Parent needing help with a velocity question including air resistance

[jackiem1075]

Homework Statement

USEFUL DATA
mass of ping-pong ball = 10g
height dropped = 3m
acceleration due to gravity on Earth = 9.81Nkg-1
acceleration due to gravity on Mars = 38% of the value on Earth
air density on Earth = 1.2kgm-3
air density on Mars = 0.6% of the value on Earth

CALCULATE THE FOLLOWING
a. The velocity of the ping-pong ball just before landing
1. On Earth
2. On Mars

Relevant Equations

KE = 0.5mv2
GPE = mgh
Air resistance = 0.5 x air density x drag coefficient x area x v2

I'm a parent and have no idea, just looking for help please

 

[berkeman]

Homework Statement:: USEFUL DATA
mass of ping-pong ball = 10g
height dropped = 3m
acceleration due to gravity on Earth = 9.81Nkg-1
acceleration due to gravity on Mars = 38% of the value on Earth
air density on Earth = 1.2kgm-3
air density on Mars = 0.6% of the value on Earth

CALCULATE THE FOLLOWING
a. The velocity of the ping-pong ball just before landing
1. On Earth
2. On Mars
Relevant Equations:: KE = 0.5mv2
GPE = mgh
Air resistance = 0.5 x air density x drag coefficient x area x v2

I'm a parent and have no idea, just looking for help please

Welcome to the PF, Jackie.

Per the PF rules, you or your student must show your best efforts to start working the problem before we can offer tutorial help. Is this for a calculus-based intro physics class? What kind of problems are solved so far in the textbook or online learning materials that you and your student are using?

 

[jbriggs444]

KE = 0.5mv2
GPE = mgh
Air resistance = 0.5 x air density x drag coefficient x area x v2

The full solution looks like a nasty differential equation. So a useful approach would be to try to figure out an answer without air resistance and to then figure out whether air resistance is significant.

Are you able to determine an answer based on gravity alone? Do you want to develop the differential equation?

 

[jackiem1075]

No my son has just left HS and the college he will be attending sent a pamphlet of work which he has completed. This is the only part he is stuck on. The reason I have asked for hep is so that when I have the answer and how to get it, and I understand it, I would be able to offer him a few ideas to try help him know where to start in answering. If this is not something you guys can help with that's ok, Ill just keep trying

 

[jackiem1075]

The full solution looks like a nasty differential equation. So a useful approach would be to try to figure out an answer without air resistance and to then figure out whether air resistance is significant.

Are you able to determine an answer based on gravity alone? Do you want to develop the differential equation?

 

[jbriggs444]

If this is not something you guys can help with that's ok

This is something we can help with. But we need to know where to aim the help. You are saying that you want help determining the impact velocity under the effect of gravity alone?

The way this forum works is that you show your best effort and we help when you get stuck. You have shown some relevant equations. How far did you get applying them? Show your work.

 

[jackiem1075]

I really don't have any idea and my son is not here to ask

 

[jbriggs444]

I really don't have any idea and my son is not here to ask

Let us start with the equation: GPE=mgh

What is the gravitational potential energy of the ping pong ball before it is dropped to the Earth?

Please, show some work this time. Otherwise I will drop out and report this thread to the mentors.

 

[etotheipi]

I really don't have any idea and my son is not here to ask

The question is essentially asking you to solve $$mg - \frac{1}{2}AC\rho \dot{x}^2 = m\ddot{x}$$as @jbriggs444 mentioned, it is a bit nasty. So to start off with, you should check that you can do the question whilst ignoring air resistance first

 

[berkeman]

No my son has just left HS and the college he will be attending sent a pamphlet of work which he has completed.

I agree with @jbriggs444 that you would need to solve a differential equation in order to include air resistance. That seems like a pretty advanced question for the transition from HS to university. Has your son taken calculus already in HS? And this class at university -- is it an honors class or otherwise pretty advanced from the get-go?

 

[jackiem1075]

Hi, no my son is 16 he has just left high school to start A level at college next month. I have absolutely no idea about physics and was just looking for someone to fully explain what he would need to do to solve the problem of what the velocity of the ping-pong ball just before landing
1. On Earth
2. On Mars
would be.

 

[PeroK]

Hi, no my son is 16 he has just left high school to start A level at college next month. I have absolutely no idea about physics and was just looking for someone to fully explain what he would need to do to solve the problem of what the velocity of the ping-pong ball just before landing
1. On Earth
2. On Mars
would be.

This isn't an A-level problem. In any case, there's no way to explain these problems if you have absolutely no idea about physics. You have to be an active student yourself.

 

[jackiem1075]

Ok, no problem, thanks anyway

 

[berkeman]

was just looking for someone to fully explain what he would need to do to solve the problem of what the velocity of the ping-pong ball just before landing
1. On Earth
2. On Mars
would be.

Given the level of your son right now, I agree that a good start is to solve the problem ignoring air resistance for now:

Are you able to determine an answer based on gravity alone?

And one of the easier ways to solve it is to equate the Potential Energy (PE) of the ball in its starting position with its Kinetic Energy (KE) when it hits the ground. In this problem without air resistance, the sum of PE+KE is a constant. Since the ball is motionless at first, KE=0, and if you take h=0 to have zero PE, then all of the energy is in the motion or KE when the ball hits the ground.

Can you use the Relevant Equations you posted to start working the simplified problem from the energy conservation angle?

http://physics.bu.edu/~duffy/py105/EnergyConservation.html

 

[jackiem1075]

Thank you so much for your help

 

[aspodkfpo]

Ok, no problem, thanks anyway

Ignore air resistance, it's beyond A-levels for sure.

KE = 1/2 m v^2
PE = m g h

When the ball reaches the ground, all the initial potential energy has become kinetic energy, i.e. PE -----> KE. Plug and chug for the symbols.

 

[jackiem1075]

Thank you I'll pass this onto my son

 

[kuruman]

I have always favored the idea that people who need help from PF should get it directly from PF and not through an intermediary. I think it's more efficient and effective that way because the message might be unintentionally garbled as it goes from point A to point C through point B.

 

[jackiem1075]

Thanks for your input

 

[Lnewqban]

...
I'm a parent and have no idea, just looking for help please

Please, don't worry much; everything is going to be alright.
As others have stated, the answer of this problem seems to be too complicated for the level.

The problem is that the velocity of the falling ball depends on the acceleration of the gravity in both places and should increase with falling height.
As velocity increases, the force that resists that movement also increases, but at a different rate (square of velocity).

If the fall lasts enough time, there is a point at which both forces (up and down directions) reach a balance point, which is named terminal velocity.
In Mars, the magnitudes of both forces should be smaller (less gravity acceleration and less drag from the atmosphere).

This information may help:
https://courses.lumenlearning.com/physics/chapter/2-7-falling-objects/

https://en.wikipedia.org/wiki/Terminal_velocity

 

[jackiem1075]

Thanks we have put something together. We did the velocity of the ping pong when it lands on earth

We found out the time it took to fall and then did time x acceleration due to gravity = Velocity

for time we did height(3m) x 2 / 9.81 = 0.6116207951 then found the square root of that answer = 0.782 seconds

We then did 0.782 x 9.81 = 7.672 m/s

Is this right for a basic level?

 

[Lnewqban]

Thanks we have put something together. We did the velocity of the ping pong when it lands on earth

We found out the time it took to fall and then did time x acceleration due to gravity = Velocity

for time we did height(3m) x 2 / 9.81 = 0.6116207951 then found the square root of that answer = 0.782 seconds

We then did 0.782 x 9.81 = 7.672 m/s

Is this right for a basic level?

You are welcome
That is the correct answer for final velocity, disregarding drag of air, for one of the planets.

That is very close to the actual terminal velocity of a pingpong ball: 8 m/s, according to this link:
https://link.springer.com/article/10.1111/j.1747-1567.2006.00017.x

 

[jackiem1075]

Yes it did give us the air density readings and an equation to find the air resistance but it was getting far too complicated so just thought at least he has something to show them that he has understood.

Thank you all for your help

 

[jackiem1075]

We used the same for Mars we just worked out the acceleration due to gravity on Mars = 38% of the value on Earth so changed it from 9.81 to 3.73

 

[jbriggs444]

Thanks we have put something together. We did the velocity of the ping pong when it lands on earth

We found out the time it took to fall and then did time x acceleration due to gravity = Velocity

for time we did height(3m) x 2 / 9.81 = 0.6116207951 then found the square root of that answer = 0.782 seconds

We then did 0.782 x 9.81 = 7.672 m/s

Is this right for a basic level?

It gives the right answer and is a correct approach, so that is good.

I have two suggestions to make.

1. It is good to work with symbols and get the result with algebra rather than numbers.

In the case at hand, you start by trying to determine time. You know distance and acceleration. The standard symbols for those quantities are "t" for time, "s" for distance or displacement and "a" for acceleration. The equation you would have started with would be:$$s=\frac{1}{2}at^2$$You solved that correctly (multiply both sides by 2, divide both by a, take the square root of both and then flip left with right) to arrive at$$t=\sqrt{2\frac{s}{a}}$$One assumes that your child did all of this correctly. The next step would to multiply acceleration (a) by time (t) to arrive at the change in velocity ($\Delta v$):$$\Delta v=at$$But since we already have the formula for t above, we can substitute that in, yielding:$$\Delta v = a \sqrt{2\frac{s}{a}}$$This is where doing things with symbols starts to shine compared to doing things with numbers. You can cancel that $\sqrt{a}$ in the denominator with the $a$ in the numerator yielding a single $\sqrt{a}$ in the numerator:$$\Delta v = \sqrt{2as}$$Now with your desired quantity alone on the left hand side, you can plug numbers into the right hand side, evaluate and done:$$\Delta v =\sqrt{2as} = \sqrt{2 \times 9.81 \times 3} = 7.672 \frac{m}{s}$$That is the first point I wanted to make: Try to use symbols instead of numbers.2. There is another approach that could have been taken. An "energy" approach was suggested. This time the starting point is:$$PE_{\text{initial}}=KE_{\text{final}}$$You plug in your formulas for potential energy ($PE=mgh$) and kinetic energy ($KE=\frac{1}{2}mv^2$) and get:$$mgh=\frac{1}{2}mv^2$$Now you can divide both sides by m (eliminating m entirely), multiply by 2, take the square root and flip left and right yielding:$$v=\sqrt{2gh}$$By no coincidence, this is the exact same result that we had obtained above. It is just that we are using "$g$" for gravity instead of "$a$" for acceleration and "$h$" for height instead of "$s$" for displacement.

There is actually a third approach that could have been taken. It is closely related to the "energy" approach. This time the starting point is the "SUVAT" equations. This is a set of equations that relate displacement ($s$), starting velocity ($u$), final velocity ($v$), acceleration ($a$) and time ($t$) in various arrangements. You can Google for "SUVAT" and find a bazillion hits. If one does so and searches for an equation with $v$ on the left hand side and variables you have as givens on the right, one quickly arrives at:$$v^2=u^2+2as$$ Substitute in starting velocity ($u$) = 0, acceleration ($a$)=9.81, displacement ($s$) = 3. Take the square root and you are done.

 

[haruspex]

I doubt there is any expectation that the student would solve the ODE. There is a rather more pragmatic approach.
For a ping pong ball in Earth's atmosphere, 3m is likely to get close to terminal velocity, so assume that and solve the force balance for zero acceleration.
In the Martian atmosphere, drag over that drop will barely make a difference, so ignore air resistance for that case.

 

[JC356]

Allow me to expand on haruspex's guidance.

For your case HS->college I don't think an exact answer is expected due to the complexity of the math. It seems you are being asked to think outside the box. In engineering it is OK to say "I can't get an exact answer without a lot of work, but I can get close enough with some simplifications.". You make an engineering decision or judgement regarding whether a term in an equation really matters. 40 years ago, when I was an engineering student, professors would take liberties with unsolvable equations and cross out the "small terms" to make the equation solvable. With good judgement you will get an answer that is "close enough". At first glance it sounds wrong but as put by haruspex, it is pragmatic. Close and quick generally beats slow and perfect, and it always beats no answer at all.

You need to calculate the velocity of the ping pong ball at impact, without air resistance for both cases. Next calculate what the air resistance would be at both impact speeds. If the air resistance is close to the mass of the ball (you may find it is more) you declare that "air resistance dominates". Keep in mind that the fastest the ball will can fall is when air resistance equals the mass. If the air resistance is very small compared to the ball mass you can declare it to be insignificant. In that case "Gravity dominates".

In either case you get to do the math, use the answers to decide what a reasonable answer is, and go with it. Don't forget to an explanation like "I computed A to be X, and B to be Y, so I ignored B because...". It is possible this is a test to see just how good you math skills are. However, after 40 years of being an engineer I ignore the complex math unless I determine it is necessary to do the math. If the person who grades the answer doesn't like it that's OK. You can't win them all and in the long run it won't matter.

Most important, give them an answer even if you approximated it. Don't say "If there was no air it would be...". That paraphrases to "I couldn't get an answer so I changed the question.". I mentor people and personally that doesn't fly. In the real world you will be the judge of whether you are close enough.

When you get an answer don't ask what I think of it. That is for you to decide.

This response is more mentoring help than physics. It may not go over well with some people but realize you are dropping a ping pong ball not landing a Mars rover.

 

[kuruman]

I doubt there is any expectation that the student would solve the ODE. There is a rather more pragmatic approach.
For a ping pong ball in Earth's atmosphere, 3m is likely to get close to terminal velocity, so assume that and solve the force balance for zero acceleration.
In the Martian atmosphere, drag over that drop will barely make a difference, so ignore air resistance for that case.

Well, I solved the ODE and obtained expressions for $v(t)$ and $x(t)$. Here is a summary of results that substantiate @haruspex's approximations.
Earth
Using the official mass [https://en.wikipedia.org/wiki/Table_tennis#Ball] of 2.7 g and radius 20 mm for the ping-pong ball, I got a terminal velocity of 8.64 m/s which agrees with the value mentioned in the link in post #22. I then solved for the time of flight numerically and went back to find the speed at that time. As expected, I got a number less than $\sqrt{2g_{Earth}h}=7.7~\mathrm{m/s}$.

Mars
Repeated the above for Mars. Terminal velocity on Mars is 69 m/s. I then solved for the time of flight numerically and got and went back to find the speed at that time. As expected I got a number extremely close to $\sqrt{2 g_{Mars}h}=4.73~\mathrm{m/s}$.

Finally, the graph below shows Drop distance vs. time on Earth and on Mars in air and in vacuum ( i.e. neglecting air resistance). On Mars, the air is so thin that it makes no difference as already explained by @haruspex.

I agree that the expectation is not that one solve the ODE. Perhaps it is a test as to how inventive and realistic one's answer is.

 

[jackiem1075]

Thanks everyone for your input