/*! 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 15 High-speed stroboscopic photogra... [FREE SOLUTION] | 91Ó°ÊÓ

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High-speed stroboscopic photographs show that the head of a golf club of mass \(200 \mathrm{g}\) is traveling at \(55.0 \mathrm{m} / \mathrm{s}\) just before it strikes a 46.0 -g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at \(40.0 \mathrm{m} / \mathrm{s} .\) Find the speed of the golf ball just after impact.

Short Answer

Expert verified
The speed of the golf ball just after impact is approximately 65.2 meters per second.

Step by step solution

01

Understand the meaning of momentum conservation

Linear momentum conservation follows the principle that the total momentum of an isolated system remains constant if no external forces act on it. The equation for momentum is defined as mass (m) times velocity (v), so the momentum before collision should be equal to the momentum after collision.
02

Represent the problem mathematically

The problem can be represented by the conservation of momentum equation as follows: \[m_{c}v_{c0}+m_{b}v_{b0}=m_{c}v_{cf}+m_{b}v_{bf}\] where \(m_{c}\) and \(v_{c0}\) denote the mass and initial velocity of the club, \(m_{b}\) and \(v_{b0}\) denote the mass and initial velocity of the ball, \(v_{cf}\) and \(v_{bf}\) denote the final velocities of the club and ball respectively.
03

Plug in known values into the equation

Knowing the initial velocities (\(v_{c0}=55.0m/s\), \(v_{b0}=0m/s\)) and the final velocity of the club (\(v_{cf}=40.0m/s\)) as well as the masses of the club (\(m_{c}=200g\) or 0.2 kg) and ball (\(m_{b}=46.0g\) or 0.046kg), we can substitute these into the equation as follows: \[0.2kg*55.0m/s+0.046kg*0m/s=0.2kg*40.0m/s+0.046kg*v_{bf}\]
04

Solve for the unknown final velocity of the ball

Now solve for \(v_{bf}\) by first simplifying the equation to get: \[11kg*m/s=8kg*m/s+0.046kg*v_{bf}\] Then, subtract \(8kg*m/s\) from both sides to get \[3kg*m/s=0.046kg*v_{bf}\], Further simplifying gives: \[v_{bf} = \frac{3kg*m/s}{0.046kg}\]

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Key Concepts

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

Linear Momentum
When we speak of linear momentum in physics, we're referring to the product of an object's mass and velocity. It's a vector quantity, meaning it has both magnitude and direction. For a moving object, the linear momentum (\( p \)) is represented mathematically as \( p = m \times v \), where \( m \) signifies mass and \( v \) is velocity. In the example of the golf club and ball, we calculate the momentum of each to understand the situation before and after the collision.

To better understand the concept in an applied context, think of momentum as a measure of 'how much motion' an object has. Heavier objects moving rapidly have a lot of momentum, which illustrates why they're harder to stop compared to lighter, slower-moving ones. The momentum of the golf club before impact is significant due to its relatively high velocity, and after it hits the golf ball, part of this momentum is transferred to the ball.
Collisions in Physics
Collisions in physics are events where two or more bodies exert forces on each other for a relatively short time. These can be either elastic, where objects bounce off each other without losing kinetic energy, or inelastic, where they might stick together or deform, losing some kinetic energy in the process. Yet, what remains constant in both types of collisions is the conservation of momentum.

The collision between the golf club and the ball is an excellent example to study. It's a common real-life application of collision concepts, where momentum is being transferred and shared between the colliding bodies. Understanding collisions helps us predict the outcome of such interactions, which is crucial in a wide range of fields from sports to automobile safety.
Momentum Conservation Principle
The momentum conservation principle is a fundamental concept in physics stating that if no external force acts on an isolated system, the total momentum of the system remains constant. This principle is grounded in Newton's third law, which posits that for every action, there’s an equal and opposite reaction.

Applying this principle to the golf club and ball situation, the 'system' consists of both the club and the ball. Since there are no external forces acting on the club-ball system at the moment of collision, the initial momentum (club's momentum) must equal the final momentum (combined momentum of both club and ball). This conservation underpins the calculations in the step-by-step solution that students are trying to understand.
Solving Momentum Problems
Solving momentum problems involves employing the principles of momentum conservation to find unknown quantities. These problems typically involve a few steps: understanding the physical situation, translating it into mathematical equations by identifying knowns and unknowns, and then solving these equations.

In the context of the above example, the problem is solved by setting up the conservation of momentum equation, substituting the known values, and then algebraically solving for the unknown value, which is the velocity of the golf ball after the collision. The ability to break down the problem into these steps and execute them is a critical skill when dealing with momentum problems in physics. This methodical approach helps in achieving a clear and logical pathway to the solution, making it easier for students to follow and comprehend the concepts at play.

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Most popular questions from this chapter

A golf ball \((m=46.0 \mathrm{g})\) is struck with a force that makes an angle of \(45.0^{\circ}\) with the horizontal. The ball lands \(200 \mathrm{m}\) away on a flat fairway. If the golf club and ball are in contact for \(7.00 \mathrm{ms},\) what is the average force of impact? (Neglect air resistance.)

Two blocks of masses \(M\) and \(3 M\) are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them (Fig. P9.4). A cord initially holding the blocks together is burned; after this, the block of mass \(3 M\) moves to the right with a speed of \(2.00 \mathrm{m} / \mathrm{s}\) (a) What is the speed of the block of mass \(M ?\) (b) Find the original elastic potential energy in the spring if \(M=0.350 \mathrm{kg}\)

A 10.0 -g bullet is fired into a stationary block of wood \((m=\) \(5.00 \mathrm{kg}) .\) The relative motion of the bullet stops inside the block. The speed of the bullet-plus-wood combination immediately after the collision is \(0.600 \mathrm{m} / \mathrm{s} .\) What was the original speed of the bullet?

A friend claims that, as long as he has his seat belt on, he can hold on to a 12.0 -kg child in a \(60.0 \mathrm{mi} / \mathrm{h}\) head-on collision with a brick wall in which the car passenger compartment comes to a stop in 0.050 0 s. Show that the violent force during the collision will tear the child from his arms. A child should always be in a toddler seat secured with a seat belt in the back seat of a car.

Most of us know intuitively that in a head-on collision between a large dump truck and a subcompact car, you are better off being in the truck than in the car. Why is this? Many people imagine that the collision force exerted on the car is much greater than that experienced by the truck. To substantiate this view, they point out that the car is crushed, whereas the truck is only dented. This idea of unequal forces, of course, is false. Newton's third law tells us that both objects experience forces of the same magnitude. The truck suffers less damage because it is made of stronger metal. But what about the two drivers? Do they experience the same forces? To answer this question, suppose that each vehicle is initially moving at \(8.00 \mathrm{m} / \mathrm{s}\) and that they undergo a perfectly inelastic head-on collision. Each driver has mass 80.0 kg. Including the drivers, the total vehicle masses are \(800 \mathrm{kg}\) for the car and 4000 kg for the truck. If the collision time is \(0.120 \mathrm{s},\) what force does the seatbelt exert on each driver?

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