The General Description of the Tennis Strokes. For math oriented players only!

[sureshs]

Originally Posted by toly

Let’s analyze collision of the resting ball with moving racquet.

The ball during impact stays on the racquet's string bed just about T = 4 msec. Assume, it leaves the racquet with speed equal to Vb=100 mph (160 km/h). The mass of the ball is mb = 57 grams.

The ball kinetic energy before impact = 0.The ball kinetic energy after impact Eb =0.5*mb*Vb ^2.

In physics, power is the rate at which energy is converted. Let’s calculate power P which produced the conversion of the ball kinetic energy from zero to Eb.

It is P = Eb /T = 20hp.
Where, hp is horsepower.

The 20 horses are working very hard during 4 msec to provide the conversion of the ball’s kinetic energy.

Should you apply during impact your own additional force to help them to do this work? Even if you add your maximum force during collision the result would be negligible. We have to collect kinetic energy during our forward swing. That’s why there is no short swing in professional tennis when pros hit hard strokes.

So, when we scrutinize real collision of racquet and ball we can ignore the additional force completely!!!

Do you think the racket moved by itself under gravity and its horizontal component was enough to keep it moving to strike the ball? No. The person is exerting a force on the racket.

[toly]

Originally Posted by sureshs

That is what I always thought and have done problems with it, till I came up with the example above (inelastic collision) when I was trying to see if toly was wrong. Can you tell me the fallacy in my argument (#32)?

The law of conservation of momentum

If the velocities of the particles are U1 and U2 before the interaction, and afterwards they are V1 and V2, then

M1U1 + M2U2 = M1V1 + M2V2 (1)

This law holds no matter how complicated the force is between particles.

M1 and M2 are constants, but U1, U2, V1, and V2 are not. They depend on an arbitrarily chosen coordinate system. If we change coordinate system, we have to rewrite equation (1), but physics of collision process would be the same.

[toly]

Originally Posted by sureshs

Do you think the racket moved by itself under gravity and its horizontal component was enough to keep it moving to strike the ball? No. The person is exerting a force on the racket.

Yes, you can and maybe should exert force on the racquet to keep proper racquet orientation, but this force practically doesn’t affect collision, because 20 horses are much, much … stronger than you.

[FrisbeeFool]

Toly. What is your math background? You're using modeling techniques from an introductory algebra based physics class. These models are designed to introduce basic concepts of physics to someone with no math background.

Have you taken any physics courses that use modeling techniques from calculus? Have you taken any engineering courses? Are you self-taught?

The situation you are attempting to describe is complex and has many variables. If you relate these variables using simplified algebra, you won't be accurately modeling the situation.

We haven't even gotten into wind/air resistance. It would be a huge factor in this situation. I still have my notes from the first week of my ordinary differential equations class. If you want I can scan them and post them. They model a simplified view of wind resistance, in order to introduce the topic. There are many ways to model air resistance with increasing complexity and sophistication. Since I only studied math at the undergraduate level, I don't know them. I'm guessing you don't either.

Quit wasting people's time. This thread isn't for math oriented players at all. You're trying to use pre-algebra and algebra to model a dynamic, constantly changing situation. Hmmm, what discipline studies rates of change???? Calculus

The two objects you're examining aren't moving in space uneffected by other forces. The have constant mass, but they don't have constant velocity. They don't have constant acceleration!! Racket head acceleration is not constant!! It's not like two billiard balls with no outside forces at the time of the collision. The force exerted by the person on the racket is effecting the collision.

I can probably scan old notes from my differential equations courses if you want. Just go to google look up differential equations, and look up simplified models of motion. You're trying to use introductory techniques to describe a dynamic event? It won't work!

[sureshs]

Originally Posted by toly

Yes, you can and maybe should exert force on the racquet to keep proper racquet orientation, but this force practically doesn’t affect collision, because 20 horses are much, much … stronger than you.

I really don't follow. You seem to be saying gravity is doing all the work.

In your example of ball leaving the racket at 100 mph, it could also be a serve, not a forehand. Are you saying that gravity is what provides the force on the serve when in reality the racquet is almost moving up or slightly down at contact? If I just hang the racket and let it fall on the ball toss, a 100 mph serve will be produced?

Doesn't seem right to me. If the above is not right, then the argument that there is no force exerted by the player on the forehand is also wrong.

[sureshs]

Do a simple experiment. Hang a racket with a thin string and also hang a ball in front of it. Take the racket back like a pendulum to the same distance that Fed takes back his racket, and let go, and position the ball so that it gets hit.

See if the ball leaves at 80 mph.

[sureshs]

Originally Posted by toly

The law of conservation of momentum

If the velocities of the particles are U1 and U2 before the interaction, and afterwards they are V1 and V2, then

M1U1 + M2U2 = M1V1 + M2V2 (1)

This law holds no matter how complicated the force is between particles.

M1 and M2 are constants, but U1, U2, V1, and V2 are not. They depend on an arbitrarily chosen coordinate system. If we change coordinate system, we have to rewrite equation (1), but physics of collision process would be the same.

That is OK, but I wanted someone to point out what was wrong in my argument

[toly]

Originally Posted by FrisbeeFool

Toly. What is your math background? You're using modeling techniques from an introductory algebra based physics class. These models are designed to introduce basic concepts of physics to someone with no math background.

Have you taken any physics courses that use modeling techniques from calculus? Have you taken any engineering courses? Are you self-taught?

I have several degrees - in math, radio electronics, computer science, and radio physics. I worked as one of the leader designer of radio locators for more than 20 years. I had few master and PhD students who tried to complicate any assigned work, unfortunately for them they didn’t get job. If you want to contribute some sophisticated and useful analysis to this thread, you are welcome.

[FrisbeeFool]

My bad, sounds like you know your stuff. I guess you're just introducing the algebraic approach to start things off?? If that's really your background, you have to realize racket head acceleration won't be constant? Why don't you explain the different types of collisions and the different models?

[bhupaes]

Originally Posted by FrisbeeFool

You could do that, but the relative speed between the racket and ball would be constantly changing because racket head acceleration is not a constant, due to the variable force being exerted by the hitter with respect to time. You couldn't calculate it using algebra, you'd need to use vector calculus, a differential equation, or some other sophisticated technique that would more accurately model the situation.

I believe the objective is to model the collision in a very small timeframe that brackets the collision. How the racquet and ball achieved their speeds and direction would be immaterial for the purposes of this exercise. I believe this is what toly is trying to do, in order to simplify the model/calculations used, and I don't see anything wrong with it...

[sureshs]

Originally Posted by bhupaes

I believe the objective is to model the collision in a very small timeframe that brackets the collision. How the racquet and ball achieved their speeds and direction would be immaterial for the purposes of this exercise. I believe this is what toly is trying to do, in order to simplify the model/calculations used, and I don't see anything wrong with it...

There is nothing wrong, but there is nothing useful either. If you remember all the collision problems you have solved, do you ever recall any mention of acceleration? No, it was always, u1, v1, u2, v2 etc. Why? Because no force was acting on those objects. When a racket head hits a ball, it is accelerating just prior to impact. So, conservation of momentum cannot be applied.

[bhupaes]

Originally Posted by sureshs

Mass 2 is resting sets one of the variables to 0. That is not what you are saying. You are saying relative velocities are all that matter. That is like replacing vx1 by vx1-vx2.

Say you have a small mass m moving horizontally to the right with speed v and colliding completely inelastically with a larger mass M. The two together will then move to the right with speed mv/(m+M). If m was stationary and M moved to the left with speed v, the two will move to the left with speed Mv/(m+M).

Going directly to your post #32...

This is a different problem, isn't it? The original problem assumed an elastic collision to simplify the calculation, which would be true if the energy lost in the collision is small (as I think it is). But in any case, even with totally inelastic collision, the relative speed between the two masses after the collision is zero in both of the cases you have mentioned. Not sure what I am missing...

[bhupaes]

Originally Posted by sureshs

There is nothing wrong, but there is nothing useful either. If you remember all the collision problems you have solved, do you ever recall any mention of acceleration? No, it was always, u1, v1, u2, v2 etc. Why? Because no force was acting on those objects. When a racket head hits a ball, it is accelerating just prior to impact. So, conservation of momentum cannot be applied.

I understand, and you are right in the large picture. But we are applying the rules differentially, as when solving a problem in numerical analysis, by dividing time into very small segments.

[sureshs]

Originally Posted by bhupaes

Going directly to your post #32...

This is a different problem, isn't it? The original problem assumed an elastic collision to simplify the calculation, which would be true if the energy lost in the collision is small (as I think it is). But in any case, even with totally inelastic collision, the relative speed between the two masses after the collision is zero in both of the cases you have mentioned. Not sure what I am missing...

I am asking why the difference in speeds if analyzed with one at rest, then with the other at rest. If m is moving to the right with v towards M, isn't it the same as M moving to the left with v towards m? Then why is there a difference in the speed of the combined mass after the collision? (direction will of course be different).

[sureshs]

Originally Posted by bhupaes

I understand, and you are right in the large picture. But we are applying the rules differentially, as when solving a problem in numerical analysis, by dividing time into very small segments.

Then you are saying the velocity is constant almost at collision, even if the racket was accelerating before? Then toly's point would be that the person applies no force, and of course he doesn't, because we just made the acceleration 0. That would be true for anything! If we say we will go by a boxer's hand's final speed only, we can claim that the boxer did not do anything at the instant of impact!

What use is that?

[bhupaes]

Originally Posted by sureshs

I am asking why the difference in speeds if analyzed with one at rest, then with the other at rest. If m is moving to the right with v towards M, isn't it the same as M moving to the left with v towards m? Then why is there a difference in the speed of the combined mass after the collision? (direction will of course be different).

In a given coordinate system or frame of reference, of course there is a difference between a small mass and a big mass moving with a given velocity. I don't think anybody is disputing that. As I recall, this argument started because someone wanted to measure the angle of reflection of a ball when it hits a racquet, and that is a relative entity which only depends on the relative movement between the racquet face and the ball.

[bhupaes]

Originally Posted by sureshs

Then you are saying the velocity is constant almost at collision, even if the racket was accelerating before? Then toly's point would be that the person applies no force, and of course he doesn't, because we just made the acceleration 0. That would be true for anything! If we say we will go by a boxer's hand's final speed only, we can claim that the boxer did not do anything at the instant of impact!

What use is that?

The racquet could be accelerating all the way to impact. But the actual impact lasts only a few milliseconds. As toly pointed out, how much energy could the force have contributed in that time? IMO, negligible for the purposes of this calculation. If someone can prove this to be wrong, I will accept it - I don't have any means other than the above logic to prove my assertion.

[toly]

I analyzed Federer inside out hard FH on the APAS System http://www.youtube.com/watch?v=pPLmCqGIotM

There are data about his arm speed and acceleration around the wrist during forward swing.



The picture demonstrates that Federer is able to increase acceleration/force of the arm from frame #1 to #9. The magnitude of the force declined after frame #9. Around impact, frame #24, acceleration decreased more than 70%.

So, he is not able to accelerate the arm significantly near impact. But, this is very hard FH and without doubt he is trying to produce maximum acceleration/force.

Question: What is wrong with Federer?
The answer will be tomorrow.

[toly]

Originally Posted by bhupaes

The racquet could be accelerating all the way to impact. But the actual impact lasts only a few milliseconds. As toly pointed out, how much energy could the force have contributed in that time? IMO, negligible for the purposes of this calculation. If someone can prove this to be wrong, I will accept it - I don't have any means other than the above logic to prove my assertion.

IMO you are right. Thanks for clarification!!!

[toly]

Originally Posted by sureshs

Mass 2 is resting sets one of the variables to 0. That is not what you are saying. You are saying relative velocities are all that matter. That is like replacing vx1 by vx1-vx2.

Say you have a small mass m moving horizontally to the right with speed v and colliding completely inelastically with a larger mass M. The two together will then move to the right with speed mv/(m+M). If m was stationary and M moved to the left with speed v, the two will move to the left with speed Mv/(m+M).

1. In first case momentum before collision was mv. After collision momentum is

(M+m)(mv/(m+M))=mv. So, mv=mv and everything is fine with law of conservation of momentum.

2. In second case momentum before collision was Mv. After collision momentum is

(M+m)(Mv/(m+M))=Mv. Again everything is fine and Mv=Mv.