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SilentSky23

Question about same accelerations

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I wanted to make sure of something here. Now, I believe it is the case when an object with lighter mass, or inertia, will travel faster and farther than one with a heavier mass or inertia under the same force. Now, consider an object that is light and one that is heavy, both going at the same acceleration, but with different forces acting, and they start moving across either the ground or the air, depending on where they are when a force acts on them. Now, with the same acceleration, which object will go farther before stopping, with either friction, air resistance, or a force exerted by something or someone that is the same force for both objects?

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Ok complete guess and I know your looking for someone more qualified, but surely the lighter one would go further due to more friction on the heavier one. 

 

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7 minutes ago, Curious layman said:

Ok complete guess and I know your looking for someone more qualified, but surely the lighter one would go further due to more friction on the heavier one. 

 

But by inertia, wouldn't the lighter object stop sooner, since it has less resistance to change in its velocity?

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If the force causing the acceleration is not removed, neither will stop.

If the force is removed, factors other than mass will cause them to stop.
Things like frontal area or surface area if on the ground, streamlining or coefficient of friction, or even the fluid/atmosphere they are immersed in.
IOW your question is not specific enough.

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Posted (edited)
1 hour ago, SilentSky23 said:

I wanted to make sure of something here. Now, I believe it is the case when an object with lighter mass, or inertia, will travel faster and farther than one with a heavier mass or inertia under the same force. Now, consider an object that is light and one that is heavy, both going at the same acceleration, but with different forces acting, and they start moving across either the ground or the air, depending on where they are when a force acts on them. Now, with the same acceleration, which object will go farther before stopping, with either friction, air resistance, or a force exerted by something or someone that is the same force for both objects?

You need  to be clearer  how long the original force(s)  is maintained, (when it starts and when it stops) and  the same  for the resisting forces.

Acceleration in not like velocity, which is why they are different properties.

Edited by studiot

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Posted (edited)
5 minutes ago, studiot said:

You need  to be clearer  how long the original force(s)  is maintained, (when it starts and when it stops) and  the same  for the resisting forces.

Acceleration in not like velocity, which is why they are different properties.

Yeah, I know, I was going to edit that part about acceleration, but I see I can't now. Still, say that the lighter object is exerted for less than a second, and the second object has a bigger force around the same amount of time. For the negative acceleration from the resistance that eventually stops them, I am not sure. Would the times of negative acceleration be different for each of them?

Edited by SilentSky23

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Posted (edited)

Just  try again and do your best to set out what you actually want to say.

Edited by studiot

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Posted (edited)
9 minutes ago, studiot said:

Just  try again and set out what you actually want to say.

Well, what I meant in my initial saying was that I believe lighter objects accelerate faster and go farther than the ones with more mass, under the same exerted force,  of course. Now, for a light object and a heavy object going at the same acceleration, under different forces acting on each of them; assuming the forces are both different in magnitude but acting in the same amount of time; with resisting forces like friction being the same on both objects in terms of magnitude, but acting for a longer time than the two forces that accelerated the objects in the first place, which one would come to a stop sooner than the other if one has the higher inertia than the other?

How is this?

EDIT: For the surface, say the surface is a wooden floor in a house.

Edited by SilentSky23

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The relationship is F=ma where

F is directional force
a is directional acceleration
m is mass

Only forces will change velocity of a mass by accelerating it; if no force is acting, masses will continue at their current velocity.
Obviously for the same force, the lesser mass will experience the greater acceleration.

If that force ( and acceleration ) are limited to a certain time, then the less massive object will have a higher final velocity.

From that point different forces come into play, that tend to slow the masses down, and accelerate it in the opposing direction ( reducing its velocity ). These can be due to surface friction, air resistance or even the 'level' of the surface.
Unless all of these factors are specified, we can't even begin to answer how long it takes for the masses to come to rest.
 

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4 minutes ago, MigL said:

The relationship is F=ma where

F is directional force
a is directional acceleration
m is mass

Only forces will change velocity of a mass by accelerating it; if no force is acting, masses will continue at their current velocity.
Obviously for the same force, the lesser mass will experience the greater acceleration.

If that force ( and acceleration ) are limited to a certain time, then the less massive object will have a higher final velocity.

From that point different forces come into play, that tend to slow the masses down, and accelerate it in the opposing direction ( reducing its velocity ). These can be due to surface friction, air resistance or even the 'level' of the surface.
Unless all of these factors are specified, we can't even begin to answer how long it takes for the masses to come to rest.
 

I did say the type of surface, which is wooden. Not to mention smooth. As for the other things, say the air resistance is normal, or the usual amount in everyday life (nothing too strong) with no wind, and the level of the surface is, well, flat.

Anything else you need to know?

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Posted (edited)

That doesn't specify anything...
Especially since the 'retarding' force is also active during the initial force application that caused the velocities.

But everything else being equal. They both feel equal retarding force ;(highly unlikely ) and if the retarding force is trivial compared to the initiating  force, the lighter mass will experience greater negative acceleration, and, since it started with a proportionately greater velocity, they should come to a stop about the same time.

Edited by MigL

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7 minutes ago, MigL said:

That doesn't specify anything...
Especially since the 'retarding' force is also active during the initial force application that caused the velocities.

But everything else being equal. They both feel equal retarding force ;(highly unlikely ) and if the retarding force is trivial compared to the initiating  force, the lighter mass will experience greater negative acceleration, and, since it started with a proportionately greater velocity, they should come to a stop about the same time.

So, are you expecting numbers? I am kinda having trouble understanding what you want from me when talking about this situation. If surface level and air resistance and such aren't going to be enough here as it may seem, tell me, what should I tell you, especially for next time when I do ask something like this, just to be safe?

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Even if you gave me numbers I couldn't give you a numerical answer ( time ), but I would be able to tell you which would stop first.
If the problem is not laid out specifically you can't expect a specific answer.

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10 minutes ago, MigL said:

Even if you gave me numbers I couldn't give you a numerical answer ( time ), but I would be able to tell you which would stop first.
If the problem is not laid out specifically you can't expect a specific answer.

To be honest, being told which would stop first; that is actually all I need.

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4 hours ago, SilentSky23 said:

To be honest, being told which would stop first; that is actually all I need.

Just a thought, but Why not try it yourself. Use a football and a beach ball. Roll them down a slope and see where they stop.

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Posted (edited)
15 hours ago, SilentSky23 said:

 

I wanted to make sure of something here. Now, I believe it is the case when an object with lighter mass, or inertia, will travel faster and farther than one with a heavier mass or inertia under the same force. Now, consider an object that is light and one that is heavy, both going at the same acceleration, but with different forces acting, and they start moving across either the ground or the air, depending on where they are when a force acts on them. Now, with the same acceleration, which object will go farther before stopping, with either friction, air resistance, or a force exerted by something or someone that is the same force for both objects?

The times it takes to stop the objects are related as their masses: [math]m_{1}/m_{2}[/math]

Since question is not very clear I do some assumptions
there are two objects  [math]m_{1}[/math] and [math]m_{2}[/math] where mass of [math]m_{1}[/math] < mass of [math]m_{2}[/math]
at time [math]t_{0}[/math]=0 [math]m_{1}[/math] and [math]m_{2}[/math]are accelerated from rest with acceleration a. 
at time [math]t_{1}[/math] acceleration of [math]m_{1}[/math] and [math]m_{2}[/math] stops
at time [math]t_{2}[/math] some kind of force F starts to affect [math]m_{1}[/math] and [math]m_{2}[/math]. 
Force [math]F[/math] is acting in opposite direction of movement of [math]m_{1}[/math] and [math]m_{2}[/math]
Force [math]F[/math] acts independently on [math]m_{1}[/math] and [math]m_{2}[/math] until each object have stopped.
Newtonian physics applies (low speeds, solid bodies)

Apply some math:
velocity [math]v[/math] for both objects [math]m_{1}[/math] and [math]m_{2}[/math] at time [math]t_{1}[/math] is then [math]v=a*t_{1}[/math]
momentum at [math]t_{1}[/math] is [math]p_{1}=m_{1}*v[/math] and [math]p_{2}=m_{2}*v[/math]
Impulse = change in momentum
Force F is applied until objects are stopped, that means change of momentum is [math]p_{1}[/math] and [math]p_{2}[/math]  for [math]m_{1}[/math] and [math]m_{2}[/math]
If we label time to stop [math]m_{1}=t_{s1}[/math] and time to stop [math]m_{2}=t_{s2}[/math] then
[math]t_{s1}=p_{1}/F=m_{1}*v/F[/math]
[math]t_{s2}=p_{2}/F=m_{2}*v/F[/math] 

Result:
[math]\frac{t_{s1}}{t_{s2}}=\frac{m_{1}}{m_{2}}[/math]

 

 

Edited by Ghideon
more format

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Not to be picky, Ghideon, and I realize the OP is poorly worded and incomplete, but...

Why would both masses have the same velocity, v, at time t1, when they were accelerated by the same initial force ?

If the retarding force slows the lesser mass more quickly, then the initiating force imparts greater acceleration to the lesser mass.
So v is not the same for both masses.

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9 minutes ago, MigL said:

Why would both masses have the same velocity, v, at time t1, when they were accelerated by the same initial force ?

 

That is a good point! I tried to account for that in my assumption, but may have failed. The acceleration is identical for both objects. That's why I defined the identical acceleration in the first phase and the identical force in the second phase of the question. I probably should have written that I assume the initial accelerating forces to be different for m1 and m2?

From OP: "Now, consider an object that is light and one that is heavy, both going at the same acceleration, but with different forces acting".

 

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Ah, OK.
I hadn't noticed the bolded part.
My mistake; carry on.

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On 3/12/2020 at 7:33 PM, SilentSky23 said:

Well, what I meant in my initial saying was that I believe lighter objects accelerate faster and go farther than the ones with more mass, under the same exerted force,  of course

I'm trying to understand this? Didn't the eagle feather and the hammer accelerate equally and didn't they hit the ground at the same time? Honestly, it's the easiest way I believe to insure that the acceleration force on both objects is equal. I believe that if you look it up experiments were done in the 17 or 18 hundreds where balls dimensionally equal except for their mass were dropped into clay then the impression depth was measured. Actually, I think several considerations were considered. One being each ball dropped from the same height, so that acceleration was consistently equal for each ball, and if memory serves the ball with greater mass left the deeper impression. So if equal amounts of force are applied in the stopping, the object of more mass goes further.

Remember that the balls were dimensionally the same except for their mass. Each presenting the same surface area upon contact with the clay. Of course you can complicate with the need for math and lateral movement, and all other things considered, if you want to.  :)

Note. If I have completely misunderstood your question, my apologies, but it did seem that you wanted everything equal except object masses.

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Posted (edited)

Here is a video that seems made to order... Maybe?

Well at least there is no math. There is lateral movement and various objects dimensionally different. His method of launch is somewhat simple to the point of antiquated and not always entirely accurate, but it may better answer your questions than anything I could find in writing. Note it doesn't say a word about which object would go furthest, but you might be surprised about what it says about your lighter object if the reason it is lighter is because it is hollow.

Edited by jajrussel

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Posted (edited)

OK. I think I understand the problem, but I concur with studiot, MigL, and Ghideon that the conditions are less than clear.

Sorry for changing your notation, Ghideon, but it's clearer to me if I set, m, M, f, F. And initial acceleration a_0, positive.

Eqs. of motion:

\[f=ma_{0}\]

\[F=Ma_{0}\]

We get,

\[\frac{f}{F}=\frac{m}{M}\]

Now comes the braking force. I'll call it script F. You don't say it's constant, you don't say anything except it's equal for both masses. I will assume the forces f and F keep acting as before.

\[f-\mathcal{F}=ma\]

\[F-\mathcal{F}=MA\]

Where a and A are the new accelerations. Using f/F=m/M, we get,

\[\frac{a_{0}-a}{a_{0}-A}=\frac{M}{m}>1\]

This leads to,

\[a_{0}-a>a_{0}-A\]

or, equivalently,
\[A>a\]
As script F is a friction, it can never overcome either f or F. So m stops further away, I believe.
 
 
Edited by joigus
Correction of discussion.

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Posted (edited)
On 3/13/2020 at 4:06 AM, Curious layman said:

Just a thought, but Why not try it yourself. Use a football and a beach ball. Roll them down a slope and see where they stop.

But first note that Curious is from the UK...

To the OP"

They will have the same velocity upon release...assuming they both start accelerating with the same initial velocity...

Both "going at the same acceleration" doesn't tell you that on it's own.

Assuming that's the case it comes down to how the extra mass effects drag. If it's linear then they go the same distance. But it could readily be greater or lesser than linear with the mass, depending on your conditions.

Using Curious' example on a hard surface a soccer ball travels further than a beachball. Same example on a surface that will yield easily to weight and the beachball goes further while the soccer ball "digs in".

For something sliding on a wood floor it depends (mostly) on how the coefficient of friction is affected by the extra weight. If it's the same they stop at the same point. (air resistance aside)

Edited by J.C.MacSwell

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38 minutes ago, joigus said:

As script F is a friction, it can never overcome either f or F. So m stops further away, I believe.

Now, if \mathcal{F} (LateX for script F) were not a friction, the discussion would be more complicated. I'm not sure that's the case you're interested in.

Ghideon is considering a force that acts just at the moment initial forces stop acting, I believe. 

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