/*! 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 3 A golfer, driving a golf ball of... [FREE SOLUTION] | 91Ó°ÊÓ

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A golfer, driving a golf ball off the tee, gives the ball a velocity of \(+38 \mathrm{~m} / \mathrm{s}\). The mass of the ball is \(0.045 \mathrm{~kg},\) and the duration of the impact with the golf club is \(3.0 \times 10^{-3} \mathrm{~s}\). (a) What is the change in momentum of the ball? (b) Determine the average force applied to the ball by the club.

Short Answer

Expert verified
Change in momentum is 1.71 kg·m/s. The average force is 570 N.

Step by step solution

01

Understand the Basic Concepts

Before we dive into calculations, it's important to recognize that the change in momentum of an object is given by the formula \( \Delta p = m \cdot \Delta v \), where \( m \) is the mass and \( \Delta v \) is the change in velocity. In this situation, the initial velocity (\( v_i \)) of the ball is 0 (at rest), and the final velocity (\( v_f \)) is \( +38 \mathrm{~m/s} \).
02

Calculate Change in Momentum

Using the formula for change in momentum, \( \Delta p = m \cdot (v_f - v_i) \), substitute in the values: \( m = 0.045 \mathrm{~kg} \), \( v_i = 0 \mathrm{~m/s} \), and \( v_f = +38 \mathrm{~m/s} \). Thus, \( \Delta p = 0.045 \cdot (38 - 0) = 1.71 \mathrm{~kg \cdot m/s} \).
03

Apply the Impulse-Momentum Theorem

The impulse-momentum theorem states that the impulse (\( J \)) is equal to the change in momentum of an object, \( J = \Delta p \). We found \( \Delta p \) to be \( 1.71 \mathrm{~kg \cdot m/s} \) in the previous step. Thus, the impulse applied to the ball is \( 1.71 \mathrm{~N \cdot s} \).
04

Determine Average Force

To find the average force (\( F_{avg} \)) applied to the ball, use the formula for impulse: \( J = F_{avg} \cdot \Delta t \). Rearrange to get \( F_{avg} = \frac{J}{\Delta t} \). Substituting the known values \( J = 1.71 \mathrm{~N \cdot s} \) and \( \Delta t = 3.0 \times 10^{-3} \mathrm{~s} \), you find \( F_{avg} = \frac{1.71}{3.0 \times 10^{-3}} = 570 \mathrm{~N} \).

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Key Concepts

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

Change in Momentum
When a golfer strikes a golf ball off the tee, an instantaneous transformation occurs known as the change in momentum. This concept is pivotal in understanding how objects move and interact in the physical world. Momentum, a product of mass and velocity, is expressed mathematically as \( p = m \cdot v \). The change in momentum, \( \Delta p \), occurs when the velocity of an object alters. This change is critical and is calculated using the formula: \( \Delta p = m \cdot (v_f - v_i) \).
Here, the ball initially rests, resulting in an initial velocity, \( v_i \), of zero. Once struck, it acquires a velocity, \( v_f \), of \(+38 \mathrm{~m/s}\).
Using the given values and the mass of the golf ball as \( 0.045 \mathrm{~kg} \), the change in momentum is calculated as:
\[ \Delta p = 0.045 \times (38 - 0) = 1.71 \mathrm{~kg \cdot m/s} \]
Essentially, this tells us that the ball's momentum has increased by \( 1.71 \mathrm{~kg \cdot m/s} \) as a result of the applied force, setting it into motion.
Average Force Calculation
Average force is a measure of how much force is applied to an object over a given time. In this context, it involves calculating the average force exerted by the golf club on the ball during the short time they are in contact. We use the impulse-momentum theorem to find this force.
This theorem states that the impulse, \( J \), equals the change in momentum of the object; thus, \( J = \Delta p \).
The formula for impulse is \( J = F_{\text{avg}} \cdot \Delta t \), where \( F_{\text{avg}} \) represents average force and \( \Delta t \) represents time duration. To solve for \( F_{\text{avg}} \), rearrange the formula as:
\[ F_{\text{avg}} = \frac{J}{\Delta t} \]
Substituting the values, \( J = 1.71 \mathrm{~N \cdot s} \) and \( \Delta t = 3.0 \times 10^{-3} \mathrm{~s} \), the calculation results in:
\[ F_{\text{avg}} = \frac{1.71}{3.0 \times 10^{-3}} = 570 \mathrm{~N} \]
This indicates that a considerable average force of 570 Newtons was exerted by the club over a very short period, showcasing the significant impact it has on the ball's motion.
Physics Problem-Solving
Physics problem-solving involves applying theoretical concepts to real-world scenarios in a structured manner. It's not merely about finding numbers but understanding the physical phenomena behind those numbers. This particular physics problem is a formulaic adventure that requires knowledge of fundamental concepts such as momentum, force, and time.
Here's a step-by-step practical approach:
  • Start by identifying what you know—values for mass, velocities, and impact time are crucial in problems like this.
  • Recognize the formulas needed—momentum change and impulse-momentum relationship are key.
  • Perform intermediate calculations slowly and double-check—as seen in momentum change calculation.
  • Ensure to apply the right units and conversion factors, keeping the problem consistent like \( \mathrm{kg \cdot m/s} \) for momentum.
  • Calculate carefully using step-by-step methods to avoid mistakes, especially with decimals and conversions like millisecond to second.
Understanding and doing the problem in pieces ensures you grasp each concept before moving to the next. Practicing physics with this methodology builds confidence and problem-solving skills that are invaluable in academics and practical physics application.

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Most popular questions from this chapter

A wagon is coasting at a speed \(v_{\mathrm{A}}\) along a straight and level road. When ten percent of the wagon's mass is thrown off the wagon, parallel to the ground and in the forward direction, the wagon is brought to a halt. If the direction in which this mass is thrown is exactly reversed, but the speed of this mass relative to the wagon remains the same, the wagon accelerates to a new speed \(v_{\mathrm{B}}\). Calculate the ratio \(v_{\mathrm{B}} / v_{\mathrm{A}}\).

Consult Interactive Solution \(7.9\) at for a review of problem-solving skills that are involved in this problem. A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The incident water stream has a velocity of \(+16.0 \mathrm{~m} / \mathrm{s}\), while the exiting water stream has a velocity of \(-16.0 \mathrm{~m} / \mathrm{s}\). The mass of water per second that strikes the blade is \(30.0 \mathrm{~kg} / \mathrm{s}\). Find the magnitude of the average force exerted on the water by the blade.

A baseball \((m=149 \mathrm{~g}\) ) approaches a bat horizontally at a speed of \(40.2 \mathrm{~m} / \mathrm{s}(90 \mathrm{mi} / \mathrm{h})\) and is hit straight back at a speed of \(45.6 \mathrm{~m} / \mathrm{s}(102 \mathrm{mi} / \mathrm{h})\). If the ball is in contact with the bat for a time of \(1.10 \mathrm{~ms}\), what is the average force exerted on the ball by the bat? Neglect the weight of the ball, since it is so much less than the force of the bat. Choose the direction of the incoming ball as the positive direction.

A student \((m=63 \mathrm{~kg})\) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of \(0.010 \mathrm{~s}\). The average force exerted on him by the ground is \(+18000 \mathrm{~N}\), where the upward direction is taken to be the positive direction. From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground.

Each of these problems consists of Concept Questions followed by a related quantitative Problem. The Concept Questions involve little or no mathematics. They focus on the concepts with which the problems deal. Recognizing the concepts is the essential initial step in any problem-solving technique. Concept Questions Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. (a) Is it possible that the two-object system has a final total momentum of zero after the collision? (b) Roughly, what is the direction of the final total momentum of the two-object system after the collision? Problem Object A has a mass of \(m_{A}=17.0 \mathrm{~kg}\) and an initial velocity of \(\overrightarrow{\mathbf{v}}_{0 \mathrm{~A}}=8.00 \mathrm{~m} / \mathrm{s},\) due east. Object \(\mathrm{B},\) however, has a mass of \(m_{\mathrm{B}}=29.0 \mathrm{~kg}\) and \(\mathrm{an}\) initial velocity of \(\overrightarrow{\mathbf{v}}_{0 \mathrm{~B}}=5.00 \mathrm{~m} / \mathrm{s},\) due north. Find the magnitude and direction of the total momentum of the two-object system after the collision. Make sure that your answers are consistent with your answers to the Concept Questions.

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