# Mass

Discussion in 'Question of the Day' started by dvsDave, Jul 5, 2018.

1. ### dvsDaveBenevolent DictatorAdministratorSenior TeamCB ModsFight Leukemia

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From @Stevens R. Miller :

Challenge to my friends who are science nerds (yes, if you are even reading this, that includes you): My son is learning physics. At the moment, he is studying Newton's laws of motion. In dealing with force and acceleration, I have been impressing upon him that, if an object maintains a constant velocity, that means it is not experiencing any acceleration. From the equation F=ma, that also means that the net force applied to that object is zero.

For a bit, he had a tough time understanding how an object might continue to move if no force acted on it. That got us into a discussion about momentum, defined as mv. At which point, he stopped me cold with this one:

"Dad, if an object is moving at constant velocity, and that means its acceleration is zero, which also means the force acting on the object is zero, doesn't that mean its mass is undefined?"

He defended his question thus:

[1] F=ma.
[2] Constant velocity implies a=0.
[3] a=0 implies that ma=0, and therefore, by [1], F=0,
[4] F=ma implies that m=F/a.
[5] By [2], [3], and [4], constant velocity implies that m=0/0.
[6] 0/0 is undefined,
[7] By all of the foregoing, the mass of an object moving at constant velocity is undefined.

At which point he demanded that I explain how an object can have a defined momentum when moving at a constant velocity, if momentum is mv and, therefore, (0/0)v.

I could use a little help here.

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2. ### dbaxterWell-Known Member

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This is like the math puzzle X^2 [x squared] - x^2 = x^2 - x^2. Which you can factor into (x+x)(x-x)= x(x-x). Divide both sides by (x-x) leaves x+x=x, or 2=1. The trick being when you struck (x-x) you divided by 0. Anytime you try to divide by 0 all bets are off. No thinking or processing after that has any validity. So as soon as you had a=0 as a denominator - stop! Any conclusions or further analysis is bogus.
At least that's what I learned 55 years ago when I took Physics and Math. But the world has changed...

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4. ### RonaldBealActive Member

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"Dad, if an object is moving at constant velocity, and that means its acceleration is zero, which also means the force acting on the object is zero, doesn't that mean its mass is undefined?"

Except, for the object to be moving, there must have been acceleration at some point. That acceleration happened for a definable mass.

The F=ma equation... put any constant in for mass... say 1Kg. F=1kg*a... if a=0, then F=1*0... = 0, which is correct. The force equation is not used to define mass, but used to derive force.

Just my jetlagged \$0.02

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5. ### porkchopWell-Known Member

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There are two ways I would explain this:

1. From the beginning days of physics class. The Conservation of Matter tells us that matter is neither created nor destroyed. The object has a known, defined, and measurable mass when we accelerate it up to our theoretical constant velocity and would continue to have a known, defined, and measurable mass if we accelerated it again beyond the point of having a constant velocity to one that is steadily changing(Sum of all forces≠0) . In this process we have done nothing to modify the mass of the object so by definition it must have the same mass it did when we were accelerating it into this equilibrium state.
2. From somewhere in calculus. Using the equation F=ma we solve for m and find m=F/a. The value of mass is defined for all accelerations where a≠0 and in fact is represented by two straight lines of constant value for all values a<0, and a>0 respectively. With this information (and the Conservation of Matter giving us that the value of mass should be continuous if you want to be really picky) we can positively state that the limit of mass as a->0 from either side is the same as the value of mass at all other values of a.
Good question. Too many students are willing to memorize and not ask why these days.

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6. ### TimMcWell-Known Member

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@porkchop I thought it was Conservation of Energy.... But I didn't pay attention to physics back in school so the possibility of mis-memory is real...

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7. ### porkchopWell-Known Member

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In a closed system both energy and mass are conserved. Energy has the tendency to be more important because it can change states and be calculated, but it depends on the fact that mass doesn't randomly change either.

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8. ### CalcActive Member

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@dbaxter has the answer above. In step 4 you divided both sides of the first equation by A to convert it into the second equation. That's where the error starts in the proof.

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9. ### rphilipActive Member

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Looking at this from a conceptual level remember that mass is a property of an object that never changes (excluding physical changes or nuclear reactions).

10. ### RickRWell-Known Member

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Fusion reactions and E=Mc^2 tells us Mass and Energy are interchangeable. Someday we may learn how to go from energy to mass.

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11. ### dvsDaveBenevolent DictatorAdministratorSenior TeamCB ModsFight Leukemia

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Isn't that the whole idea behind the food replicators on Star Trek?

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12. ### TNastyActive Member

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Ah, let's bring in the mechanical physics learning assistant here. This'll be fun.

From what I understand, this seems like an issue stemming from core concepts; basically we need to figure out what we agree upon, and what we don't.

So we all agree that (net)force is equal to mass times acceleration. The key part there is net. If you were to draw a free-body diagram of the system (if you're not certain what this is, it's pretty much where your system is a dot, and you draw a vector/arrow going from the point in the direction of each force), you'll draw an arrow pointing down (gravity). If this object is not accelerating, there would be some force with an equally large vertical component (if the force is straight up, it'll be the exact same size), and thus there is no vertical net force- and consequentially no vertical acceleration.

A perfect example is an object that is dropped, and reaches terminal velocity. If this object has a mass of 'M' kilograms, and gravity is the constant 'g' m/(ss), then the force of gravity on this object is Mg. Once this object has been falling (and accelerating) for long enough, it reaches terminal velocity. At terminal velocity, the object stops accelerating, and thus the net force on it is zero- the force of drag is equal to Mg (force of gravity), but in the opposite direction. If we were to say "up is positive", we could do force of drag - force of gravity (drag in positive direction + gravity in negative direction, so drag + negative gravity), we would wind up with the expression Mg-Mg=0, confirming that there is no net force (and no acceleration), but there still are forces (such as gravity and drag).

This is just like how gravity is always acting on you, but we don't "fall through Earth". This is because at rest, the normal force from whatever surface you are standing on is equal (and opposite) to the force of gravity on you. There are some interesting things that happen when you jump, such as the normal force being larger when you jump up, and being smaller as you're landing, but will go back to being equal once you're at rest.

Not sure how much math your son knows, but it should be fairly easy to explain that the slope (derivative) of a displacement (distance) graph is the velocity graph, the slope (derivative) of the velocity graph is the acceleration graph, and conversely the area (integral) of the acceleration graph is the delta (change in) velocity, and the area (integral) of the velocity graph is the delta (change in) displacement graph. Then you can point out how the area on the acceleration graph only shows us the change in velocity, and that the object could have any velocity before that; just like how we can't find out where an object is only with velocity, but we can tell how far it has moved using the velocity graph.

I'll be happy to explain even more if he doesn't get it, or something isn't clear. I had students come to me during my office hours this past semester with issues such as this one, some more complicated issues, or even things just as simple as not realizing that their calculator was interpreting 50/2π as (50/2)π instead of 50/(2π).

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13. ### bobgaggleWell-Known Member

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The way I see it:

Just because you can't determine the mass with the given formula due to current circumstance doesn't mean the rock doesn't have mass. You could weigh the rock before you launch it, or you could use the given formula after the rock crashes into something. You just can't use the formula while the rock is in flight with constant velocity...

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@bobgaggle I like your logical thinking and find it somewhat akin to why I feel Prince Phillip always follows 6 paces behind Queen Elizabeth; If the Queen were to inadvertently step on an IED, Phillip may survive his injuries.
Toodleoo!
Ron Hebbard

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I feel Mass should always be performed in Latin.

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16. ### bobgaggleWell-Known Member

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Isn't that what the LHC is doing? Making particles and such...

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17. ### RickRWell-Known Member

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My limited understanding is that they are using matter. Smashing small bits into smaller bits to see how it's built. That some of those particles seem to be more like energy than matter is a side note.

If you subscribe to the notion that matter is vibrations or fields then everything is energy.

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18. ### jonlilesWell-Known MemberFight Leukemia

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On Earth, or even near Earth, Mass ALWAYS has a acceleration in the form of gravity. Even in the deep vacuum of space, you have the interstellar gravity acting on mass. Then again, my degree is in Nuclear Engineering.

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19. ### JDWell-Known Member

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"Mass undefined" is the correct statement. It doesn't mean that an object is without mass, it only means we do not have enough data to define it.
Let's say an unknown orb is headed towards earth. Is it a solid rock which would do a lot of damage, or is it a thin shell that would have no effect on us? Mass undefined.
We have to wait until there is some interaction, some force that acts on it prior to determining the mass. We know it has some, but it is undefined until some known action interacts with it.

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