Energy Conservation: Kinetic Energy After Collision

[newTonn]

It's kind of hard to understand what you're trying to say, but I'll take a stab at it.

I don't know what you mean by the first sentence at all, but as for the other parts of your post:

Yes, the ball requires a time interval to reach its final velocity, and yes there is a displacement necessary to acquire this velocity.

The ball is in contact with the club for a small but definite interval of time. It appears to us that the ball is only in contact with the club instantaneously, but in reality (as was shown by a video posted earlier), the ball is in contact with the club for a few hundredths of a second. There is no conflict with the laws of physics here. The ball is accelerated very quickly over a short distance and a short time period.

Even the distance and interval is short,we have to accept that a force is acting on the ball during this interval.while the club was not in contact with the ball.

 

[rcgldr]

Even the distance and interval is short,we have to accept that a force is acting on the ball during this interval.while the club was not in contact with the ball.

The only forces involved when the club is not in contact with the ball is aerodynamic drag and gravity. After contact is completed, the ball is decelerating (both backwards and downwards, one it reaches peak height, it starts accelerating downwards).

Look at that video again, the ball is long gone before the club gets about 1/3rd over the tee. The club isn't going to exert any aerodynamic push on the ball.

Perhaps what you're referring to is the visual illusion that makes a golf ball seem to accelerate well after contact. This is due to a tracking problem by a person. The person overshoots the visual tracking then slows it down, and the ball appears to accelerate as it returns to the center of vision.

 

[newTonn]

The only forces involved when the club is not in contact with the ball is aerodynamic drag and gravity. After contact is completed, the ball is decelerating (both backwards and downwards, one it reaches peak height, it starts accelerating downwards).

Look at that video again, the ball is long gone before the club gets about 1/3rd over the tee. The club isn't going to exert any aerodynamic push on the ball.

Perhaps what you're referring to is the visual illusion that makes a golf ball seem to accelerate well after contact. This is due to a tracking problem by a person. The person overshoots the visual tracking then slows it down, and the ball appears to accelerate as it returns to the center of vision.

Then tell me when does the ball accelerated,due to the force exerted by club?

 

[rcgldr]

Then tell me when is the ball accelerated. The force exerted by club?

Yes it's accelerated by the force from the club. As posted before this force only last for about 1/200th of a second, but it's enough to accelerate the golf ball from 0 to 170mph. In case you're wondering, that's an average acceleration of 1550 g's, it's no wonder the ball deforms so much from the rate of acceleration.

 

[Doc Al]

Al together you are saying ball is attaining its final velocity (in the absense of other forces)from the same position(where it was staying at rest) and at the same instance(no interval),when the club hit the ball.
still i cannot digest,how a velocity (and thus an acceleration)is acquired without displacement of ball?

Where in the world did you get the idea that the ball attains its maximum velocity (from being hit with the club) without moving or in zero time? Of course there is a displacement of the ball as it's being smashed by the club.

 

[HallsofIvy]

The face of the golf club deforms a very slight amount, the golf ball deforms much more and the shaft of the golf club bends. All of that takes place in a very short time (according to Jeff Reid, above, about 1/200 sec.) and it is during that time that the acceleration takes place- while the face of the golf club is in contact with the ball.

 

[newTonn]

Where in the world did you get the idea that the ball attains its maximum velocity (from being hit with the club) without moving or in zero time? Of course there is a displacement of the ball as it's being smashed by the club.

Sorry if anybody got frustrated on my arguments.But please go through some points from one of your previous post to find the answers for your question.(text was made bold-by me)

A free body diagram of the golf ball is easily drawn, but realize--as has been explained and illustrated already--that the force exerted by the club is not a constant force: when the club first touches the ball, the force is small; then it peaks as it smashes the ball; then, as the ball leaves contact with the club, the force goes to zero again.

the club was in contact with the ball ,when ball was at rest.ball moved a millionth of an inch means no contact and hence force on it is zero.

Again, if the ball is accelerating a net force must be acting on it.

consider the ball tooks a millionth of second to move millionth of an inch,(acceleration from zero to final velocity).what is the force on body in between this interval? to be specific,the interval between billionth of a second after club left its contact and billionth of second before ball reaches its final velocity-For acceleration,of course ball has to change its velocity from zero to final velocity-it has to be displaced-may be fractions of our standard units,but those fractions are again divisible.

 

[Doc Al]

the club was in contact with the ball ,when ball was at rest.ball moved a millionth of an inch means no contact and hence force on it is zero.

I'm not sure what you are saying here. When the club first makes contact with the ball, of course the ball is at rest. During the interaction, both club and ball move a bit. Once the ball shoots off the face of the club, there is no longer any force exerted on the ball by the club. What is so hard to understand about this?

consider the ball tooks a millionth of second to move millionth of an inch,(acceleration from zero to final velocity).what is the force on body in between this interval? to be specific,the interval between billionth of a second after club left its contact and billionth of second before ball reaches its final velocity-For acceleration,of course ball has to change its velocity from zero to final velocity-it has to be displaced-may be fractions of our standard units,but those fractions are again divisible.

Again you repeat the same misconception. For some reason you think the ball reaches maximum speed some time after it breaks contact with the club. No! As soon as the club loses contact with the ball, the only forces on the ball are gravity and air drag: these forces act to slow down the ball.

 

[Ariste]

what is the force on body in between this interval? to be specific,the interval between billionth of a second after club left its contact and billionth of second before ball reaches its final velocity-For acceleration,of course ball has to change its velocity from zero to final velocity-it has to be displaced-may be fractions of our standard units,but those fractions are again divisible.

I don't understand why you keep saying this. In those couple of hundredths (not billionths) of a second that the ball is in contact with the club, the ball is accelerated from rest to around 170 mph. It attains ALL of this velocity in the time that it is in contact with the club. After the ball leaves the club, it is no longer accelerating forward. At the instant that the ball leaves the club, it is traveling at 170mph. There is no additional acceleration afterwards, except in the backwards direction due to air resistance.

 

[newTonn]

I don't understand why you keep saying this. In those couple of hundredths (not billionths) of a second that the ball is in contact with the club, the ball is accelerated from rest to around 170 mph. It attains ALL of this velocity in the time that it is in contact with the club. After the ball leaves the club, it is no longer accelerating forward. At the instant that the ball leaves the club, it is traveling at 170mph. There is no additional acceleration afterwards, except in the backwards direction due to air resistance.

I am saying again that ,as i understand,velocity is the rate of displacement.
So while the club was in contact with ball,since ball didn't displaced.how somebody can say there is a velocity for ball ,without being displaced?
If you corrected your statement as ,while the ball is in contact with the club,it is attaining a potential(kinetic energy),to move at 170mph,i will agree.yes. throughout the path of the ball this potential is reduced until it becomes zero,due to the interaction of other forces.
Any how if you can represent this as a vector.you can clearly draw a freebody diagram of the ball,at any point in the path of ball,justifying the movement of ball (yes i understand ,it is accelerating ,negatively because the net force(resultant is in forward direction,but getting reduced)
My point is that if you consider this potential/kinetic energy as force,you have to consider same for a body moving with uniform velocity also.

 

[Ariste]

I am saying again that ,as i understand,velocity is the rate of displacement.So while the club was in contact with ball,since ball didn't displaced.how somebody can say there is a velocity for ball ,without being displaced?

The ball is displaced while it is in contact with the club. It's only displaced by a small amount before it leaves the club, but it is definitely displaced.

Here, I'll give you some mathematical figures. Assume that the ball weighs .05kg and that it reaches a final velocity of 80 m/s. Thus the club gives the ball .5(.05)(80)^2 = 160J of kinetic energy. This means that the club did 160J of work on the ball. Let's assume that the club exerts an average force of 3200N on the ball. Using the formula W=Fd we conclude that the ball was in contact with the club for 160/3200 = .05m.

Obviously these numbers are all made up, and I think they're only valid when the ball is hit perfectly horizontally, but you can see that there most certainly is a displacement while the ball is in contact with the club, and thus a velocity.

 

[newTonn]

The ball is displaced while it is in contact with the club. It's only displaced by a small amount before it leaves the club, but it is definitely displaced.

Here, I'll give you some mathematical figures. Assume that the ball weighs .05kg and that it reaches a final velocity of 80 m/s. Thus the club gives the ball .5(.05)(80)^2 = 160J of kinetic energy. This means that the club did 160J of work on the ball. Let's assume that the club exerts an average force of 3200N on the ball. Using the formula W=Fd we conclude that the ball was in contact with the club for 160/3200 = .05m.

Obviously these numbers are all made up, and I think they're only valid when the ball is hit perfectly horizontally, but you can see that there most certainly is a displacement while the ball is in contact with the club, and thus a velocity.

Really i am confused by by your last statement.Any how i think you are explaining that ball was in contact with the club until it reach 0.05m (in your example).
if you assume that the club exerts an average force of 160N(instead of 3200N) on the ball,the club will be in contact with the ball until it reach 1 m.This i feel has no meaning.

Let us look at the problem in another way.
Let the weight of ball remains same-0.05 kg.
Let the club exerts a force of 10 kg m/s2.
So acceleration a =f/m = 10/.05 = 200 m/s2.
since a = (v-0)/t ; Final velocity v= 200m/s;
now kinetic energy = 0.5 (0.05) * 200^2 =1000 kg m2/s2
Now d = W/F = 1000/ 200 = 5 meter (does it make sense that the club was in contact with the ball for 5m?)
Please check .i am not sure about my calculations

Note:Logic and concept of the problem is Not mine-it belongs to Ariste.
Approach and new figures are the only contribution from my side

 

[Cthugha]

Let us look at the problem in another way.
Let the weight of ball remains same-0.05 kg.
Let the club exerts a force of 10 kg m/s2.
So acceleration a =f/m = 10/.05 = 200 m/s2.
since a = (v-0)/t ; Final velocity v= 200m/s;
now kinetic energy = 0.5 (0.05) * 200^2 =1000 kg m2/s2
Now d = W/F = 1000/ 200 = 5 meter (does it make sense that the club was in contact with the ball for 5m?)
Please check .i am not sure about my calculations

Note:Logic and concept of the problem is Not mine-it belongs to Ariste.
Approach and new figures are the only contribution from my side

Your example is just nonsense.

Usual forces acting on golf balls range between 3000N and 15000N.
http://hypertextbook.com/facts/2001/EmilyAccamando.shtml

Hitting a ball with just 10N will never get the ball to 200m/s...unless the force works on the ball for a long time, which means long contact.

In reality you will just have much less than 200 m/s. Using 10N, you won't even reach 15 m/s, I suppose.

Using silly numbers produces ridiculous results.

 

[newTonn]

Your example is just nonsense.

Usual forces acting on golf balls range between 3000N and 15000N.
http://hypertextbook.com/facts/2001/EmilyAccamando.shtml

Hitting a ball with just 10N will never get the ball to 200m/s...unless the force works on the ball for a long time, which means long contact.

In reality you will just have much less than 200 m/s. Using 10N, you won't even reach 15 m/s, I suppose.

Using silly numbers produces ridiculous results.

Yes may be practically you have to use more force.because here in my calculations,i didnt consider any other forces like gravity or air resistance.(Moreover my unit is meter/second -not miles per second).But i think for force acting on the ball (due to club),i don't have to bother about other forces.

Also at the end of article,i find somebody has calculated a mere 22.5 N of force but the author ignored it in a democratic way.(here anybody can get a doubt that wheather the force was calculated by scientist's or golf player's?- I am just joking)

Even Using some relevant figures ,i am getting some ridiculous results,here it is.

if you use the fundamental equation of net force ,F = ma; for a force of 3000N,the ball should accelerate by, a = F/m = 3000/.05 =60,000 m /sec2 or we can say this force can accelerate a shot put ball(15 kg) to 200 m/s2.

And to be frank,I never read(ignorance) anywhere in physics textbooks ,and i din't find in any equations for force about the relevence of contact period of force.
If you can help me on this ,giving me an equation for force,in which,we can use contact period as a variable,then may be i can agree with you.

 

[Cthugha]

if you use the fundamental equation of net force ,F = ma; for a force of 3000N,the ball should accelerate by, a = F/m = 3000/.05 =60,000 m /sec2 or we can say this force can accelerate a shot put ball(15 kg) to 200 m/s2.

Why not? This sounds ok to me.
An accelaration of 60 km/sec^2 means a velocity of 60 m/sec, if the contact time is 1 ms (using v=at). That sounds sensible.

An acceleration of 200 m/sec^2 concerning a 15 kg ball sounds ok as well. Using 1 ms as contact time again gives a velocity of 20 cm/s. I don't know the exact contact time in real situations, but the ms time range seems plausible.

This page provides some further reading about the physics of baseball, including citations of some reviewed publications:
http://www.kettering.edu/~drussell/bats-new/impulse.htm

 

[K.J.Healey]

Really i am confused by by your last statement.Any how i think you are explaining that ball was in contact with the club until it reach 0.05m (in your example).
if you assume that the club exerts an average force of 160N(instead of 3200N) on the ball,the club will be in contact with the ball until it reach 1 m.This i feel has no meaning.

Let us look at the problem in another way.
Let the weight of ball remains same-0.05 kg.
Let the club exerts a force of 10 kg m/s2.
So acceleration a =f/m = 10/.05 = 200 m/s2.
since a = (v-0)/t ; Final velocity v= 200m/s;
now kinetic energy = 0.5 (0.05) * 200^2 =1000 kg m2/s2
Now d = W/F = 1000/ 200 = 5 meter (does it make sense that the club was in contact with the ball for 5m?)
Please check .i am not sure about my calculations

Note:Logic and concept of the problem is Not mine-it belongs to Ariste.
Approach and new figures are the only contribution from my side

You did that calculation correct, just bad numbers.
Lets change that force the club exerts to 6000N which is more reasonable.
Why? Well imagine the deformation the ball goes through. What sort of "weight" would be needed in a static world to get it to deform that much? You couldn't put 10N of force on it and expect it to deform. It takes quite a bit.

So with 6k as the new number.
"
Let the weight of ball remains same-0.05 kg.
Let the club exerts a force of 6000 kg m/s2.
So acceleration a =f/m = 6000/.05 = 120,000 m/s2.
since a = (v-0)/t ; Final velocity v= 200m/s;
now kinetic energy = 0.5 (0.05) * 200^2 =1000 kg m2/s2
Now d = W/F = 1000/ 120,000 = 8.3mm (does it make sense that the club was in contact with the ball for 8.3mm?)
"
Yes, with a realistic force the distances becmoe much more realistic for the displacement of the ball during acceleration.
The time would then be
dt = dv/a
dt = 200 / (6000/0.05) = 200/120000 = 0.00166 seconds

So it accelerates to 200m/s in 0.0016 seconds over a distance of 8.3mm

You'll see that the more force you give, for a given final velocity, the less distance the force can be applied. This makes sense because the object accelerates only when it has a force, so its very "touchy" where a huge force has to be applied for only a split second for it to accelerate to that speed.

 

[K.J.Healey]

This page provides some further reading about the physics of baseball, including citations of some reviewed publications:
http://www.kettering.edu/~drussell/bats-new/impulse.htm

Haha, I remember all of that. I went to KU and worked/learned with/from Dr. Russell, though I was never big into acoustics.

 

[Ariste]

Really i am confused by by your last statement.Any how i think you are explaining that ball was in contact with the club until it reach 0.05m (in your example).
if you assume that the club exerts an average force of 160N(instead of 3200N) on the ball,the club will be in contact with the ball until it reach 1 m.This i feel has no meaning.

Let us look at the problem in another way.
Let the weight of ball remains same-0.05 kg.
Let the club exerts a force of 10 kg m/s2.
So acceleration a =f/m = 10/.05 = 200 m/s2.
since a = (v-0)/t ; Final velocity v= 200m/s;
now kinetic energy = 0.5 (0.05) * 200^2 =1000 kg m2/s2
Now d = W/F = 1000/ 200 = 5 meter (does it make sense that the club was in contact with the ball for 5m?)
Please check .i am not sure about my calculations

Note:Logic and concept of the problem is Not mine-it belongs to Ariste.
Approach and new figures are the only contribution from my side

No, it doesn't, but you chose terrible numbers. A club will never exert merely 10N.

Honestly, though, the numbers are irrelevant. You can see that, choosing ANY number that you like, the ball is in contact with the club for a finite displacement. I have no idea what you are trying to argue anymore.

 

[newTonn]

Why not? This sounds ok to me.
An accelaration of 60 km/sec^2 means a velocity of 60 m/sec, if the contact time is 1 ms (using v=at). That sounds sensible.

An acceleration of 200 m/sec^2 concerning a 15 kg ball sounds ok as well. Using 1 ms as contact time again gives a velocity of 20 cm/s. I don't know the exact contact time in real situations, but the ms time range seems plausible.

This page provides some further reading about the physics of baseball, including citations of some reviewed publications:
http://www.kettering.edu/~drussell/bats-new/impulse.htm

Please note that you are calculating velocity per millisecond( 60meter/millisecond),when you are using 1 millisecond contact time,so the velocity is as i calculated 60,000 meter/sec.So it makes no sense to me.

 

[newTonn]

You did that calculation correct, just bad numbers.
Lets change that force the club exerts to 6000N which is more reasonable.
Why? Well imagine the deformation the ball goes through. What sort of "weight" would be needed in a static world to get it to deform that much? You couldn't put 10N of force on it and expect it to deform. It takes quite a bit.

So with 6k as the new number.
"
Let the weight of ball remains same-0.05 kg.
Let the club exerts a force of 6000 kg m/s2.
So acceleration a =f/m = 6000/.05 = 120,000 m/s2.
since a = (v-0)/t ; Final velocity v= 200m/s;now kinetic energy = 0.5 (0.05) * 200^2 =1000 kg m2/s2
Now d = W/F = 1000/ 120,000 = 8.3mm (does it make sense that the club was in contact with the ball for 8.3mm?)
"
Yes, with a realistic force the distances becmoe much more realistic for the displacement of the ball during acceleration.
The time would then be
dt = dv/a
dt = 200 / (6000/0.05) = 200/120000 = 0.00166 seconds

So it accelerates to 200m/s in 0.0016 seconds over a distance of 8.3mm

You'll see that the more force you give, for a given final velocity, the less distance the force can be applied. This makes sense because the object accelerates only when it has a force, so its very "touchy" where a huge force has to be applied for only a split second for it to accelerate to that speed.

In your calculation (significant areas bolded by me) is incorrect.

since acceleration is 120000 m/s2 final velocity will be 120000 m/s

Substituting this will give a contact of 600 m which is impossible.

 

[Ariste]

In your calculation (significant areas bolded by me) is incorrect.

since acceleration is 120000 m/s2 final velocity will be 120000 m/s

Substituting this will give a contact of 600 m which is impossible.

Look, the numbers are irrelevant. They can be tweaked such that you get realistic values. The point is that there is an interval of contact and the ball is displaced while in contact with the club and thus there is a velocity while the ball is in contact with the club.

What are you trying to prove newTonn?

 

[newTonn]

Look, the numbers are irrelevant. They can be tweaked such that you get realistic values. The point is that there is an interval of contact and the ball is displaced while in contact with the club and thus there is a velocity while the ball is in contact with the club.

What are you trying to prove newTonn?

Please substantiate your argument with some numbers to get some realistic values.

 

[Cthugha]

Please note that you are calculating velocity per millisecond( 60meter/millisecond),when you are using 1 millisecond contact time,so the velocity is as i calculated 60,000 meter/sec.So it makes no sense to me.

Sorry. I can't help you, if simple math is too complicated for you.

v = a*t= 60*10^3 m/sec *10^(-3)sec= 60 m/sec.

 

[newTonn]

Sorry. I can't help you, if simple math is too complicated for you.

v = a*t= 60*10^3 m/sec *10^(-3)sec= 60 m/sec.

Yes you cannot help.
please help me on this.
a car was at rest.
car started and moved 60 meter in 1 second.
what is the velocity of car?
and please tell me what is the acceleration of the car in this time interval?
please don't tell me acceleration is 60000m/sec.

And please note that if you reduce the time interval to calculate the acceleration,you have to reduce the distance (because car reach 60 meter only at the end of 1 second).

 

[Cthugha]

Oh, come on. Please tell me, that you are just kidding and trying to fool us all.

car started and moved 60 meter in 1 second.
what is the velocity of car?
and please tell me what is the acceleration of the car in this time interval?
please don't tell me acceleration is 60000m/sec.

mean velocity: 60 m/sec, velocity after 1 sec: 120 m/sec.
The acceleration (assumed as constant): 120 m/sec^2

Calculate it yourself using s=1/2 a t^2 and v=a*t

By the way: An acceleration of 60000 m/sec^2, which lasts for 1 ms would give (almost) the same result. An acceleration of 60000000 m/sec^2, which lasts for 1 \mus as well. And so on.

And please note that if you reduce the time interval to calculate the acceleration,you have to reduce the distance (because car reach 60 meter only at the end of 1 second).

I don't see your problem. You seem to hang on to a belief, that forces have to act continuously (or at least 1 second), which is pure nonsense.
As long as an acceleration acts on the car, it speeds up. Afterwards it continues at constant speed. Do you understand that?

Is it clear that an acceleration of 60 m/sec^2 and an acceleration of 60 \mum/ms^2 are exactly the same?

Is it clear, that an acceleration of 10 m/sec^2, which acts for 1 second and an acceleration of 20 m/sec^2, which acts for 0,5 seconds lead to the same final velocity?

 

[newTonn]

Oh, come on. Please tell me, that you are just kidding and trying to fool us all.

A very big same to you.

mean velocity: 60 m/sec, velocity after 1 sec: 120 m/sec.
The acceleration (assumed as constant): 120 m/sec^2

Calculate it yourself using s=1/2 a t^2 and v=a*t

By the way: An acceleration of 60000 m/sec^2, which lasts for 1 ms would give (almost) the same result. An acceleration of 60000000 m/sec^2, which lasts for 1 \mus as well. And so on.



I don't see your problem. You seem to hang on to a belief, that forces have to act continuously (or at least 1 second), which is pure nonsense.
As long as an acceleration acts on the car, it speeds up. Afterwards it continues at constant speed. Do you understand that?

Is it clear that an acceleration of 60 m/sec^2 and an acceleration of 60 \mum/ms^2 are exactly the same?

Fantastic.What is that new unit of acceleration?

Is it clear, that an acceleration of 10 m/sec, which acts for 1 second and an acceleration of 20 m/sec, which acts for 0,5 seconds lead to the same final velocity?

Please anybody else out there?

Now i understand why you din't see any problem.

 

[Cthugha]

Fantastic.What is that new unit of acceleration?

New unit? I am just using micrometers and milliseconds.

By the way: Please quote me correctly.

In the second quote, there should be 10 m/sec^2 and 20 m/sec^2.

 

[newTonn]

New unit? I am just using micrometers and milliseconds..

Unfortunately Only 'Milli' (or micro)will be there because meter will be canceled by the meter in denominator.

By the way: Please quote me correctly.

In the second quote, there should be 10 m/sec^2 and 20 m/sec^2.

So you mean basically all the accelerations are 10m/sec^2 (we will say 20m/sec^2 if it acts only 0.5 second and we will call 40m/sec^2 if it acts 0.25 seconds only and so.on...
By the way ; i cannot quote more correctly than this

 

[Cthugha]

Unfortunately Only 'Milli' (or micro)will be there because meter will be canceled by the meter in denominator.

There is no meter in the denominator. ms is short for millisecond. It is just microseconds/millisecond^2

So you mean basically all the accelerations are 10m/sec^2 (we will say 20m/sec^2 if it acts only 0.5 second and we will call 40m/sec^2 if it acts 0.25 seconds only and so.on...

The mean accelerations averaged over 1 second are in this case always 10 m/sec^2. The instantaneous accelerations (and of course the time span, in which they act) do vary.

By the way ; i cannot quote more correctly than this

The ^2 was missing. Maybe it is a problem of the quotation function. I don't know.

 

[newTonn]

There is no meter in the denominator. ms is short for millisecond. It is just microseconds/millisecond^2.

then at least cancel one second from denominator.
And you should use same time interval for calculating accelerations.

for example, a = v/t (since initial velocity is zero)
or a = d / t^2
here you used time for velocity as 1 second
and the other t as 1 * 10^-3 seconds
if we are calculating acceleration for 10^-3 seconds ,then we have to correct it for 1 second (as you say in the case of 10 m/s^2 ;is same as telling

 

[Cthugha]

then at least cancel one second from denominator.
And you should use same time interval for calculating accelerations.

for example, a = v/t (since initial velocity is zero)
or a = d / t^2
here you used time for velocity as 1 second
and the other t as 1 * 10^-3 seconds
if we are calculating acceleration for 10^-3 seconds ,then we have to correct it for 1 second (as you say in the case of 10 m/s^2 ;is same as telling

You are again confusing the concepts of mean acceleration and instantaneous acceleration.
a= v/t is only true for constant acceleration. Otherwise that formula is not valid. Instantaneous acceleration is \frac{dv}{dt}, so you have to use the differential form.

There is no need to correct anything for 1 second, if the acceleration does not act for 1 second. You just need to know the acceleration and its time dependence. Then you get:

v= \int_{t_0}^{t_1} a(t) \ dt

So only intervals, when a is not 0, contribute to the final velocity.

In the case of the golf ball, we have a=const (let's say 5000N divided by the mass of the golf ball) for about 1 millisecond and a=0 afterwards.

Is it clear, that this situation leads to the same final velocity as an acceleration of 5N divided by the mass of the golf ball, which acts for 1 second.
Is it also clear, that the necessary contact time is much shorter in the first case (1 millisecond) than it is in the second case (1 second)?

 

[Ariste]

Please substantiate your argument with some numbers to get some realistic values.

I already did! Yet you ignore that post. This whole argument is getting ridiculous. You are trying to challenge the fundamentals of Newtonian physics, and you're just plain wrong. You've been proved wrong multiple times in this thread already, and yet you seem to just ignore good advice. I don't know what you're after here, but look man - Newton was right and you are wrong. End of story.

P.S. -

The ball is displaced while it is in contact with the club. It's only displaced by a small amount before it leaves the club, but it is definitely displaced.

Here, I'll give you some mathematical figures. Assume that the ball weighs .05kg and that it reaches a final velocity of 80 m/s. Thus the club gives the ball .5(.05)(80)^2 = 160J of kinetic energy. This means that the club did 160J of work on the ball. Let's assume that the club exerts an average force of 3200N on the ball. Using the formula W=Fd we conclude that the ball was in contact with the club for 160/3200 = .05m.

 

[K.J.Healey]

In your calculation (significant areas bolded by me) is incorrect.

since acceleration is 120000 m/s2 final velocity will be 120000 m/s

Substituting this will give a contact of 600 m which is impossible.

It is you who are not correct here.

WHY is your velocity the same as your Accel? That makes no sense.

Velocity = Acceleration * TIME
are you assuming we keep that acceleration over a WHOLE SECOND? Cause in my post I mention a specific time. (0.00166 seconds) as the contact time.
So do that V=AT calculation please with my acceleration and my time and tell me what velocity you get for the golfball.

I think you have a misunderstanding of units. a = dv/dt. If you know your acceleration (from knowing the force applied and the mass of the ball) then you have A. If you know how fast the ball is after contact, you have V. You assume the ball is at rest first.
If you have A and V, T=V/A

Change in Velocity = Acceleration * Time that object is Accelerating

 

[K.J.Healey]

I already did! Yet you ignore that post. This whole argument is getting ridiculous. You are trying to challenge the fundamentals of Newtonian physics, and you're just plain wrong. You've been proved wrong multiple times in this thread already, and yet you seem to just ignore good advice. I don't know what you're after here, but look man - Newton was right and you are wrong. End of story.

P.S. -

It doesn't sound like he's trying to challenge anything (other than our patience :) i kid, i kid) He just misunderstands a little physics, and is having a tough time changing how he views the physical world.
NewTonn, please believe what we're telling you. We're right about this, and once you realize that, you can understand WHY we're right, and then you'll agree. You have to use our definitions of physical properties (like force, acceleration, mass) if you're going to talk about them. Theres no other way. Thats what they are.

Each object has a position. X.
If that position changes constantly over some period of time, you have a velocity. V.
If that position changes NOT contantly over some period of time, that means the velocity V is constantly changing over some period of time. You then have a constant acceleration, A.
If your acceleration is in constant change over a period of time you have what some engineers refer to as "Jerk".

Momentum is a massful object traveling at some speed V. P=MV
A force is a change in momentum, dp/dt, which really is a change in velocity times the mass. A change in velocity is acceleration. So a force becomes a mass in acceleration.

If you apply a force over a period of time, you have a term called Impulse. Impulse = Force*Time Applied

If you apply a foce over a distance, you have performed Work, W = Force*Distance Applied

During the Golfball example, there is an IMPULSE on the ball. That is, a force acting on the ball over some period of time.
When I say "over" i do not mean "divided by" but rather multiplied by, so that the longer you apply the force, or the greater the force, the more the Impulse.

This impulse would then be N*Seconds, or N*milliseconds, or any Force Unit * Time Unit.

As for these units, really that works with any term.

Velocity is Distance Unit divided by Time Unit. It does not matter which units you chose so long as you convert it during operations to a common unit with respect to other velocities.

Acceleration is Velocity Unit divided by Time Unit.
For accel you could even have something as crazy as:
m/(micros*s)
meters per microsecond per second.
Its just a measure of change over some period of time.

 

[asamaid1]

The thread is about the conservation of energy, but i don't find anything about energy or conservation, Newton, pls state ur idea clearly, what is ur objection? What is bugging u about these three laws of Newton?

 

[K.J.Healey]

If you read early on he had the misconception that there was energy being done for an object with a constant velocity because some force had to keep it going.

 

[newTonn]

If you apply a force over a period of time, you have a term called Impulse. Impulse = Force*Time Applied


This impulse would then be N*Seconds, or N*milliseconds, or any Force Unit * Time Unit.

Thank you for your descriptive post.Still if your patience is not lost.

My dis agreement starts with the above portion of your post.

Let us see the right hand side of equation.

=Force * Time applied

= Mass * acceleration * Time applied

let us see the units of this

= (Kg) * (m/s^2) * s

= kg m /s

This is momentum.

Ultimately you are using impulse(mometum) in the place of Force (at least in case of the ball).

Is there something wrong or not?

 

[newTonn]

Change in Velocity = Acceleration * Time that object is Accelerating

Please change your equation as
rate of Change in velocity = Acceleration

 

[Doc Al]

My dis agreement starts with the above portion of your post.

Let us see the right hand side of equation.

=Force * Time applied

= Mass * acceleration * Time applied

let us see the units of this

= (Kg) * (m/s^2) * s

= kg m /s

This is momentum.

What "disagreement"? Healey01 was describing impulse, which equals the change in momentum. Of course it has units of momentum.

Ultimately you are using impulse(mometum) in the place of Force (at least in case of the ball).

Is there something wrong or not?

Sure there's something wrong: You desperately need to sit down with a basic physics text and learn something about Newton's laws, impulse, and momentum. (I gave you links on Newton's laws; here's one on impulse: http://hyperphysics.phy-astr.gsu.edu/hbase/impulse.html" )

This thread has gone on long enough. Despite the valiant efforts of many, this thread is just going in circles.