/*! 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 10 A bat strikes a \(0.145 \mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

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A bat strikes a \(0.145 \mathrm{~kg}\) baseball. Just before impact. the ball is traveling horizontally to the right at \(40.0 \mathrm{~m} / \mathrm{s} ;\) when it lcaves the hat, the ball is traveling to the left at an angle of \(30^{\circ}\) above horizontal with a speed of \(52.0 \mathrm{~m} / \mathrm{s}\). If the ball and hat are in contact for \(1.75 \mathrm{~ms},\) find the horicontal and vertical components of the average force on the ball.

Short Answer

Expert verified
The horizontal and vertical components of the average force on the ball are \(-7014 \mathrm{~N}\) and \(2153 \mathrm{~N}\) respectively.

Step by step solution

01

Finding velocity after impact

The ball is moving at an angle of \(30^{\circ}\) above the horizontal level after being hit which indicates use of vector components. The velocity of the ball after impact in horizontal direction (\(v_{fx}\)) will be \(52 \mathrm{~m/s} \times \cos(30^{\circ}) = 45 \mathrm{~m/s}\) towards the left (considering left as negative). The velocity of the ball after impact in vertical direction (\(v_{fy}\)) will be \(52 \mathrm{~m/s} \times \sin(30^{\circ}) = 26 \mathrm{~m/s}\) upwards (considering upwards as positive).
02

Finding change in velocity

The horizontal change in velocity (\( \Delta v_x\)) is the final horizontal speed minus the initial horizontal speed, which is \(-45 \mathrm{~m/s} - 40.0 \mathrm{~m/s} = -85 \mathrm{~m/s}\). The vertical change in velocity (\(\Delta v_y\)) is the final vertical speed minus the initial vertical speed, which is \(26 \mathrm{~m/s} - 0 = 26 \mathrm{~m/s}\), since the initial vertical speed of the ball was 0.
03

Calculate average force

By using the second law of motion (force = mass × acceleration), calculate the horizontal and vertical components of the average force. The average force in the horizontal direction (\(f_{avx}\)) is the mass of the ball times the horizontal change in speed divided by the contact time, which is \(0.145 \mathrm{~kg} \times -85 \mathrm{~m/s} ÷ 1.75 \mathrm{~ms} = -7014 \mathrm{~N}\). Similarly, the average vertical force (\(f_{avy}\)) is the mass of the ball times the vertical change in speed divided by the contact time, which is \(0.145 \mathrm{~kg} \times 26 \mathrm{~m/s} ÷ 1.75 \mathrm{~ms} = 2153 \mathrm{~N}\).
04

Present the results

The horizontal component of the average force on the ball is \(-7014 \mathrm{~N}\) and the vertical component of the average force on the ball is \(2153 \mathrm{~N}\). Negative sign for the horizontal force component shows that it is directed towards the left.

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Key Concepts

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

Force Components
When we discuss forces in physics, it is important to break them down into understandable parts, called force components. This is especially crucial when dealing with objects moving at an angle, as seen in our baseball example. In this exercise, the baseball moves at a slanted angle after being hit, requiring us to separate its motion into horizontal and vertical components.
- **Horizontal Component**: After the ball is hit, it moves at an angle of 30 degrees. To find the horizontal component of its velocity, apply the cosine function: \[ v_{fx} = 52 \times \cos(30^{\circ}) = 45 \, \text{m/s} \] This indicates the velocity heading leftwards, hence considered negative in direction.- **Vertical Component**: Similarly, the vertical component uses the sine function: \[ v_{fy} = 52 \times \sin(30^{\circ}) = 26 \, \text{m/s} \] Indicating an upward direction. These components simplify the calculation of forces acting on the ball, allowing us to analyze each direction separately.
Change in Velocity
Understanding how velocity changes is key to solving problems involving motion. In this scenario, we observe both horizontal and vertical changes in the baseball's velocity. Let's break it down:- **Horizontal Change**: The initial horizontal speed is 40 m/s to the right. After being struck, it becomes 45 m/s to the left. The change in the horizontal velocity is computed as: \[ \Delta v_x = -45 \, \text{m/s} - 40 \, \text{m/s} = -85 \, \text{m/s} \] The negative sign signifies a reversal in direction.- **Vertical Change**: Initially, the vertical velocity is zero (since there's no vertical motion before impact). After impact, it becomes 26 m/s upwards: \[ \Delta v_y = 26 \, \text{m/s} - 0 = 26 \, \text{m/s} \] This positive change reflects an upward movement introduced post-impact.These changes are crucial when applying Newton's Second Law to find the forces exerted.
Vector Analysis
Vector analysis brings clarity to situations involving direction and magnitude, like the baseball's motion after the bat hit. Through the step-by-step process, we apply vector analysis to decompose the ball's strike.1. **Components as Vectors**: By separating the motion into horizontal and vertical components, we essentially treat the problem through vector addition. Each component represents a vector that is perpendicular, simplifying calculations.2. **Force Calculation via Vectors**: Newton's Second Law relates force, mass, and acceleration using vectors. Here, the changes in velocity (horizontal and vertical) determine the resultant force using the formula: - Horizontal force: \[ f_{avx} = \frac{0.145 \, \text{kg} \times (-85) \, \text{m/s}}{1.75 \, \text{ms}} = -7014 \, \text{N} \] - Vertical force: \[ f_{avy} = \frac{0.145 \, \text{kg} \times 26 \, \text{m/s}}{1.75 \, \text{ms}} = 2153 \, \text{N} \] These equations calculate the force components, which show the magnitude and direction of force reacting within the ball.By considering vectors, this allows comprehensive understanding of both direction and influence during interactions, enhancing our ability to map real-world scenarios mathematically.

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Most popular questions from this chapter

Two ice skaters, Daniel (mass \(65.0 \mathrm{~kg}\) ) and Rebecca (mass \(45.0 \mathrm{~kg}\) ) are practicing. Danicl stops to tic his shoclace and, while at rest, is struck by Rebecca, who is moving at \(13.0 \mathrm{~m} / \mathrm{s}\) before she collides with him. After the collision, Rchecca has a velocity of magnitude \(8.00 \mathrm{~m} / \mathrm{s}\) at an angle of \(53.1^{\circ}\) from her initial direction, Both skaters move on the fric. tionless, horizontal surface of the rink. (a) What are the magnitude and direction of Danicl 's velocity after the collision? (b) What is the change in total kinetic energy of the two skaters as a result of the collision?

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You are standing on a large shect of frictionless ice and holding a large rock. In orver to get off the ice. you throw the rock so it has velocity \(12.0 \mathrm{~m} / \mathrm{s}\) relutive to the eurth at an angle of \(35.0^{\circ}\) above the horizontal. If your mass is \(70.0 \mathrm{~kg}\) and the rock's mass is \(3.00 \mathrm{~kg}\), what is your speed after you throw the rock? (See Discussion Question \(Q 8.7 .)\)

An apple with mass \(M\) is hanging at rest from the lower end of a light vertical rope. A dart of mass \(M / 4\) is shot vertically upward, strikes the bottom of the apple, and remains cmbedded in it. If the specd of the dart is \(\mathrm{c}_{0}\) just before it strikes the apple, how high docs the apple move upward because of its collision with the dart?

Two identical \(0.900 \mathrm{~kg}\) masses are pressed against opposite ends of a light spring of force constant \(1.75 \mathrm{~N} / \mathrm{cm}\), compressing the spring by \(20.0 \mathrm{~cm}\) from its normal length. Find the speed of each mass when it has moved free of the spring on a frictionless, horizontal table.

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