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Chapter 8: Problem 7

Force of a Golf Swing. \(A 0.0450-\mathrm{kg}\) golf ball initially at rest is given a speed of 25.0 \(\mathrm{m} / \mathrm{s}\) when a club strikes. If the club and ball are in contact for \(2.00 \mathrm{ms},\) what average force acts on the ball? Is the effect of the ball's weight during the time of contact significant? Why or why not?

Short Answer

Expert verified
The average force on the ball is 562.5 N. The weight is insignificant compared to the force from the swing.

Step by step solution

01

Identify the Given Values

Identify the mass of the golf ball as \( m = 0.0450 \, \text{kg} \), the final speed \( v = 25.0 \, \text{m/s} \), and the time of contact \( t = 2.00 \, \text{ms} = 2.00 \times 10^{-3} \, \text{s} \).
02

Determine the Change in Velocity

As the golf ball is initially at rest, its initial velocity \( u = 0 \). Thus, the change in velocity \( \Delta v = v - u = 25.0 \, \text{m/s} \).
03

Calculate the Acceleration

Using the formula for acceleration, \( a = \frac{\Delta v}{t} \). Substitute \( \Delta v = 25.0 \, \text{m/s} \) and \( t = 2.00 \times 10^{-3} \, \text{s} \): \[ a = \frac{25.0}{2.00 \times 10^{-3}} = 12500 \, \text{m/s}^2 \].
04

Calculate the Average Force

Use the formula for force, \( F = ma \), where \( m = 0.0450 \, \text{kg} \) and \( a = 12500 \, \text{m/s}^2 \): \[ F = 0.0450 \times 12500 = 562.5 \, \text{N} \].
05

Consider the Ball's Weight Effect

Compare the gravitational force on the ball, \( F_g = mg \), where \( g = 9.81 \, \text{m/s}^2 \): \[ F_g = 0.0450 \times 9.81 = 0.44145 \, \text{N} \]. Since \( 0.44145 \, \text{N} \) is negligible compared to \( 562.5 \, \text{N} \), the weight is insignificant during contact.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Average force
Average force is an important concept in dynamics, which helps us understand the effect of a force over a short duration. When a force is applied to an object, it usually doesn鈥檛 remain constant over time. Instead, we see a variation in magnitude. To make things simpler, we calculate the average force, which gives us a useful approximation of the overall impact during the time of contact.
In the exercise, the average force was calculated when a golf club hits a ball. We used the equation for force:
  • Average Force (\( F \)) = Mass (\( m \)) x Acceleration (\( a \))
We used the mass of the golf ball and the acceleration it experienced to find that the average force was 562.5 Newtons. The average force essentially tells us how hard the club hit the ball, giving an overview of the motion factor during that short contact time.
Constant acceleration
In many physics problems, especially those involving motion, constant acceleration plays a critical role. It assumes that acceleration does not change over time.In the context of the golf ball problem, after determining the ball speeds up to 25 m/s in 2 milliseconds, we calculated acceleration using the formula:
  • Acceleration (\( a \)) = Change in velocity (\( \Delta v \)) / Time (\( t \))
We found that the acceleration was a massive 12500 m/s虏. This large value happens due to the very short duration the force acts upon the ball, showing how swiftly the velocity changes when struck by the club. Constant acceleration simplifies calculations since the kinematic equations can be straightforwardly applied, assuming uniform motion.
Momentum
Momentum is a key concept in dynamics representing the quantity of motion an object has. It depends on two factors: mass and velocity. In formula terms:
  • Momentum (\( p \)) = Mass (\( m \)) x Velocity (\( v \))
When the club strikes, it changes the momentum of the ball significantly. Initially at rest, the ball's momentum is zero. But once struck, it gains a velocity of 25 m/s, boosting its momentum. This change in momentum is directly related to the force applied and the time duration, described by the impulse-momentum theorem:
  • Impulse = Change in momentum (\( \Delta p \))
The impulse provided by the club corresponds to the product of average force and contact time. In the problem, this impulse led to a fast-moving golf ball, showing how forces exert over even tiny times can hugely influence momentum.

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Most popular questions from this chapter

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