Irritating question

Say you throw a ball up in the air, and it lands. Say you repeat the throw, and this time your model takes air resistance into account. Does the entire journey take less time, more time, or the same amount of time?

This was on an Oxford interview last year, the guy mentioned something about using the equations of motion 'backwards' but I can't remember (or figure out) how to do it. Anyone got any ideas??

john

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Reply 1

forkwise

I THINK it would be the same time; it wouldn't go as far upwards because of air resistance, but it would take longer to come back down because of air resistance. But I don't know if the differences are equal.

Reply 2

john-boro

yeah, i've been trying to work it out with proportionality, but I can't work it out. I remember when the guy explained it, it seemed pretty simple, and intuitively it seems like it should take the same time...

Reply 3

forkwise

I tried it just now but air resistance changes with velocity... maybe use average air resistance?? I dunno what to do anyway.

Reply 4

uk6458

a resistive force allways opposes the direction of motion, this would apply to air resistance, no matter which direction. On the way up the air resistance force will be against the motion, it will slow it down, as well as gravity decelerating it. On the way down air resistance is still against the motion, gravity accerates the ball down.

So the total time would be longer, i think........... thats could be completly wrong :redface:

anyway........ :smile:

Reply 5

john-boro

this is tough as anything. Seems a really simple question at first too! Someone must know what to do?

Reply 6

Shmerm

The ball must lose energy between the time it is first thrown and the time it lands again because it is constantly being resisted and no potential energy is gained from start to finish. It has also lost momentum.

This means it is moving slower when it arrives at the same point it left from than when it left. This means the impulse of the air is directed downwards if we take the time being from when the ball leaves the hand to when it arrives again, because:

I = mv - mu (downwards is positive, u must be negative)

EDIT: to clear up the above: Impulse with no air resistance would be 2mv over the motion, as v=-u because the kinetic energy at the start and finish are the same. Above, u has a magnitude larger than v, so I > 2mv . This means the impulse of the air must be directed downwards to account for the extra impulse.

If the air is acting as a force downwards over the whole motion in general, the average acceleration due to air resistance must be downwards, meaning acceleration downwards (gravity and air resistance combined) is increased.

s = ut + a(t^2)/2 but s=0
0 = t (u + at/2)

t = 0 or u + at/2 = 0

u = -at/2

acceleration and initial velocity are in opposite directions, so:

initial speed upwards = (acceleration downwards x time)/2

initial speed remains the same, yet acceleration downwards is increased, time must decrease.

OMG I can't believe I've put together a point that might be right! I hope it is! Though I've been doing this for an hour so someone might have beaten me to it! :biggrin:

EDIT: also:

if you imagine the ball rising then falling. It is constantly being resisted by air, meaning it is constantly losing energy. This means at any one height it has less energy when it is falling than when it is rising. This means it has less speed, meaning the force resisting it is less when it is falling than when rising. This means the force of air resistance is higher when it is rising are more than when it is falling, meaning in general the total force of air resistance is directed downwards. This can then lead you into my previous description starting at the bit where I talk about acceleration downwards being larger than gravity over the whole motion.

If this is right I think AQA should give me an A right now.

Reply 7

john-boro

The last paragraph and the first bit too seems to make sense to me. Just not totally clear about the rest of the reasoning (it is early morning after all :smile:

)

john

Reply 8

Shmerm

hmmm... I'll try again... not sure how

The average force of air resistance is downwards... F=ma, so the average acceleration because of air resistance is downwards.

This means acceleration downwards is larger than the force of gravity over the whole motion.

using s=ut + (at^2)/2

it starts and finishes in the same place:

so 0 = t(u + at/2) that makes sense right?

so it is at the start/finish point when t=0 or u + at/2 = 0

u = -at/2

initial speed is direceted upwards, and acceleration is directed downwards, so:

initial speed upwards = acceleration downwards x time / 2

but initial speed upwards is the same. Acceleration downwards is increased, so time must decrease for the equation to work.

Was that any better? I feel like I repeated myself a bit..!

Reply 9

john-boro

yeah it seems to make sense to me. I agree with the effect of acceleration being downwards overall. However, don't the equations of motion only apply to instances of constant acceleration?

Sorry if I'm being thick here, the basic ideas behind your argument seem to make sense, and the conclusion may well be right (I can't remember), it just doesn't seem mathematically 'all there' if you get me.

Still, the best idea so far, and as I said, it may well be right,

john

Reply 10

Shmerm

I'm saying there's an average force downwards, so you can imagine this is a constant force over the time. Therefore you can model it as a constant acceleration downwards.

Reply 11

john-boro

ahh I see. I think you might be right :smile:

Reply 12

teachercol

Its not trivial since the ball wont go as high with air resistance so the total distance travelled is less.

Reply 13

john-boro

so what does this mean regarding the time taken?

I'm still a bit confused here.

Reply 14

Shmerm

total distance travelled doesn't matter because it's displacement that I've used in my explanation, not distance travelled. Displacement is 0 in both cases.

I might be wrong but I can't find any flaws in what I've written. I'm sure someone more intelligent than me might though!

Reply 15

harr

Shmerm

as v=-u because the kinetic energy at the start and finish are the same
...
This means at any one height it has less energy when it is falling than when it is rising. This means it has less speed, meaning the force resisting it is less when it is falling than when rising.

I'm having trouble following part of your argument. It's probably just stupidity on my part, but I just can't get my head around your logic to show that the net effect of the air resistance is downwards.

How can the kinetic energy at the start and finish be the same if it has less energy at any height when falling than at the same height while rising?

I also don't understand why you are working with energy instead of velocity. You are assuming that the ball is always losing energy and therefore must be slowing down. However as the ball falls the velocity increases even though you argue it is losing energy, this shows that energy and velocity cannot be considered to be the same thing. If it is losing energy but gaining velocity how can you claim that losing energy implies slower movement?

On our timescale the ball appears to be constantly colliding with gas particles, but this is not true over smaller periods of time. The ball will travel for a while before colliding with a particle and then will lose energy in the collision. Now go back to the example without air resistance. The speed is the same whatever the direction at any given height. Now introduce one particle. When the ball hits this particle it loses momentum. Because it is travelling the at the same speed whether the velocity is positive or negative the collision will have the same effect, so the ball will lose the same amount of momentum whichever direction it is travelling. If you intoduce more particles the same effect occurs, this seems to me to suggest that the effect of air resistance is the same falling as rising, so the net effect of air resistance would be neither upwards nor downwards. Of course there could be flaws in my model.

Edit: You may want to note that I have not yet started A level physics, so I may make mistakes due to a lack of knowledge.

Reply 16

Shmerm

harr

How can the kinetic energy at the start and finish be the same if it has less energy at any height when falling than at the same height while rising?

because the "v = -u" part is for no air resistance, as I said in my initial description. When we account for air resistance, that's when there's less energy at any height while falling compared to rising.

harr

I also don't understand why you are working with energy instead of velocity. You are assuming that the ball is always losing energy and therefore must be slowing down. However as the ball falls the velocity increases even though you argue it is losing energy, this shows that energy and velocity cannot be considered to be the same thing. If it is losing energy but gaining velocity how can you claim that losing energy implies slower movement?

There is a resistive force throughout the entire motion (I hope that's clear) which means it is constantly losing energy. At any given height, Potential energy is the same while rising compared to falling. This means that kinetic energy while falling at a given height must be less than while rising. Which means velocity must be less. Velocity and Energy are very clearly linked, you just seem to be forgetting about potential energy.

harr

On our timescale the ball appears to be constantly colliding with gas particles, but this is not true over smaller periods of time. The ball will travel for a while before colliding with a particle and then will lose energy in the collision. Now go back to the example without air resistance. The speed is the same whatever the direction at any given height. Now introduce one particle. When the ball hits this particle it loses momentum. Because it is travelling the at the same speed whether the velocity is positive or negative the collision will have the same effect, so the ball will lose the same amount of momentum whichever direction it is travelling. If you intoduce more particles the same effect occurs, this seems to me to suggest that the effect of air resistance is the same falling as rising, so the net effect of air resistance would be neither upwards nor downwards. Of course there could be flaws in my model.

I believe the bit I have put in bold is the flaw in your point here. It's a good thought, but after the ball collides with the particle after it rises, it leaves with a speed, and arrives at that point again with the speed it left with, not with the speed it arrived with.

Let's say the ball has mass 10m, and the particle has mass m. The collision can be assumed to be elastic. Let's say also the ball it moving with speed 5 m/s, and the particle is at rest.

Momentum: 5x10m + 0xm = 10mV(ball) + mV(particle)

50 = 10V(ball) + V(particle)

collision is elastic, so speed of seperation = speed of approach

V(particle) - V(ball) = 5

so 50 = 10V(ball) + 5 + V(ball)

45 = 11V(ball)

V(Ball) = 45/11 m/s

Ball started with KE = 125m J and now has KE = (10125/121)m J so has lost exactly (5000/121)m J of energy in the collision.

now, when the ball falls to the same height it must have velocity 45/11 m/s based on this single particle model you are using. Also, we shall assume that the particle has come to rest, because perhaps it has collided with other particles and lost all it's energy. Or perhaps there is now a new particle in the way. It doesn't really matter as long as the particle is at rest. By the same principles used above (Vball and Vparticle are now different to above):

Momentum = 10m x 45/11 (<- the speed the ball arrives at) + 0 x m = 10mVball + mVparticle

450/11 = 10Vball + Vparticle

elastic collision

45/11 = Vparticle - Vball

so:

450/11 = 10Vball + 45/11 + Vball

405/11 = 11Vball

Vball = 405/121 m/s

so now the KE = 5m x 164025/14641 = 820125/14641 m J

before it had 10125m/121 J

10125m/121 - 820125m/14641 = 405000m/14641 J lost = 3347.1...m/121 J lost.

This is not the same as the Energy lost when it collided on the way up. Therefore the idea you had I don't think can be valid.

Now I'm not saying I'm definitely right, and if anyone comes along to disprove me then that's great, but I will try to protect my argument if people question it and I feel I am able to do so :smile:

Reply 17

harr

Shmerm

I've changed my mind. :smile:

The flaw in my model was similar to the one you suggest, I made an assumption (without realising it, but I really should have noticed it) when converting it to a two particle system (one collision on the way up and one on the way down) which is necessary for it to work with n particles.

I will think about it again tomorrow (my brain isn't working properly at the moment and I always find sleep gives me ideas), but it does look like my model actually implies that the ball travels faster on the way up than the way down, which is not what I initially thought and would prove you correct.

This is why I like discussions like this really. Much more difficult than what we are normally expected to do in school and I am quite often wrong (for some reason I quite like realising I am wrong about something in science, I suppose it is just more exciting than being correct all the time).

Now I'm saying I'm definitely right, and if anyone comes along to disprove me then that's great, but I will try to protect my argument if people question it and I feel I am able to do so