# A=F/m with a plastic and real golf ball

1. Apr 30, 2010

### MichTeacher

I discussed with my students that if force remains constant and mass decreases then acceleration increases (a=F/m). I also explained that if force remains constant and mass increases, acceleration decreases.

I know that there are a number of variables that play into this next situation and I did my best to explain this to my students without going further than my present knowledge will allow. I am interested to here your take on this coming question.

If you hit a real golf ball and a plastic golf ball with a club with the same force, angle, etc. The plastic ball should have a higher initial acceleration, than the real golf ball because the force is the same but the mass is less. After both balls are hit the travel of the plastic ball will not go as far do to air resistance. The plastic ball will have less momentum (p=mv) due to its lower mass. Therefore the plastic golf will be effected more my an equal force delivered by air resistance. I could also talk about the time that the balls are in contact with the club and the longer contact and stored energy in a real golf ball due to its elasticity. Please if you have time give me your take and I will share it with the kids. I don't believe we would know everything about this situation unless we conduct a study.

2. Apr 30, 2010

### xxChrisxx

What age are you aiming this at?

I've only skim read what you put, but it all seems ok.

3. Apr 30, 2010

### MichTeacher

4. Apr 30, 2010

### Stonebridge

I think you need to be careful when saying the plastic ball has "less momentum due to its lower mass", if you have also implied that it has a greater velocity because the acceleration was greater. In this case, the two balls should have the same momentum. One has a higher mass but the other a higher velocity.
The key is the impulse from the club. This is F times time. If you argue that the force is the same for both golf balls, and the time of contact between club and ball is the same for both, you have to say that their momentum is also the same.
At that level I think students would accept that the force and time of contact could be the same - and probably is near enough!
The effect of air resistance seems ok. It's intuitive for the students that the lighter ball will be more influenced by this, and F=ma makes sense here where F is the drag force.
It's also true that the drag force depends on the velocity of the ball, and the lighter ball would start with a higher velocity according to our previous statements.

5. Apr 30, 2010

### MichTeacher

Stonebridge,

I looked up drag force and understand what you mean there.

Are you saying that since F=ma and the mass of a plastic golf ball is low so it is likely to deccelerate faster than the golf ball.

I believe what you are saying about drag force is that the faster the velocity of the object the higher the drag force. How do you calculate drag force of a projectile?

6. Apr 30, 2010

### Brilliant

Yeah that makes sense to me. If the plastic ball has a higher velocity, then it will have more drag. I also think that even if they were moving at the same velocity, and therefor experiencing the same force of drag, the plastic ball will be more affected because of its lesser mass. The drag will more rapidly slow it down. It gets complicated then because as velocity decreases so does the drag force.

To do any calculations calculus will probably be required, but you can still find the force of drag at a specific velocity with the equation: FD=1/4Av2

This model for drag only works if:
• The object's diameter is between a few milimeters and a few meters
• The object's speed is not more then a few hundred meters per second
• The object is moving through air near the earth's surface

Yeah, and the more I think about trying to calculate an actual displacement the more my head hurts. I think you have some kind of differential equation, I don't know. I've never had to do anything like that before.

7. Apr 30, 2010

### Feldoh

Well here's the math...

$$\vec{F} = mg \hat{y} - cv^2\hat{v}$$

Splitting it into x & y components:

$$m\ddot{x} = -cv_x\sqrt{\dot{x}^2+\dot{y}^2}$$

$$m\ddot{y} = -mg - cv_y\sqrt{\dot{x}^2+\dot{y}^2}$$

Where c is the coefficient for the drag force: $$\vec{F}_{drag} = -cv^2\hat{v}$$ which at standard temp and pressure is like $$~0.25 \frac{N^2 s^2}{m^4}* diameter^2$$

These equations can't be solved analytically I dont think

Edit:
I actually did it out in Mathematica using a golf ball and ping pong ball both starting with the same velocity and direction and here's what I got

http://filer.case.edu/pal25/projectile/ProjectileGolfBall.pdf [Broken] -- (x,y) in meters plot of a golf ball
http://filer.case.edu/pal25/projectile/ProjectilePingPong.pdf [Broken] -- (x,y) in meters plot of ping pong ball

Heres the mathematica file in case your interested: http://filer.case.edu/pal25/projectile/Projectile.nb [Broken]

Last edited by a moderator: May 4, 2017
8. May 1, 2010

### Stonebridge

Yes. The drag force depends on the shape and size of the object as well as the velocity. As the two balls have the same shape and size, only the velocity matters.
Whatever the drag force then is, its effect on the ball is determined by a=F/m so the lighter ball will be decelerated more.

I'm not sure how old "9th graders" are in the US, so I'm not sure how far it's possible to push this because, as others have said, the maths gets rather complicated once you introduce drag. It's usually best to consider the ideal case first, where there is no air resistance. This establishes the fundamentals involved (force mass momentum velocity acceleration) without adding too many distractions and complications; moving on the the drag force qualitatively rather than quantitatively at first.