Why the large ball falls farther than the smaller one?

[YMMMA]

Homework Statement

A large and a small ball, made of the same uniform density foam, are released at the same time from a point 100 meters above Earth. After 1 second the large ball is observed to have fallen farther than the small ball. Which of the following explains this observation?

(A) The large ball weighs more than the small ball.
(B) The gravitational force on the large ball is greater.
(C) The buoyant force of the air on the large ball is greater.
(D) The large ball has a greater ratio of mass to surface area.
(E) Water vapor in the air adheres to the small ball more readily and thus increases its mass.

Homework Equations

Concept-based question.

The Attempt at a Solution

They have the same density or mass-to-volume ratio. What could that hint at? I thought their mass could be different, they fall with the same acceleration due to gravity, and so their weights differ. The smaller ball might have greater mass. Or is it air resistance? Can someone help me in understanding the choices?

 

[BvU]

Sure. In vacuo the would fall equally fast, so it must have something to do with the air. What relevant equations can you think of (or pick up from your lecture notes/textbook) ? List them under 2. relevant equations. The template is there for a good reason. Its sue is compulsory, as you should know by now.

And a tip: approach from the other end, i.e. assume (A) is correct -- would that lead to the symptoms described ?

 

[YMMMA]

I didn’t write down any equations because I thought it doesn’t depend on any. But for choices A and B it’s all about gravitational force and weights, which are essentially the same Fg=W=mg. But it has to be the other way around for the larger ball to fall farther; the smaller ball should then have larger mass so larger weight.

I am not sure what it means to have a buoyant force of air greater on one object. But I think that might lead the larger object not to fall farther. Same for choice D, I don’t guess this would affect the ball’s motion.

For the last choice, that is the only choice stating the smaller ball has the larger mass.

 

[PeroK]

I didn’t write down any equations because I thought it doesn’t depend on any. But for choices A and B it’s all about gravitational force and weights, which are essentially the same Fg=W=mg. But it has to be the other way around for the larger ball to fall farther; the smaller ball should then have larger mass so larger weight.

I am not sure what it means to have a buoyant force of air greater on one object. But I think that might lead the larger object not to fall farther. Same for choice D, I don’t guess this would affect the ball’s motion.

For the last choice, that is the only choice stating the smaller ball has the larger mass.

What do you know about air resistance and buoyancy forces?

Also, the key factor in whether something moves further than something else is not force, it's acceleration; which is the ratio of force to mass. So, one equation you could have written is: $a = \frac{F}{m}$.

 

[Chestermiller]

I didn’t write down any equations because I thought it doesn’t depend on any.

Think again.

 

[YMMMA]

What do you know about air resistance and buoyancy forces?

Also, the key factor in whether something moves further than something else is not force, it's acceleration; which is the ratio of force to mass. So, one equation you could have written is: $a = \frac{F}{m}$.

I just know that air resistance slows down motion. For buyonat forces, It is not in any chapter of an SAT physics book that I studied from. But I searched anyway and got that If the byoyant force in the air is greater than the density of the object, that will slow the motion. If it is less than the density of the object, the objects moves faster down.

 

[CWatters]

Think about how mass, area and volume change as you increase the size of a sphere.

 

[YMMMA]

Think again.

Isn’t it just a=F/m as @PeroK said? The greater the mass the less it will accelerate.

 

[YMMMA]

Think about how mass, area and volume change as you increase the size of a sphere.

If one ball has a larger size than the other, and both balls have the same density or mass-to- volume ratio, that means the larger ball has the greater area.

 

[Chestermiller]

Isn’t it just a=F/m as @PeroK said? The greater the mass the less it will accelerate.

Because of air resistance, F depends on the surface area of the sphere. Which sphere has a larger surface to volume ratio? This is why you have to actually write out the equation.

 

[YMMMA]

Because of air resistance, F depends on the surface area of the sphere. Which sphere has a larger surface to volume ratio? This is why you have to actually write out the equation.

Ahh, I didn’t know that F depends on the surface area. Now I can tell that the larger ball has a larger surface area. That means it has greater force exerted on it (air resistance) so it accelerates faster.

 

[CWatters]

Ahh, I didn’t know that F depends on the surface area. Now I can tell that the larger ball has a larger surface area. That means it has greater force exerted on it (air resistance) so it accelerates faster.

No. Air resistance (drag) acts upwards in this problem. If it was that simple the larger sphere would fall slower.

I suggest you think more carefully about how area and mass change as you enlarge a sphere. For example if you double the volume the mass doubles but what about the area?

 

[Chestermiller]

Ahh, I didn’t know that F depends on the surface area. Now I can tell that the larger ball has a larger surface area. That means it has greater force exerted on it (air resistance) so it accelerates faster.

You are not going to be able to do this without writing down the force balance equation on the falling ball(s). Are you not able to do that? Let's see your free body diagram.

 

[Delta2]

As already pointed out by others, you have to write down the force equation and solve that equation for the acceleration of the ball. Then you ll see which ball has the greatest acceleration and why and so revealing which is the correct answer.

However if you want to think in a more tricky way, using an exclusion logic system, you can exclude all (except one) of the answers because they are wrong. For example as you already write at post #3, answers A and B essentially mean the same thing, so they are either both wrong or both correct, and because in this type of questions we know that only one answer is correct, we can conclude that A and B are both wrong. It turns out that you can also exclude the other two of the three answers because they are also wrong, so that will leave you at the end with only one possible choice.

EDIT: When I say that A and B are wrong I mean that they just don't explain the observation, as the question says. A and B are correct if viewed as stand alone statements.

 

[BvU]

If the byoyant force in the air is greater than the density of the object, that will slow the motion. If it is less than the density of the object, the objects moves faster down.

Sorry to be late with this, but I really have to object: you cannot compare a force with a density and say that one or the other is greater.

Please oblige and make a free body diagram, if you want for each of the balls. Then, with $F_{\text net} = ma$ see how things scale with the size of the ball.

 

[YMMMA]

I am not sure about my free body diagram, but here it is anyway...

The net force is downward for the large ball to have it accelerate more thant the smaller one. So again it has less mass. Still I am confused between E and D. Can’t understand them either.

I can only think of E as it says that the smaller ball has larger mass regardless of the reason that I don’t understand. But it should have larger mass for it to accelerate less. I am nit sure where its net force will point, it is either constant or that it point upward meaning that air resistance is dragging it upward and slowing its motion.

 

[Delta2]

What do you think about E), is it correct as a standalone statement? Normally I would expect the increase of mass due to the water vapor to be proportional to the surface of the ball, and the larger surface is that of the larger ball, so normally I would expect water vapor to increase more the mass of the large ball.

On the other hand D) is correct as a standalone statement but you got to do some math to prove it. But is D) this which explains the observed result? You have to work it again with math by working on the force equation. Can you write a force balance equation for the large ball? I see your FBD diagram page does not contain any equations.

 

[PeroK]

I am not sure about my free body diagram, but here it is anyway...

I can only think of E ...

E is what I would call a curve-ball answer. It sounds complicated and might tempt you but it can't possible be the correct answer. Who says there is enough water vapour in the air to make a difference? What have you learned in class about water vapour adhering to a falling object? Why would the small ball getting heavier slow it down under gravity? Are you told the relative density of water and the foam?

As has already been more than hinted at, all of A-D are true statements. But, which explains that the large ball falls faster?

 

[YMMMA]

This rule has nothing to do with surface area, I mean it doesn’t contribute to the acceleration of the balls.
I doubt what I did, though.

 

[YMMMA]

E is what I would call a curve-ball answer. It sounds complicated and might tempt you but it can't possible be the correct answer. Who says there is enough water vapour in the air to make a difference? What have you learned in class about water vapour adhering to a falling object? Why would the small ball getting heavier slow it down under gravity? Are you told the relative density of water and the foam?

As has already been more than hinted at, all of A-D are true statements. But, which explains that the large ball falls faster?

Well, I don’t remember seeing anything related to water vapor in my book. That test sometimes asks general questions that can be understood using logic but I am struggling with those..

But I did understand E now that @Delta² explained.

 

[Delta2]

Ok great, I see your equations, they are correct, let me expand them just a bit so the acceleration of a falling ball is

$a=\frac{F_w}{m}-\frac{F_{air}}{m}=g-\frac{F_{air}}{m}$ (can see that $\frac{F_w}{m}=\frac{mg}{m}=g$)

So the acceleration of a falling ball is $g$ MINUS a fraction $F_{air}/m$ . For which ball, this fraction becomes smaller, so the overall acceleration becomes bigger?
What do you know about the air resistance on a moving object, how it depends on its surface, velocity, density e.t.c?

 

[YMMMA]

This fraction is smaller for the larger ball.

Actually I don’t know. Is that a rule or something?

 

[YMMMA]

Ah, I got that. So, the larger ball has more area and thus more force of drag.

 

[Delta2]

Ah, I got that. So, the larger ball has more area and thus more force of drag.

Yes more drag for the larger ball, however the fraction $\frac{F_{air}}{m}$ is smaller for the larger ball.

 

[YMMMA]

So, D) is also not correct because it must be smaller for the larger ball. That leaves C.

 

[Delta2]

No D is correct, D doesn't say that drag is smaller for the larger ball, it says something else.

 

[Delta2]

Essentially what D) tell us is that the fraction $F_{air}/m$ is smaller for the larger ball, but you got to do the math behind it to understand it. Substitute $F_{air}$ =($D$ drag in the latest picture you gave us) with the formula you have from the latest picture you gave us and do some math.

 

[YMMMA]

I am able to prove D is true now after doing the math.

I appreciate your help all. Milion thanks!

 

[FactChecker]

There are a combination of forces and the resulting acceleration of the total force. You should separately consider the forces related to each proposed answer. Famously, all masses fall at the same rate under the influence of gravity alone. The force of gravity, which increases proportional to mass, does not increase the falling acceleration, which is inversely proportional to mass.

 

[PeroK]

So, D) is also not correct because it must be smaller for the larger ball. That leaves C.

You still seem to be working to a variation of Newton's second law:

$F = a$

Where you ignore the mass of object. For example, a car accelerates faster than a train:

A) because the car's engine produces a greater force
B) because the train is more massive

It's neither of these. The answer is:

C) the ratio of the force to mass of the car is greater than the ratio of force to mass of the train.

For air resistance it is the ratio of the resistance to mass that is critical. Not just the magnitude of the force.

 

[Chestermiller]

The actual force balance equation is $$ma=mg\left(1-\frac{\rho_a}{\rho}\right)-\rho_a\frac{v^2}{2}\pi R^2C_D$$where $\rho_a$ is the density of air, $\rho$ is the density of the ball material, v is the falling velocity of the ball, R is the radius of the ball, and $C_D$ is the air drag coefficient (assumed constant for the present development). The first term on the right hand side is the net downward force of gravity and buoyancy. The second term on the right hand side is the air resistance force.

If we divide both sides of this equation by the mass of the ball, we have
$$a=g\left(1-\frac{\rho_a}{\rho}\right)-\rho_a\frac{v^2}{2}\left(\frac{\pi R^2}{m}\right)C_D$$Note that the first term on the right hand side of this equation is independent of the mass of the ball. But the second term is not. The term $\left(\frac{\pi R^2}{m}\right)$ is proportional to the surface area to mass ratio of the ball. Since the mass of the ball is equal to its volume $\frac{4}{3}\pi R^3$ times its density $\rho$, we have that $$\left(\frac{\pi R^2}{m}\right)=\frac{3}{4\rho R}$$Since R is smaller for the smaller ball than for the larger ball, the expression is larger for the smaller ball than for the larger ball. That means the the effect of air resistance on the ball acceleration is greater for the smaller ball than the larger ball.

 

[YMMMA]

The actual force balance equation is $$ma=mg\left(1-\frac{\rho_a}{\rho}\right)-\rho_a\frac{v^2}{2}\pi R^2C_D$$where $\rho_a$ is the density of air, $\rho$ is the density of the ball material, v is the falling velocity of the ball, R is the radius of the ball, and $C_D$ is the air drag coefficient (assumed constant for the present development). The first term on the right hand side is the net downward force of gravity and buoyancy. The second term on the right hand side is the air resistance force.

If we divide both sides of this equation by the mass of the ball, we have
$$a=g\left(1-\frac{\rho_a}{\rho}\right)-\rho_a\frac{v^2}{2}\left(\frac{\pi R^2}{m}\right)C_D$$Note that the first term on the right hand side of this equation is independent of the mass of the ball. But the second term is not. The term $\left(\frac{\pi R^2}{m}\right)$ is proportional to the surface area to mass ratio of the ball. Since the mass of the ball is equal to its volume $\frac{4}{3}\pi R^3$ times its density $\rho$, we have that $$\left(\frac{\pi R^2}{m}\right)=\frac{3}{4\rho R}$$Since R is smaller for the smaller ball than for the larger ball, the expression is larger for the smaller ball than for the larger ball. That means the the effect of air resistance on the ball acceleration is greater for the smaller ball than the larger ball.

It’s very clear now, Thank you!