What is the force exerted by a golf club on a ball in a typical swing?

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Homework Help Overview

The discussion revolves around estimating the force exerted by a golf club on a ball during a typical swing, specifically focusing on the brief contact time and the resulting speed of the ball. The subject area includes concepts from mechanics, particularly relating to motion and forces.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the use of basic equations of motion, specifically questioning the application of collision formulas and conservation of momentum. There is discussion on how to calculate acceleration and its relation to force.

Discussion Status

The conversation is progressing with participants clarifying concepts related to force and acceleration. Some guidance has been provided regarding the appropriate equations to use, and there is an acknowledgment of the correct approach to calculating acceleration.

Contextual Notes

Participants are working within the constraints of a homework problem, focusing on the parameters given, such as the mass of the ball and the time of contact. There is an implicit understanding that assumptions about ideal conditions may be in play.

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Homework Statement



In a typical golf swing, the club is in contact with the ball for about .001s. If the 45g ball acquires a speed of 53 m/s, estimate the magnitude of the force exerted by the club on the ball.

Homework Equations



do i use a collision formula for this?

The Attempt at a Solution



pi=pf
p1i+p2i=p1f+p2f
vf=m1v1i/m1+m2
 
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Okay, let's step back a minute to our basic equation of motion;
F=ma
You want to solve for a force, so this would be the natural equation to use.
Now, you know enough to solve for F, you know the mass and you can work out the acceleration from what you have!

Using conservation of momentum is a handy thing to use but not in this kind of problem.
 
so acceleration is rate of change of velocity over time so would i divide 53 by .001?
 
That is correct!
 

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