Energy Conservation: Kinetic Energy After Collision

[rcgldr]

A rocket is moving at uniform velocity with respect to you.

If "you"'re not acceleration then that rocket with uniform velocity isn't accelerating either.

Another rocket passed you with an acceleration. What will be the force(net) acting on these rockets?

Zero on the first rocket. You'd have to know the mass of the second rocket, and if so, then force equals mass times acceleration.

I'll ingore the fact that the accelerating rocket is using fuel for it's thrust and therefore it's mass is decreasing over time as it accelerates.

 

[newTonn]

If "you"'re not acceleration then that rocket with uniform velocity isn't accelerating either.

Zero on the first rocket. You'd have to know the mass of the second rocket, and if so, then force equals mass times acceleration.

.

In the case of second rocket,
If you can use the equation force equals mass times acceleration(without knowing the time of application of thrust(force) ,it is equaly valid in the case of ball in the previous example.
So if you really mean what you say,you are confirming that a force is acting on an accelerating body,irrespective of the time of application of force(if it is accelerating).
you see the ball coming accelerating.[you din't see wheather it was hit by a club or thrown by somebody] .But you see the ball is accelerating,so of course as in the case of the second rocket,there should be a force acting on ball or not?
if still you say no ,there is no force acting (the force was acted only at the instant of club hit the ball).then you are challenging the equation or you are going to put some conditions(ie,force can be calculated only if acceleration is very quick or acceleration will be considered only at the instance of the application of force etc.etc)

 

[Doc Al]

Could anybody draw a freebody diagram of the ball ,just after the club hits the ball.

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. This all takes place in a very small--but certainly nonzero--time interval.

The freebody diagram of the ball would show:
Before the club hits: gravity acting down; normal force acting up.
As the club hits: gravity acting down; huge impulsive force of the club acting at some angle.
After the club loses contact: gravity acting down; air resistance acting in the opposite direction to its motion.

You didn't answer all my questions
let me ask it in a simple manner?
consider you are at rest or in uniform motion.
A rocket is moving at uniform velocity with respect to you.
Another rocket passed you with an acceleration.
(consider you don't know,wheather a thrust is applied on the rockets during its journey past you-you have been assingned to find out the force acting on these rockets).
What will be the force(net) acting on these rockets?
Do you use f = m*a in the case of second rocket? or do you say a force was acting on that rocket in past.
Please explain me how you will confirm,the acceleration was due to some action in past or it is attained during it was passing you.

I think you are confusing velocity with acceleration. If the rocket is accelerating now, then there's a force acting on it now. If the rocket has some velocity now, and you know it had a different velocity at some earlier point, then you can deduce that at some time in the past a force must have acted on it.

regarding golf club really accelerates quickly.it depends on how hard you hit it.forgetting all other forces,we can say ,acceleration takes a time equal to the time required for ball to reach its final velocity from zero velocity.it never can be zero.

If a net force acts for some time interval, then a change in velocity will result.

In the case of second rocket,
If you can use the equation force equals mass times acceleration(without knowing the time of application of thrust(force) ,it is equaly valid in the case of ball in the previous example.
So if you really mean what you say,you are confirming that a force is acting on an accelerating body,irrespective of the time of application of force(if it is accelerating).
you see the ball coming accelerating.[you din't see wheather it was hit by a club or thrown by somebody] .But you see the ball is accelerating,so of course as in the case of the second rocket,there should be a force acting on ball or not?

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

if still you say no ,there is no force acting (the force was acted only at the instant of club hit the ball).then you are challenging the equation or you are going to put some conditions(ie,force can be calculated only if acceleration is very quick or acceleration will be considered only at the instance of the application of force etc.etc)

Once the ball loses contact with the club, of course the club no longer exerts a force on the ball.

Again, you seem to think that by examining the speed of something you can figure out its acceleration or deduce that some force is acting. Again, it's quite simple: Is the velocity changing now? If so, then there is a force acting on it now. If the velocity is steady now, then there's no net force acting on it now.

Of course, if you wish to measure the change in velocity, you'll need to make observations over some time interval.

 

[newTonn]

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. This all takes place in a very small--but certainly nonzero--time interval.

The freebody diagram of the ball would show:
Before the club hits: gravity acting down; normal force acting up.
As the club hits: gravity acting down; huge impulsive force of the club acting at some angle.
After the club loses contact: gravity acting down; air resistance acting in the opposite direction to its motion..

of course ,it is very easy to draw free body diagram.
yes .first case resultant is zero.so no motion-agreed.
second case,of course resultant is in the direction of motion of club,fine the ball will move in that direction.
But in the third case ,you can see the direction of resultant is opposite.so where should the ball move.?
in any freebody diagram,the object should move in the direction and with a magnitude equal to the resultant of all forces.
So don't you feel something is missing?At least don't you think the motion of ball has to be considered as a force in that direction?

 

[Doc Al]

of course ,it is very easy to draw free body diagram.
yes .first case resultant is zero.so no motion-agreed.

The net force is zero means that there's no change in motion, not that there's no motion. (In this case it's true that the ball is at rest.) A force is not required to maintain velocity: Newton's 1st law.

second case,of course resultant is in the direction of motion of club,fine the ball will move in that direction.

The direction of the change in velocity will be in the direction of the net force. In this case the ball starts from rest, so it will begin moving in the direction of the net force.

But in the third case ,you can see the direction of resultant is opposite.so where should the ball move.?

What do you mean by "direction of resultant"? The club is no longer exerting a force on the ball once they break contact. But now the ball has a high speed due to the impulse provided by the club, so it will just continue moving (of course gravity and air resistance will change its velocity). Again: Newton's 1st law.

in any freebody diagram,the object should move in the direction and with a magnitude equal to the resultant of all forces.

Incorrect. The direction of the change in velocity will be in the direction of the net force, not the velocity. Simple example: Toss a ball in the air. It follows a curved path, yet gravity always acts down.

So don't you feel something is missing?

Nope.

At least don't you think the motion of ball has to be considered as a force in that direction?

Of course not.

I urge you to study Newton's laws of motion. Start here: http://hyperphysics.phy-astr.gsu.edu/hbase/newt.html"

 

[Ariste]

In the case of second rocket,
If you can use the equation force equals mass times acceleration(without knowing the time of application of thrust(force) ,it is equaly valid in the case of ball in the previous example.
So if you really mean what you say,you are confirming that a force is acting on an accelerating body,irrespective of the time of application of force(if it is accelerating).
you see the ball coming accelerating.[you din't see wheather it was hit by a club or thrown by somebody] .But you see the ball is accelerating,so of course as in the case of the second rocket,there should be a force acting on ball or not?
if still you say no ,there is no force acting (the force was acted only at the instant of club hit the ball).then you are challenging the equation or you are going to put some conditions(ie,force can be calculated only if acceleration is very quick or acceleration will be considered only at the instance of the application of force etc.etc)

Important bits bolded.

Are you implying here that the ball is accelerating in the forward direction through the air after it leaves the club? That seems to me what you're trying to say, and it's completely wrong.

Once the ball leaves the club, it has no forces acting on it besides air resistance and gravity. So if I'm hanging out in the air and see this ball fly past me, I would see that it has a substantial velocity in the forward direction and a small acceleration in the backward direction. In other words, the ball would be slowing down, not speeding up. I would not see the ball going faster and faster as it approaches me. I would see it going slower and slower. The only conclusion that I could correctly draw from this is that the net force on the ball is in the backward direction. This is the correct conclusion.

But in the third case ,you can see the direction of resultant is opposite.so where should the ball move.?
in any freebody diagram,the object should move in the direction and with a magnitude equal to the resultant of all forces.

No, no, no. The direction of the net force is opposite in this diagram, yes. But you're misinterpreting what a net force does. A body does not necessarily move in the same direction as the net force that is applied to it. It will accelerate in that direction, and, if given an adequate period of time, will eventually move in that direction. It does not immediately have to move in that direction, though.

Consider the case of a car. When a car is moving forward at a constant velocity, all of the forces acting on the car - the engine's thrust in the forward direction, rolling friction and air resistance in the backward direction, gravity in the downward direction, normal force in the upward direction - are balanced. Say this car is moving at 90mph North.

What happens when the driver of this car takes his foot off the gas and applies the brakes? Now what would the free-body diagram for this car look like? All forces would be directed South, right? But does this car immediately stop and begin moving South? No, it begins to decelerate. If, say, the brakes were applied for 2 seconds, perhaps the car would slow down to 70mph, but it would still be moving North, even though, for a certain time interval, all of the forces on the car were pushing South.

This is the same as the case of the ball. When the ball is in the air, say it is moving at 150mph North. All of the forces on the ball will be acting in the South direction, but this does not mean that the ball moves South. It simply accelerates to the South. It accelerates in the opposite direction of the its motion; in other words, it slows down. It does not immediately stop and turn around. This is directly analogous to the car situation.

What is it that you're not understanding here? Maybe it's the definition and behavior of a force? A force does not cause motion, it causes acceleration. Don't confuse the two.

 

[Danger]

Alright... enough is enough.
I'm staying 'way the hell back from any of that math stuff. Even without it, it's bloody obvious that a relatively moving body has kinetic energy. If you don't believe it, try standing in front of a bullet.
As for the cause of inertia, it's just the reluctance of mass to change its current state of motion. (And the last that I read, we don't yet know what 'causes' mass.)

edit: Wow... a tonne of responses in there while I was composing this. I'd love to read them all, but since I'm at work, I'm too drunk to follow them.

 

[jambaugh]

newTonn,
I'm going to add some points to these other good posts. You have a confused notion about force and inertia which I can't quite follow from your posts.

Inertia is purely the conservation of momentum.

Kinetic momentum (mass times velocity) is conserved in the absence of forces.

Kinetic energy (approximately half mass times squared velocity) is conserved in the absence of forces.

A force applied to an object changes its kinetic energy (Newtonian scalar) in proportion to how far the object travels in the direction of the force,
and changes its kinetic momentum (Newtonian vector) in proportion to how long it is applied.

When considering "instantaneous" forces we are ignoring the time and distance element and the magnitude of the force and only looking at these product components of change in momentum and change in kinetic energy.

$\Delta$ here means "change in", $T$ means kinetic energy and $\mathbf{P}$ means (vector) momentum.
$$\Delta KE = \mathbf{F}\cdot \Delta \mathbf{X}$$ (force constant throughout the displacement)
$$\Delta \mathbf{P} = \mathbf{F}\Delta t$$

Once you understand momentum and energy and how forces change these, then consider acceleration as change in velocity and thus how forces accelerate by changing the momentum which is proportional to the velocity.

Finally let me say that you are trying to analyze scenarios verbally without breaking down and doing the vector math. You really really really need to dig through the mathematics to see how everything fits together.

Start with two massive balls colliding in two scenarios: bouncing off each other (elastic collision where energy is conserved) and hitting and sticking to each other (inelastic collision where energy will not be conserved).

Try first one dimensional and then two dimensional scenarios.

In fact you should work out billiard ball collisions on the table, identifying where the cue must hit the object ball to make a shot, and then where the cue ball ends up hitting the bank (to see if you will scratch). This is a good practical exercise and you can test your calculations at the nearest pool hall.

To figure the angles assume the force between the balls is perpendicular to their surface at the point of contact (i.e. ignore ball-ball friction).

Remember that you can hypothesize and deduce all day but none of this has scientific meaning until you test your hypotheses with empirical experiments. The pool shot exercises will be both instructive and fun.

 

[Danger]

Oh, sure... go and get math involved again...
Nice post, James.

 

[K.J.Healey]

I think the major issue is Newtonn is picturing the golfball still accelerating FORWARD after it leaves the club. Which it isnt.

But he IS right in that , the free body diagram, the forces are not net zero.

There is a force that is due to air resistance, this pushes say, to the left if the ball was hit to the right.

There is a force downward due to gravity.

So if you took a static picture of the ball, you would say that it would move down and left, correct?

BUT the ball had an initial velocity, so that velocity WILL slow down. But remember that wind resistance is a FUNCTION of its RIGHTWARD VELOCITY. So once it slwos down to zero to the right, it won't keep going left. Theres no more force on it leftward.

As for the gravity, since it was hit UP, the gravity SLOWS it first to Zero (its peak) then it DOES move DOWN. And it would keep accelerating if it could never hit the earth.
As for your rocket example. The second rocket that is ACCELERATING past you MUST have a current thrust. Once the thrust stops, the acceleration stops. They are the same thing. You cannot thrust some rocket, stop the burners/jets, and expect it to continue accelerating. It stops accelerating the INSTANT that thrust stops, and continues at its current velocity.

Please respond with your questions.

So, remember, that "acceleration" can mean going from 100 m/s to 50 m/s. Slowing is acceleration too. Don't forget that. Accel is the change in speed, in any direction. So it has to change its current velocity.

 

[newTonn]

The net force is zero means that there's no change in motion, not that there's no motion. (In this case it's true that the ball is at rest.) A force is not required to maintain velocity: Newton's 1st law.


The direction of the change in velocity will be in the direction of the net force. In this case the ball starts from rest, so it will begin moving in the direction of the net force.

What do you mean by "direction of resultant"? The club is no longer exerting a force on the ball once they break contact. But now the ball has a high speed due to the impulse provided by the club, so it will just continue moving (of course gravity and air resistance will change its velocity). Again: Newton's 1st law.


Incorrect. The direction of the change in velocity will be in the direction of the net force, not the velocity. Simple example: Toss a ball in the air. It follows a curved path, yet gravity always acts down.

Nope.

Of course not.

I urge you to study Newton's laws of motion. Start here: http://hyperphysics.phy-astr.gsu.edu/hbase/newt.html"

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?

 

[Ariste]

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?

Why don't you reply to any of my responses? I've given you answers to all the questions you are asking, as have others, yet you refuse to acknowledge them.

I get the sense that you are more interested in being right than in being correct.

 

[newTonn]

Important bits bolded.

Are you implying here that the ball is accelerating in the forward direction through the air after it leaves the club? That seems to me what you're trying to say, and it's completely wrong.

Once the ball leaves the club, it has no forces acting on it besides air resistance and gravity. So if I'm hanging out in the air and see this ball fly past me, I would see that it has a substantial velocity in the forward direction and a small acceleration in the backward direction. In other words, the ball would be slowing down, not speeding up. I would not see the ball going faster and faster as it approaches me. I would see it going slower and slower. The only conclusion that I could correctly draw from this is that the net force on the ball is in the backward direction. This is the correct conclusion..

Again you are saying the same thing.
The ball attains its final velocity from same position and same instance.
how can it be possible? forget about all other forces exept that from club.Does the ball require a time interval to reach its final velocity?and is a displacement necessary ,to acquire a velocity?

 

[newTonn]

Why don't you reply to any of my responses? I've given you answers to all the questions you are asking, as have others, yet you refuse to acknowledge them.

I get the sense that you are more interested in being right than in being correct.

Sorry Ariste,it is 8.54 am in this part of the world.While you were sending this message,i was writting a reply for you.
Not at all,i would like to be corrected in a correct manner.

 

[Ariste]

Again you are saying the same thing.
The ball attains its final velocity from same position and same instance.
how can it be possible? forget about all other forces exept that from club.Does the ball require a time interval to reach its final velocity?and is a displacement necessary ,to acquire a velocity?

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.

 

[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.