Question

Take a look at the following link: http://www.learnapphysics.com/solution.php?pc=9eb25d1d92d30335f8a80c27356e4af4

There appears to be some slightly convoluted reasoning here...

Answer 1

Anyone willing to take a stab at this interesting problem?

I'm just not convinced that the question writer has his physics completely right. Looking at the diagram, and taking away the "forces" due to air resistance on each of the spheres, it would appear by the writer's reasoning that the heavier ball would still land first - since it still has a greater force.

I think the solution actually has to do with the air resistance slowing the acceleration of the ping-pong ball more than the acceleration of the steel ball.

Answer 2

The heavier ball feel the greater force, but simliar accleration because it has bigger mass as well.

Answer 3

True - and then the air friction modifies the acceleration of the lighter ball.

Answer 4

air friction force= k v right?

Answer 5

It's obvious that the mass comes into play when dealing with air friction - but how? The given solution doesn't make sense.

Answer 6

my first guess would be that the metal ball has greater momentum then the ping-pong so when you imagine a molecule hitting a ball from below, the ping-pong ball will loose more momentum then a metal ball, ofc you have to imagine a lot of molecules hitting the balls but the idea is the same...

Answer 7

Mass doesn't have affect . take a two identicle piece of paper. Roll one and keep other as it is. Throw is from same height. Rolled one will fall faster then flat one even though they have same mass

Answer 8

yes but if they are the same shape, different mass?

Answer 9

i mean 2 same balls with different masses

Answer 10

mass*

Answer 11

Change of momentum is force (impulse), so since the heavier object gets a smaller impulse it's velocity changes less (in the upward direction) then that of a ping-pong ball. That just my guess, It's not from a textbook so it is probably wrong...

Answer 12

actually the impulse they both get is the same, but the velocity of the heavier object changes less then that of a lighter object...Impulse should be the same since they are both hit by a molecule with the same mass and velocity...

Answer 13

The logic makes sense, and here is why.

First, let's balance the forces.
Ping-pong ball: \(F_y = F_{d1} - m_1 g\)
Steel ball : \( F_y = F_{d2} - m_2 g\)

Now, we need an expression for the drag force, this is expressed as\[F_d = {c_d \rho v^2 A \over 2}\]where \(c_d\) is the coefficient of drag, \(\rho\) is the density of air, \(v\) is the velocity of the ball, and \(A\) is the cross-sectional area of the ball perpendicular to the direction of travel. Both balls are of the same size and shape, therefore the drag coefficients are the same, and both balls travel through the same air so the density is the same. Therefore, the drag coefficient depends only on the velocity, when we compare both balls.

The force of gravity is much greater than the force due to drag. When the balls are initially dropped, their velocities are initially zero, therefore gravity is the only force. The balls begin to fall. As they accelerate, the drag force becomes larger so their acceleration rate decreases. If we set up a ratio of gravitational force to drag force, we will see two things. A) The ratios will always be greater than 1, until terminal velocity is reached (which it won't since we are only falling 45 m), and B) the ratio for the steel ball will always be greater than the ping-pong ball. These being said, the steel ball will have the greatest acceleration throughout the 45 m drop, and therefore will hit the ground first.

Answer 14

Thanks Eashmore - I have one question. You note that the drag forces due to air are the same. Let's now say that the balls are in a vacuum. What would you change in your calculations?

Answer 15

in vacumm, there wouldn't be a drag force

Answer 16

So... according to Eashmore's calculations, how would that affect the respective accelerations of the spheres?

Answer 17

well , we could ignore all affect of drag, and just focus on gravitational forces

Answer 18

Easmore says this: "the ratio for the steel ball will always be greater than the ping-pong ball."

Would this change in a vacuum?

Answer 19

yes they would

Answer 20

I'm confused; if the drag force is dependent on the velocity of the sphere, then what changes the velocity of the sphere?

Answer 21

gravity

Answer 22

Here's where I'm coming from:

Let's start with the known facts. If we drop the spheres in a vacuum, they will land at the same time. If we drop them in air, the steel ball will land first. There is one difference between the two situations - air.

Also, the only difference between the two spheres is mass - or density - since the spheres are the same volume, mass and density are interchangeable here.

Therefore, the only factor affecting which will fall first is air, and the only factor that allows air to do that is mass. Air and mass.

Visualizing the situation, it's clear that somehow, air resistance keeps the lighter ball from falling as fast as the steel ball - and its clear that the greater mass of the steel ball allows it to overcome air resistance better than the lighter ball.

Both are slowed by air resistance - the steel ball less than the ping-pong ball. If the spheres fell through water, the steel ball would still fall faster, but both balls would be slowed significantly.

Our thought experiments thus seem to indicate an equation in which both the density of the falling substance and the density of the medium through which the substance falls come into play.

Does this make sense?

Answer 23

I might replace the post with different one soon, stand by

Answer 24

OK, I see. You're using conservation of momentum and having the heavy object and the air molecule stick together upon collision. Then you're determining the velocity of the combined mass...

Answer 25

Hello, very interesting topic.

First of all, is any one capable of performing the actual experiment? I would like to know what the actual observation yields, but I am afraid I don't have the balls to do it :-(

Second, I am trying to follow eashmore's response but I am not quite sure I do; I keep running into the same question as Underhill.

Let us say each sphere is subject to a vertical force Fy that is the sum of its weight (m * g) and the drag force (Fd) acting in opposite directions, with Fd being a function of the square of the sphere's velocity through air:\[F _{y}=F _{d} - mg\]\[F _{d}=\frac{c _{d}ρv ^{2}A}{2}\]As can be seen, Fd does not depend on the object's mass and must therefore be equal in both spheres. If Fy is the only source of acceleration for each ball and they effectively arrive on the ground at different times, the difference must surely rely on mg, since it is the only factor that differs between both spheres.

However, let us now take the same situation to a vacuum, where Fd = 0. If the logic applied in the previous paragraph is logic, why would both spheres now fall at the same speed? After all, their mass is different regardless of the composition of the surrounding medium.

Answer 26

yes, but I may have oversimplified it by having it collide with just one molecule

Answer 27

imranmeah91, not sure how to get from your 2nd equation to your 3rd equation - shouldn't there be a \[V_{air}\] in place of the

1

in the 3rd equation?

Answer 28

I try to simulate your problem in following fashion. There will be ten time step in which velocity increase by 9.8 due to gravity and I do momentum calculation. I list velocity at each time steps

use this formula
\[v_{new}=\frac{\left(m_{\text{air}}V_{\text{air}}+m_hV_h\right)}{m_{\text{air}}+m_h}\]

let m_air=1
V_air=1

so we get

\[v_i=\frac{1+m_h\left( v_{i-1}\right)}{1+m_h}+9.8\]

add velocity each step for gravity

if m_h=1
v_0=9.8
we get

at time step at 1 to 10
we get velocity of

\[\{15.2,17.9,19.25,19.925,20.2625,20.4313,20.5156,20.5578,20.5789,20.5895\}\]

compared to without drag(in vaccum)
\[\{19.6,29.4,39.2,49.,58.8,68.6,78.4,88.2,98.,107.8\}\]

now let's increase our mass of ball(m_h) to 10

\[\{18.8,26.9818,34.4198,41.1817,47.3288,52.9171,57.9973,62.6158,66.8143,70.6312\}\]

m_h=100
\[\{19.5129,29.1296,38.6511,48.0783,57.4122,66.6536,75.8036,84.863,93.8326,102.713\}\]

m_h=1000
\[\{19.5912,29.3726,39.1443,48.9062,58.6583,68.4007,78.1334,87.8563,97.5696,107.273\}\]

notice as we increase the mass of the mass, velocity as each point is comparatively bigger . but never bigger than if it were in vacuum

Answer 29

Underhill. Let me clarify some things. First, the drag forces are not the same, the drag coefficient, \(c_d\), are the same. Meaning at the same velocity, the drag force will be the same. However, as I established later, the steel ball will be travelling faster, and will, as such, have a higher drag force. This is accurately depicted on the website. (N.B.: The larger upward arrow on the steel ball.)

In a vacuum, a feather and a hammer will hit the ground at the same time. They have difference forces from gravity, but they also have different inertia's. They accelerate at the same rate. This is an important concept in Newtonian physics. Gravity exerts different forces on different objects, but gravity always accelerates objects at the same rate, in a vacuum. Air-resistance always complicates things, this is why it is often ignored in basic physics.

Now, if we step outside of a vacuum, gravity cannot be treated as a constant source of acceleration, it must be treated as a force. Initially, both balls accelerate at the same rate. If we observe them falling on a highspeed camera, we will see they will be within a few millimeters for the first couple cm. However, as drag force increases, the steel ball overtakes because the force of gravity to drag force ratio is larger for the steel ball than for the ping-pong ball. \[{m_1 g \over F_{d_1}} \gt {m_2 g \over F_{d_2}}\]

Answer 30

[Reading posts...] By the way, thanks, Erik and Imran, for your detailed replies.

Answer 31

User69. You ask some good questions. Drag force does not depend on mass, but since gravity is the only means by which the balls accelerate. The drag force can be related to gravity. \[F_g = mg\]\[v = at\]\[F_d = {c_d \rho v^2 A \over 2} = {c_d \rho (g t)^2 A \over 2}\]Clearly we can see that as time goes by, the drag force will increase, while the gravitational force remains constant.

Answer 32

Thanks, eashmore.

I see how Fd is a function of time through v = gt, but I do not quite see how Fd would be different for each ball, since g -and therefore the evolution of v over time- does not depend on the object's mass but on its distance from the Earth and the Earth's mass. If both balls do fall at different rates and the system is fully described with the equations provided, then the difference must surely rely on their weights (m * g) being different, because the drag coefficient is the same (they are both spheres), their cross-sectional area is the same (they have the same radius), the medium's density and viscosity is the same for both (they are traveling through the same air), and 2 is 2 in any case.\[F _{d}(t)=\frac{c _{d}ρ(gt) ^{2}A}{2}\]\[g _{1} = g _{2} \approx 9.81 m/s ^{2}\]\[F _{d1}(t) = F _{d2}(t)\]
However, if that is the case, the inequality should hold also in vacuum, since Fd would equal zero and therefore all the vertical force Fy would correspond to the weight, which is proportional to the object's mass. But observation suggests that the fall of a hammer and a feather -with different shapes and masses altogether- in the absence of an atmosphere -in the moon- will occur at the same rate.

So what is missing here?

Answer 33

hey how's going?

Answer 34

User69. This is why air-resistance is often ignored in physics classes - it makes things complicated and confusing, and requires we take a different approach to studying motion.

Let me present you with this equation, which is the velocity of the falling object. Notice how it does, in-fact, depend on mass:
\[v(t) = \sqrt{2 mg \over \rho A c_d} \tanh \left( t \sqrt{g \rho c_d A \over 2m} \right)\]
This equation is the solution to the differential equation that represent the balance of forces. You can find the derivation here: http://en.wikipedia.org/wiki/Terminal_velocity#Derivation_for_terminal_velocity
Click "expand"

Answer 35

Thanks as always, eashmore; you've been very helpful. Can I try to put into words what I have "understood" (if anything) regarding this topic?

As you mentioned, an object falling freely through a viscous medium is subject to two forces: its weight (Fg), pulling it downwards towards the center of the planet, and a drag (Fd) opposing the object's motion (and therefore, in this case, pushing it upwards). We can describe this opposition mathematically by giving each force an opposite sign:\[F_{d}=\frac{c_{d}ρv^{2}A}{2}\space\space\space\space\space(1)\]\[F_{g}=-mg\space\space\space\space\space\space\space\space\space\space(2)\]\[F=F_{d}+F_{g}\space\space\space\space\space\space\space(3)\]The total force (F) on the object is the sum of both individual forces. When F=0, the object stops accelerating and the remaining distance is covered at a constant velocity, called "terminal velocity". We can calculate its value for a given shape (Cd), size (A), mass (m) and medium (ρ) by substituting (1) and (2) into (3), equating F to zero and solving for v:\[0=\frac{c_{d}ρv^{2}A}{2}-mg\]\[2mg=c_{d}ρv^{2}A\]\[\frac{2mg}{c_{d}ρA}=v^{2}\]\[v=\sqrt{\frac{2mg}{c_{d}ρA}}\]The relationship is not linear, but an object's terminal velocity through a medium is directly proportional to its mass, meaning that the larger the object's mass, the larger its terminal velocity if the other factors are kept constant.

So yes, both spheres are subject to the same gravitational pull, but the heavier one reaches a higher terminal velocity due to its higher mass, and therefore arrives to the ground before the lighter one. If we now perform the experiment in the absence of an atmosphere, no drag force will develop and both spheres will fall at the same rate, according to the common local gravitational attraction.

Is that a reasonable explanation for the observation? If so, thanks again for your clicking remark.

Answer 36

User69. You nailed it. However, be careful in making the statement that, "...the heavier one reaches a higher terminal velocity due to its higher mass, and therefore arrives to the ground before the ligher one." The transient motion that occurs before terminal velocity is reaches is equally important to consider, especially when different masses and shapes are concerned.

Consider objects that have the same terminal velocity, but reach their terminal velocities at different times. The object that reaches its terminal velocity first, will be the first to hit the ground.

Answer 37

I think I see what you mean. The arrival at the terminal velocity is smooth rather than sudden, and therefore the falling rates are different long before the arrival at terminal velocity; possibly from the beginning of the fall. It is not that both objects fall accelerating at the same rate, and then the lighter one ceases accelerating at its terminal velocity. Rather, it slowly slows down its approximation to terminal velocity. Is that more or less what you mean?

Answer 38

Yes. The asymptotic behavior of the tanh function is very smooth. Look here: http://www.wolframalpha.com/input/?i=tanh%28x%29+and+tanh%280.5+x%29

Referencing the above plot, note that the derivative of tanh(x) is greater than the derivative of tanh(0.5x), at any x before we reach the asymptote. Where x is the ping-pong ball and 0.5x is the steel ball (refer to the earlier equation I gave you to see why). Now, consider this derivative to be the rate at which the falling object's acceleration approaches zero. With this, I hope you can see a steel ball continues to accelerate at a greater rate than the ping-pong ball for a longer time. Therefore, the steel ball reaches a faster terminal velocity AND accelerates ahead of the ping-pong ball. These two observations are key.

Answer 39

It is very clear now, thank you very much.

Answer 40

Excellent.

Answer 41

Great, guys - this seems to have turned out to be a very profitable discussion for everyone.

Thanks, imranmeah and eashmore, for explaining everything so clearly, and thanks, gogind and User69, for posting your reasoning along the way - it was very helpful to me and, I'm sure, others as well.