B Understanding Moving Object Formulas: F=ma, Ke=1/2mv2, and p=mv Explained

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The discussion focuses on three key physics formulas related to moving objects: F=ma (force), Ke=1/2mv² (kinetic energy), and p=mv (momentum). The user seeks clarity on how these formulas apply to practical scenarios, particularly in baseball, and how to calculate force when a bat strikes a ball. It is explained that force is not the primary concern; instead, the coefficient of restitution, which measures energy retention after a collision, is more relevant. Additionally, acceleration (a) is defined as the change in velocity over time, and understanding impulse—force multiplied by the duration of a collision—is crucial for analyzing impacts. The user expresses gratitude for the insights, indicating a better understanding of the concepts discussed.
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I am a beginner. First post. My question regards 3 formulas that relate to moving objects and I don't understand their differences. F = ma. Ke = 1/2mv2. And, p = mv.
They all relate to a mass that is moving. How do they differ?
 
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One of them relates an object's acceleration to the force applied on it, another gives the kinetic energy of a moving object, and the third is the momentum of a moving object.
 
axmls said:
One of them relates an object's acceleration to the force applied on it, another gives the kinetic energy of a moving object, and the third is the momentum of a moving object.

Thank you. I understand that much. What I don't understand is how they are used. For instance; I am a baseball fan and I am interested in the basic physics of bat and/or ball movements and force. If a ball is moving at 90mph and weighs 5 ounces it has 20250 of Kinetic energy. It has 450 of momentum. How are these used to determine Force? If at all. And where does the "a" in F= ma come from?
 
beginner952 said:
Thank you. I understand that much. What I don't understand is how they are used. For instance; I am a baseball fan and I am interested in the basic physics of bat and/or ball movements and force. If a ball is moving at 90mph and weighs 5 ounces it has 20250 of Kinetic energy. It has 450 of momentum. How are these used to determine Force? If at all. And where does the "a" in F= ma come from?

What I'm asking is how do I determine the force that will be applied to a stationary ball sitting on a tee, if the bat has 20250 of Ke and 450 of momentum?
 
beginner952 said:
What I'm asking is how do I determine the force that will be applied to a stationary ball sitting on a tee, if the bat has 20250 of Ke and 450 of momentum?
You cannot. And, in the usual case, the force is irrelevant. What you care about is the coefficient of restitution.
 
jbriggs444 said:
You cannot. And, in the usual case, the force is irrelevant. What you care about is the coefficient of restitution.
OK. How do I calculate that? And, on the show Sports Science, and in books I've read, they talk about the ball hitting a wall with "x" amount of force. What are they talking about, and how did they determine that?
 
beginner952 said:
OK. How do I calculate that? And, on the show Sports Science, and in books I've read, they talk about the ball hitting a wall with "x" amount of force. What are they talking about, and how did they determine that?
Sports Science is a joke. Force is not the almighty be all and end all of measurements that it is made out to be in such shows. It takes no particular strength and skill to generate multiple tons of force with a hammer and an anvil -- though the resulting force will be brief.

The coefficient of restitution is determined by experiment. You bounce a ball off a golf club head at right angles and see what fraction of the initial energy is present after the rebound. This will usually be reasonably constant over a range of collisions. Given the impact velocity, head mass, impact angle and the coefficient of restitution, you can compute how the ball will move after the collision. You never have to know the force of impact.
 
jbriggs444 said:
Sports Science is a joke. Force is not the almighty be all and end all of measurements that it is made out to be in such shows.

The coefficient of restitution is determined by experiment. You bounce a ball off a golf club head at right angles and see what fraction of the initial energy is present after the rebound. This will usually be reasonably constant over a range of collisions. Given the impact velocity, head mass, impact angle and the coefficient of restitution, you can compute how the ball will move after the collision. You never have to know the force of impact.
Ok, thank you, I'm learning. So, when does F = ma apply? And how is "a" determined? If the ball is traveling at a constant rate of 90mph is "a" = 90? I'm having difficulty understanding F = ma.
 
beginner952 said:
Ok, thank you, I'm learning. So, when does F = ma apply? And how is "a" determined? If the ball is traveling at a constant rate of 90mph is "a" = 90? I'm having difficulty understanding F = ma.

The a is the acceleration of the ball for the duration that the force is applied. That is, if you hit a ball with a bat, while the bat is pushing the ball, a force is applied, and while the force is applied, the ball accelerates. The ball stops accelerating when the force is no longer applied. An acceleration is a change in velocity, so the force increases the velocity of the ball (accelerates it), and when the ball flies off the bat, it does so at a constant velocity, since the force is no longer acting on it.
 
  • #10
axmls said:
The a is the acceleration of the ball for the duration that the force is applied. That is, if you hit a ball with a bat, while the bat is pushing the ball, a force is applied, and while the force is applied, the ball accelerates. The ball stops accelerating when the force is no longer applied. An acceleration is a change in velocity, so the force increases the velocity of the ball (accelerates it), and when the ball flies off the bat, it does so at a constant velocity, since the force is no longer acting on it.
I'm confused. The kinetic energy in the moving bat applies a force to the ball, which makes the ball accelerate. But, my question is; how much force is the bat applying to the ball? How does one calculate how much force is needed to make a given mass accelerate, or gain momentum? If the ball weighs 5 ounces how much force is needed to make it move? 5 times "a" = F. I don't know what "a" is, so how do I get F?
 
  • #11
beginner952 said:
I'm confused. The kinetic energy in the moving bat applies a force to the ball, which makes the ball accelerate. But, my question is; how much force is the bat applying to the ball? How does one calculate how much force is needed to make a given mass accelerate, or gain momentum? If the ball weighs 5 ounces how much force is needed to make it move? 5 times "a" = F. I don't know what "a" is, so how do I get F?
You also need to know that acceleration ≠ velocity. Acceleration is the change in velocity per unit amount of time.

If something is traveling at constant velocity, then its acceleration = 0, by definition.
 
  • #12
beginner952 said:
I'm confused. The kinetic energy in the moving bat applies a force to the ball, which makes the ball accelerate. But, my question is; how much force is the bat applying to the ball? How does one calculate how much force is needed to make a given mass accelerate, or gain momentum? If the ball weighs 5 ounces how much force is needed to make it move? 5 times "a" = F. I don't know what "a" is, so how do I get F?
"a" acceleration -- the rate of change of velocity over time. If you have a change in velocity that occurs over a brief time interval then the acceleration will be correspondingly large. The shorter the interval, the larger the acceleration. Accordingly, the shorter the interval the larger the force.

It is difficult to measure very large forces that are applied over very short intervals. By comparison, it is easy to measure the resulting change in velocity. The important thing about a collision is how much "impulse" is transferred. Impulse is [roughly speaking] the product of the force of the collision multiplied by the duration of the collision. That is, in turn, equal to the change in momentum of each of the two participants in the collision.

If both bat and ball are very hard (e.g. steel on steel) the collision may be over in a very short time and involve a very large but brief force. If you coat both with rubber, the collision will last longer, and the force will be lower, even though the same impulse may be delivered and the same ball speed may result.
 
  • #13
I want to thank everyone who responded. Between these responses and doing some reading of older posts on the subject, I think I finally understand.
 
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