/*! 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 21 High-speed stroboscopic photogra... [FREE SOLUTION] | 91Ó°ÊÓ

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High-speed stroboscopic photographs show that the head of a \(200-\mathrm{g}\) golf club is traveling at \(55 \mathrm{~m} / \mathrm{s}\) just before it strikes a 46 -g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at \(40 \mathrm{~m} / \mathrm{s}\). Find the speed of the golf ball just after impact.

Short Answer

Expert verified
The speed of the golf ball just after the impact is 65.22 m/s

Step by step solution

01

Determine initial momentum

By using the formula of momentum \(P = mv\), calculate the initial momentum, which is the sum of the momentum of the golf ball and the golf club before the impact. Given that the golf ball is initially at rest, it contributes nothing to the initial momentum. Therefore, the initial momentum is \(P_{initial} = (0.2kg * 55m/s) = 11kg*m/s\)
02

Determine final momentum

The final momentum is the sum of the momentum of the golf ball and the golf club after the impact. It is expressed as \(P_{final} = (0.2kg * 40m/s) + (0.046kg * V_b)\). \(V_b\) is the velocity of the golf ball we are trying to find.
03

Apply the conservation of momentum principle

As per the conservation law, the total momentum before the collision would be equal to the total momentum after the collision. In terms of the problem at hand, it can be expressed as: \(P_{initial} = P_{final}\). When you substitute from the previous steps, you get \(11 kg*m/s = 8 kg*m/s + 0.046kg * V_b\). Now, solve for \(V_b\)
04

Calculate the final speed of the golf ball

Take the equation from step 3 and solve for \(V_b\). This gives \(V_b = (11kg*m/s - 8kg*m/s) / 0.046kg = 65.22 m/s\). So the speed of the golf ball just after impact is 65.22 m/s

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Key Concepts

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

Momentum in Collisions
Momentum, a key concept in collisions, is a measure of the amount of motion an object possesses. It's determined by an object's mass and velocity, described by the formula: \( p = mv \), where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity.
In collisions, the total momentum of the objects involved is an essential factor in determining the outcome. When two objects collide, as in the case of our golf club and golf ball, their momentums before and after impact are crucial to understanding the motion that results from the collision.
Here, before the collision, the stationary golf ball has no momentum, and the moving golf club has a significant amount, calculated by its mass and velocity. After the impact, the momentum of each object will have changed, but the key is that the total system momentum will remain constant if there are no external forces acting on it, a principle known as the conservation of momentum.
Linear Momentum
Linear momentum is momentum that occurs in a straight line. In physics, we often deal with linear motion and hence linear momentum, particularly when analyzing impacts between objects such as balls or vehicles. It's linear because the movement is along a line, which can be horizontal, vertical, or tilted at any angle as long as it's straight.
In the example given, the golf club head is traveling in a linear fashion toward the golf ball, which leads us to calculate the linear momentum. While computing, recall that linear momentum is a vector quantity, meaning it carries both magnitude and direction. This is why the direction of motion post-collision is also taken into account when assessing the momentum change.
Physics Problem Solving
Physics problem solving often involves a step-by-step approach to break down complex scenarios into manageable calculations. The key is to identify the relevant principles, such as the conservation of momentum, and apply known formulas.
In solving our golf ball problem, firstly, we figured out the initial momentum. Then we expressed the final momentum, keeping in mind to incorporate the unknowns such as the golf ball's velocity after impact. Followed by applying the conservation of momentum principle to equate and solve for these unknowns.To enhance the understanding of the problem, it's important to visualize the situation. Imagine the club striking the ball and understand that their interactions should respect the momentum conservation law. Recognizing the relation between mass, velocity, and momentum is vital. Furthermore, practice helps in reinforcing the approach and developing the ability to quickly identify the right physical concepts to apply.
Inelastic Collision
An inelastic collision occurs when the colliding objects do not bounce away from each other completely, but instead, some kinetic energy is converted to other forms, like sound, heat, or deformation. However, even in inelastic collisions, the law of conservation of momentum always holds true.
In our golf example, we are not told whether the collision is perfectly elastic or inelastic. What's essential is to recognize that, regardless of the type of collision, the total momentum before impact will be equal to the total momentum after impact.Although we calculate the speed of the golf ball post-collision using momentum conservation, the details of energy transformation are not needed. However, in the real world, the actual collision would likely result in some loss of kinetic energy to sound and the deformation of the ball and, hence, would be considered inelastic. Understanding this concept highlights that momentum and energy are distinct entities; momentum is conserved in all closed-system collisions, but kinetic energy is only conserved in perfectly elastic ones.

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Most popular questions from this chapter

GP A space probe, initially at rest, undergoes an internal mechanical malfunction and breaks into three pieces. One piece of mass \(m_{1}=48.0 \mathrm{~kg}\) travels in the positive \(x\) -direction at \(12.0 \mathrm{~m} / \mathrm{s}\), and a second piece of mass \(m_{2}=62.0 \mathrm{~kg}\) travels in the \(x y\) -plane at an angle of \(105^{\circ}\) at \(15.0 \mathrm{~m} / \mathrm{s}\). The third piece has mass \(m_{\mathrm{g}}=112 \mathrm{~kg}\). (a) Sketch a diagram of the situation, labeling the different masses and their velocities. (b) Write the general expression for conservation of momentum in the \(x\) - and \(y\) -directions in terms of \(m_{1}, m_{2}, m_{3}, v_{1}, v_{2}\), and \(v_{3}\) and the sines and cosines of the angles, taking \(\theta\) to be the unknown angle. (c) Calculate the final \(x\) -components of the momenta of \(m_{1}\) and \(m_{2}\). (d) Calculate the final \(y\) -components of the momenta of \(m_{1}\) and \(m_{2} .\) (e) Substitute the known momentum components into the general equations of momentum for the \(x\) - and \(y\) -directions, along with the known mass \(m_{3} .\) (f) Solve the two momentum equations for \(v_{3} \cos \theta\) and \(v_{3} \sin \theta\), respectively, and use the identity \(\cos ^{2} \theta+\sin ^{2} \theta=1\) to obtain \(v_{3}\). (g) Divide the equation for \(v_{3} \sin \theta\) by that for \(v_{3} \cos \theta\) to obtain \(\tan \theta\), then obtain the angle by taking the inverse tangent of both sides. (h) In general, would three such pieces necessarily have to move in the same plane? Why?

An archer shoots an arrow toward a \(300-\mathrm{g}\) target that is sliding in her direction at a speed of \(2.50 \mathrm{~m} / \mathrm{s}\) on a smooth, slippery surface. The \(22.5-\mathrm{g}\) arrow is shot with a speed of \(35.0 \mathrm{~m} / \mathrm{s}\) and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target?

A \(25.0-g\) object moving to the right at \(20.0 \mathrm{~cm} / \mathrm{s}\) overtakes and collides elastically with a \(10.0-\mathrm{g}\) object moving in the same direction at \(15.0 \mathrm{~cm} / \mathrm{s}\). Find the velocity of each object after the collision.

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