/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 140 A \(1.62-\) oz golf ball is hit ... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(1.62-\) oz golf ball is hit with a golf club and leaves it with a velocity of \(100 \mathrm{mi} / \mathrm{h}\). We assume that for \(0 \leq t \leq t_{0},\) where \(t_{0}\) is the duration of the impact, the magnitude \(F\) of the force exerted on the ball can be expressed as \(F=F_{m} \sin \left(\pi t / t_{0}\right) .\) Knowing that \(t_{0}=0.5 \mathrm{ms}\), determine the maximum value \(F_{m}\) of the force exerted on the ball.

Short Answer

Expert verified
The maximum force exerted on the ball is approximately 6490 N.

Step by step solution

01

Convert Units

First, convert the velocity of the golf ball from miles per hour to meters per second. The conversion factors are: 1 mile = 1609.34 meters and 1 hour = 3600 seconds.Convert 100 mi/h: \[100 \text{ mi/h} \times \frac{1609.34 \text{ m}}{1 \text{ mile}} \times \frac{1 \text{ hour}}{3600 \text{ seconds}} = 44.70 \text{ m/s}\]
02

Calculate Mass

Convert the mass of the golf ball from ounces to kilograms. The conversion factor is: 1 ounce = 0.0283495 kilograms.Convert 1.62 oz:\[1.62 \text{ oz} \times 0.0283495 \text{ kg/oz} = 0.0459 \text{ kg}\]
03

Use Impulse-Momentum Theorem

The impulse-momentum theorem relates the impulse on an object to its change in momentum:\[F_{m} \Delta t = \Delta p = m \Delta v\] Substitute \(m = 0.0459 \text{ kg}\), \(\Delta v = 44.70 \text{ m/s}\), and \(\Delta t = 0.5 \times 10^{-3} \text{ s}\).
04

Expression for Force and Time

Given force function: \[F = F_{m} \sin\left(\frac{\pi t}{t_{0}}\right)\]The peak force occurs when \(\sin\left(\frac{\pi t}{t_{0}}\right) = 1\), meaning \(F = F_{m}\). We can use this to calculate \(F_{m}\).
05

Solve for Maximum Force \(F_m\)

Using impulse-momentum theorem:\[F_{m} \times \int_{0}^{t_0} \sin\left(\frac{\pi t}{t_0}\right) dt = \Delta p\]The integral of \(\sin\left(\frac{\pi t}{t_0}\right)\) over [0, \(t_0\)] is \(\frac{2}{\pi}t_0\). So:\[F_{m} \times \frac{2}{\pi} \times 0.5 \times 10^{-3} \text{ s} = 0.0459 \text{ kg} \times 44.70 \text{ m/s}\]Calculate for \(F_{m}\):\[F_{m} = \frac{0.0459 \times 44.70}{0.5 \times 10^{-3} \times \frac{2}{\pi}}\approx 6490 \text{ N}\]
06

Conclusion

The maximum force \(F_m\) exerted on the golf ball by the club is approximately 6490 N.

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Key Concepts

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

Impulse Calculation
The impulse-momentum theorem is a key concept in physics, which states that the change in momentum of an object is equal to the impulse applied to it. In mathematical terms, this is expressed as:
\[ F \cdot \Delta t = \Delta p \]where \( F \) is the average force applied, \( \Delta t \) is the time duration over which the force is applied, and \( \Delta p \) is the change in momentum.

To understand impulse, consider it as the effect of a force applied over a certain period of time. This concept is crucial when calculating how a force, such as a golf club's strike, affects an object like a golf ball. The impulse can be calculated by integrating the force function over the time interval of interest. Thus, knowing the force and the duration, you can determine how the velocity of the object changes, highlighting the interchange between force and motion.
Force Function Sinusoidal
In the given exercise, the force applied to the golf ball is represented by a sinusoidal function:
\[ F(t) = F_m \cdot \sin \left(\frac{\pi t}{t_0}\right) \]where \( F_m \) is the maximum force, \( t \) is the time, and \( t_0 \) is the total duration of the force application.

This sinusoidal pattern of the force is typical in many real-world situations where the force gradually increases, peaks, and then decreases. Think of a gentle push that gains intensity before slowly fading away. A sinusoidal function is useful as it smoothly transitions force values, making the calculation of the overall impact more realistic.

The maximum force, \( F_m \), is achieved when the sine factor is at its peak value of 1. This situation occurs at a certain instant during the impact. Understanding sinusoidal force helps us model real-world forces accurately and predict outcomes using mathematical analysis.
Momentum Change
Momentum is the product of an object's mass and velocity. A change in momentum, expressed as \( \Delta p \), represents how an object's motion is altered due to external forces. In the impulse-momentum theorem context, momentum change is the result of impulse applied over time:
\[ \Delta p = m \cdot \Delta v \]where \( m \) is the mass and \( \Delta v \) is the change in velocity.

In the golf ball exercise, the initial velocity is zero before the hit, making the change in velocity directly the final velocity. The momentum change can thus be calculated using the determined velocity from the units conversion section.

Recognizing how momentum changes allow us to understand the dynamics of an impact, such as how fast the golf ball will travel after being hit. Momentum conservation is a fundamental principle in both simple and complex physics problems.
Unit Conversion
Unit conversion is crucial in physics to ensure consistency and understanding of given measurements. In this exercise, several conversions are necessary:
  • Velocity from miles per hour to meters per second
  • Mass from ounces to kilograms
Understanding these conversions allows us to apply the correct units in our calculations, ensuring accuracy.

For velocity, use conversion factors like 1 mile equals 1609.34 meters and 1 hour equals 3600 seconds:
\[ 100 \text{ mi/h} \times \frac{1609.34 \text{ m}}{1 \text{ mile}} \times \frac{1 \text{ hour}}{3600 \text{ seconds}} = 44.70 \text{ m/s} \]

For mass, use the factor 1 ounce equals 0.0283495 kilograms:
\[ 1.62 \text{ oz} \times 0.0283495 \text{ kg/oz} = 0.0459 \text{ kg} \]

Once fully converted, these values can be seamlessly integrated into further calculations, making unit conversion a foundational step in problem-solving.

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Most popular questions from this chapter

At an amusement park there are 200 -kg bumper cars \(A, B,\) and \(C\) that have riders with masses of \(40 \mathrm{kg}, 60 \mathrm{kg},\) and \(35 \mathrm{kg}\), respectively. Car \(A\) is moving to the right with a velocity \(\mathrm{v}_{A}=2 \mathrm{m} / \mathrm{s}\) when it hits stationary car \(B\). The coefficient of restitution between each car is \(0.8 .\) Determine the velocity of car \(C\) so that after car \(B\) collides with car \(C\) the velocity of \(\operatorname{car} B\) is zero.

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