How Do Mass and Weight Differ?

[Quantum55151]

Homework Statement

A large rock falls on your toe. Which of the concepts is most important in determining how much it hurts?

(A) The mass of the rock.
(B) The weight of the rock.
(C) Both the mass and the weight of the rock are important.
(D) Either the mass or the weight, as the two are related by a single multiplicative constant g.

A large rock sits on your toe. Which of the concepts is most important in determining how much it hurts?

(A) The mass of the rock.
(B) The weight of the rock.
(C) Both the mass and the weight of the rock are important.
(D) Either the mass or the weight, as the two are related by a single multiplicative constant g.

Relevant Equations

W = mg

In the 2nd question, I definitely think it's D, because the force the rock exerts on your toe is equal in magnitude to the normal force the toe exerts on the rock which in turn is equal to the rock's weight, which is related to m by the constant g. In the 1st question, I am not as certain, but I would still say D, because the more massive the rock, the more it hurts, and so logically weight = mg is equally important.

 

[haruspex]

Can you be sure of g? Maybe this is on the moon. And it is not the same at a pole as at the equator or up a mountain.

 

[Quantum55151]

Can you be sure of g? Maybe this is on the moon. And it is not the same at a pole as at the equator or up a mountain.

Of course, the value of g changes depending on the location. But the relationship between weight and mass remains the same...

 

[hutchphd]

But the relationship between weight and mass remains the same...

Semantic question. To my ear, this means that the value of g cannot be changed, but one could argue it the other way....consensis? The perils of doing descriptive physics rather than actual mathematical physics.

 

[Quantum55151]

Semantic question.

So you reckon the question is bad from the start?

 

[haruspex]

Of course, the value of g changes depending on the location. But the relationship between weight and mass remains the same...

No, the relationship, in terms of weight per unit mass, changes. If you are told a 10kg boulder will be placed on your foot then you cannot predict the pain without knowing whether this will be on Earth or on the Moon, but if you are told the weight of the boulder there's no more you need to know.

 

[PeroK]

No, the relationship, in terms of weight per unit mass, changes. If you are told a 10kg boulder will be placed on your foot then you cannot predict the pain without knowing whether this will be on Earth or on the Moon, but if you are told the weight of the boulder there's no more you need to know.

It depends on your footwear.

 

[PeroK]

So you reckon the question is bad from the start?

It's an idiotic question, IMO.

 

[hutchphd]

Of course, the value of g changes depending on the location. But the relationship between weight and mass remains the same...

This is very badly specified because "relationship" can be interpreted at different levels (do we assume g fixed and uniform or not etc)

So you reckon the question is bad from the start?

It is a very bad examination question, but perhaps not so bad as a "discuss it in lecture" question. Seems overly complicated and too cute to me.

 

[Halc]

Homework Statement: Which of the concepts is most important in determining how much it hurts?

For both questions (dropped-on and resting-on), what is the primary factor that determines the level of pain?

It's not the color of the rock for instance. I suppose the color of the toe matters somewhat, but that's correlation, not causation.
What specific physical factor is primarily responsible for the pain? The questions as worded ask you to select the respective option that is primarily responsible for that specific physical factor.

Relevant Equations: W = mg

To illustrate my point, that isn't the only relevant equation. It's quite relevant to only one of the two questions.

 

[DaveE]

Yes, yet another confusing exam question.
I would focus on the difference between the two questions, at least for a discussion session.
What is the velocity of the rock in each case? How does that relate to force, momentum, energy, etc.?

 

[Quantum55151]

No, the relationship, in terms of weight per unit mass, changes. If you are told a 10kg boulder will be placed on your foot then you cannot predict the pain without knowing whether this will be on Earth or on the Moon, but if you are told the weight of the boulder there's no more you need to know.

Yes, I agree. So then the answer to the 2nd question would be B.

However, the answer to the 1st question, according to the answer key, is C. But what exactly is the fundamental difference between a rock falling and a rock sitting?

The only thing that comes to mind is that the sitting rock is a statics problem, whereas the falling rock could be viewed as a collision between a moving object (the rock) and a stationary object (your toe), in which case momentum considerations could come into play. However, this question comes from the chapter on Newton's laws, long before the author has introduced momentum, so momentum shouldn't really be necessary...

 

[Quantum55151]

It's an idiotic question, IMO.

Gotta thank Halliday and Resnick :)

 

[Quantum55151]

This is very badly specified because "relationship" can be interpreted at different levels (do we assume g fixed and uniform or not etc)

By "relationship", I meant the W = mg formula, as in regardless where you are, at the poles, on the Equator, on Mt. Everest, on the Moon, etc., W and m are related by the "single multiplicative constant g" (which obviously varies with location).

However, as haruspex has noted, knowing the mass without knowing the value of g won't lead you anywhere, so for all practical purposes, the weight is a more practical piece piece of information.

It is a very bad examination question, but perhaps not so bad as a "discuss it in lecture" question. Seems overly complicated and too cute to me.

It's a conceptual question from a textbook so would fall in the second category you mentioned.

 

[Quantum55151]

To illustrate my point, that isn't the only relevant equation. It's quite relevant to only one of the two questions.

It would only be relevant to the second question, right?

 

[Quantum55151]

What is the velocity of the rock in each case? How does that relate to force, momentum, energy, etc.?

Strictly speaking, none of those concepts should be invoked or at least they should not be necessary for answering the question. As I said in a previous comment, the question is from a chapter on Newton's laws, long before momentum, let alone energy, is introduced.

 

[haruspex]

the answer to the 1st question, according to the answer key, is C. But what exactly is the fundamental difference between a rock falling and a rock sitting?

Question 2 is ho-hum, but question 1 is worse.
In 2, the pain is determined by m and g, specifically in the combination mg, weight.
In 1, we can consider it depends on m, g and h as $mgh$, or on m and v as $\frac 12 mv^2$. None of the offered choices mention h or v.
You could argue for mass, if v were a given, or for weight if h were a given.

 

[DaveE]

"most important" is a big problem IMO.
Compare a 2000Lb block of lead dropped 1mm vs. a 10g meteorite.

The question is meaninglessly vague and best ignored.

My guess is that we all understand Newton. Perhaps a discussion about gravity, impulse, momentum, etc. if anyone still has questions?

 

[Halc]

It would only be relevant to the second question, right?

W=mg is relevant mostly to the second question, yes.

What equation is most relevant to the first problem question?


You also didn't answer my first question. What physical property (in this case) is most responsible for pain? I mean, it isn't temperature in this case, even if that would be very relevant to pain if the problem involved a super hot mass.
The answer to this will make it clear what is needed for the problem's first question.

 

[haruspex]

The answer to this will make it clear what is needed for the problem's first question.

I disagree; see post #17.

 

[Halc]

I disagree; see post #17.

Post 17 says:

In 2, the pain is determined by m and g, specifically in the combination mg, weight.
In 1, we can consider it depends on m, g and h as $mgh$, or on m and v as $\frac 12 mv^2$. None of the offered choices mention h or v.
You could argue for mass, if v were a given, or for weight if h were a given.

The singular value I was getting at is Force. In problem 2, the force is the weight, so W=mg is most relevant.
Now force in the first case is dependent on the height from which it was dropped, unspecified, but the question doesn't ask for specific numbers. The force on the toe is not so much a function of how long it falls, but rather the mass combined with the rate it must accelerate to be able to halt in some fraction of the distance of the thickness of a toe.

$F = \frac {.5 mv^2}{d}$

This is an expanded version of F=ma, one that works on the moon as well as on Earth or in a space station where there's no weight at all.

 

[haruspex]

Now force in the first case is dependent on the height from which it was dropped,

Quite so. Neither the mass nor the weight by itself is enough. We either need the height from which it was dropped, in which case mg gives the answer, or the impact velocity, in which case m gives the answer. There is no basis for choosing mass or weight in preference to the other.

 

[Halc]

We either need the height from which it was dropped, in which case mg gives the answer, or the impact velocity, in which case m gives the answer

Quite so. My formula takes velocity as input, and the value of g is irrelevant.
If height is input, then velocity becomes a function of that and g, yielding the same thing.

Either formula would do to answer the question asked in the OP, but the OP question admittedly doesn't identify what is held constant (height of drop? g? impact v?).

Still, the answer seems pretty clear since even a low weight object can deliver a huge force if it has a lot of momentum to dissipate in a very short time, so weight never seems to come into play in any of the relevant formulas.

 

[pbuk]

To me it is clear:

• When an object falls on your toe the amount it hurts is determined by its kinetic energy, $\frac 1 2 mv^2$, which depends directly on its mass and so this is the more 'important' concept.
• When an object is sitting on your toe the amount it hurts is determined by its normal force which is the definition of its weight and so this is the more 'important' concept.

 

[jbriggs444]

To me it is clear:

• When an object falls on your toe the amount it hurts is determined by its kinetic energy, $\frac 1 2 mv^2$, which depends directly on its mass and so this is the more 'important' concept.

As @haruspex has pointed out, this line of reasoning assumes that velocity is given and held fixed while mass is the parameter that one would want to measure.

If, on the other hand, the height of the drop is given and held fixed then $mgh$ becomes the formula for kinetic energy with $mg$ as the parameter that one would want to measure.

It is not completely clear that kinetic energy is the correct figure of merit for pain and injury. Given 160 joules of impact energy, would you rather take it as a 16 kg block falling one meter onto your shoe or as a .22 long rifle round fired through your shoe?

 

[pbuk]

As @haruspex has pointed out, this line of reasoning assumes that velocity is given and held fixed while mass is the parameter that one would want to measure.

No it doesn't, it simply observes that $\frac 1 2 m v^2$ is directly dependent only on mass and the square of velocity. If we were asked whether mass or velocity is more important in this relationship then I agree it would not be possible to give an answer, however that is not the question. Even if you were to assert (wrongly) that the velocity of a falling object depends on its weight then it would still by definition be only of secondary importance.

If, on the other hand, the height of the drop is given and held fixed then $mgh$ becomes the formula for kinetic energy with $mg$ as the parameter that one would want to measure.

So are you saying that a question that is posed in general terms cannot be answered because you can add your own restricting criteria to the question to come up with a different answer to the one intended? If you adopt that position then no question can be asked without specifiying every possible parameter; noone would ever learn anything.

It is not completely clear that kinetic energy is the correct figure of merit for pain and injury. Given 160 joules of impact energy, would you rather take it as a 16 kg block falling one meter onto your shoe or as a .22 long rifle round fired through your shoe?

I'm not sure, however I am certain that I wouldn't give a damn about the weight of the .22 round in any particular reference frame - only its mass. And a block in freefall has no weight so are you saying that it is not going to hurt?

 

[PeterDonis]

As @haruspex has pointed out, this line of reasoning assumes that velocity is given and held fixed while mass is the parameter that one would want to measure.

If, on the other hand, the height of the drop is given and held fixed then $mgh$ becomes the formula for kinetic energy with $mg$ as the parameter that one would want to measure.

I don't think either of these is the right way to look at the question. The question only says the rock falls on your toe, it doesn't specify how the details are determined. If all I know is that a rock is going to fall on my toe, I'm going to be concerned about its mass, not its weight, because, first, kinetic energy and momentum both depend only on mass, not weight, and second, as @pbuk pointed out, if the rock is in free fall it has zero weight, but it certainly is not going to do zero harm to my toe.

It is not completely clear that kinetic energy is the correct figure of merit for pain and injury.

I think there are reasonable arguments that both kinetic energy and momentum are relevant. But both of those depend only on mass, not weight, as noted above.

 

[PeterDonis]

So then the answer to the 2nd question would be B.

Is that the answer the answer key gives?

the answer to the 1st question, according to the answer key, is C.

Is any rationale given for the answer?

 

[DaveE]

However, the answer to the 1st question, according to the answer key, is C. But what exactly is the fundamental difference between a rock falling and a rock sitting?

A falling rock has stored kinetic energy. The impulse required by your toe to stop it, or rearrange your toe anatomy is an issue. A rock resting on your toe is continually exerting a force due to it's weight, even after it has stopped moving. Either or both could cause you pain depending on the mass, velocity, and gravity.

1) Suppose your working under your car and the jack fails, causing it to fall on you. When you initially scream "Ouch!", that's probably due to it's mass. Then a few seconds later, when you yell to your buddy "Get it off of me!", that's probably due to its weight.

2) Imagine you are floating in space and an asteroid hits your toe. That might hurt. Assuming the same velocities, you would rather be hit with a tiny asteroid than a massive one. But it has no weight because you and the asteroid are both in free fall ("zero G"). If your spacewalking buddy gently places a hammer on your toe, you might not even know it's there.

 

[jbriggs444]

No it doesn't, it simply observes that $\frac 1 2 m v^2$ is directly dependent only on mass and the square of velocity.

Again, that assumes that impact velocity $v$ is fixed or known.

The point is that the information that is "important" to measure depends on what you already know.

If I know impact velocity then I agree with you that mass is important to calculate impact energy. Weight won't do it since $g$ could vary.
If I know drop height then weight is important to calculate impact energy. Mass won't do it since $g$ could vary.
If I know both drop height and local g then either mass or weight will suffice to perform the calculation.
If I know both impact velocity and local g then either mass or weight will suffice to perform the calculation.

So are you saying that a question that is posed in general terms cannot be answered because you can add your own restricting criteria to the question to come up with a different answer to the one intended?

I am saying that this question as posed has no one correct answer. Yes, the only way to answer it properly is to add assumptions about what one already knows. As a test of understanding, the question is bad.

As a generator for discussion, the question may be worthwhile.