • #9
A dummy progression said:
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##:$$h = h_0 -\frac 1 2 gt^2$$
 
  • #10
BvU said:
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 !

:welcome: !

Teaser: what if ##m_2 = 2 m_1 ## ?

##\ ##

PeroK said:
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.
 
  • #11
A dummy progression said:
Oh well, at least it's legible now. Much better than what I achieved with my expensive image enhancement software :wink: .

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 !

##\ ##
 
  • #12
BvU said:
Oh well, at least it's legible now. Much better than what I achieved with my expensive image enhancement software :wink: .

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.
 
  • #13
Let me save you some time: do you see an ##m## anywhere in the calculations to get the answer ?

:biggrin:

##\ ##
 
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Likes PeroK
  • #14
BvU said:
Let me save you some time: do you see an ##m## anywhere in the calculations to get the answer ?

:biggrin:

##\ ##
No, and i think that the m is of no use
 
  • #15
A dummy progression said:
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##.
 
  • #16
PeroK said:
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😆😆
 

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