/*! 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 12 A bat strikes a 0.145 -kg baseba... [FREE SOLUTION] | 91Ó°ÊÓ

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A bat strikes a 0.145 -kg baseball. Just before impact, the ball is traveling horizontally to the right at \(50.0 \mathrm{m} / \mathrm{s},\) and it leaves the bat traveling to the left at an angle of \(30^{\circ}\) above horizontal with a speed of 65.0 \(\mathrm{m} / \mathrm{s}\) . If the ball and bat are in contact for 1.75 \(\mathrm{ms}\) find the horizontal and vertical components of the average force on the ball.

Short Answer

Expert verified
Horizontal Force = -6521 N, Vertical Force = 5399 N.

Step by step solution

01

Identify Components of Initial and Final Velocities

Initially, the ball is moving horizontally at 50 m/s to the right, so the initial velocity component is \( v_{i,x} = 50 \, \text{m/s} \) and \( v_{i,y} = 0 \, \text{m/s} \). After impact, the ball is moving at 65 m/s at a 30-degree angle to the left. Calculate the components using trigonometry: \( v_{f,x} = -65 \cos(30^{\circ}) \) and \( v_{f,y} = 65 \sin(30^{\circ}) \).
02

Calculate Change in Velocity

Compute the change in each velocity component: \( \Delta v_x = v_{f,x} - v_{i,x} = -65 \cos(30^{\circ}) - 50 \) and \( \Delta v_y = v_{f,y} = 65 \sin(30^{\circ}) - 0 \).
03

Calculate Acceleration Components

Use the change in velocity to find the acceleration components. Given the contact time \( t = 1.75 \, \text{ms} = 1.75 \times 10^{-3} \, \text{s} \), \( a_x = \frac{\Delta v_x}{t} \) and \( a_y = \frac{\Delta v_y}{t} \).
04

Calculate Force Components

Use Newton's Second Law to find the force components. The horizontal force is \( F_x = m a_x \) and the vertical force is \( F_y = m a_y \), where \( m = 0.145 \, \text{kg} \) is the mass of the ball.

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a cornerstone of physics. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. Let's break it down in the context of our problem:

- **The Formula**: The law is expressed as \( F = m a \), where \( F \) is the force applied, \( m \) is the mass of the object, and \( a \) is the acceleration. In the baseball problem, we calculate the horizontal and vertical components of force using this law.
- **Application**: We use the mass of the baseball (0.145 kg) and the calculated acceleration components to find the force. For instance, if the acceleration in the x-direction is known, the horizontal force \( F_x \) is calculated by \( F_x = m \, a_x \). Similar calculations are done for the vertical direction.

This powerful law helps understand how different forces affect the motion of objects and is fundamental in solving complex problems involving multiple force components.
Trigonometry in Physics
Trigonometry plays a pivotal role in physics, especially when solving problems with angled velocities. Here's how it's utilized:

- **Velocity Components**: For an object moving at an angle, trigonometry helps break the velocity into horizontal and vertical components using sine and cosine functions.
- **Our Problem**: In our baseball problem, the ball leaves at an angle of \(30^{\circ}\) from horizontal. We use trigonometry to find: - The horizontal component \( v_{f,x} = -65 \cos(30^{\circ}) \) - The vertical component \( v_{f,y} = 65 \sin(30^{\circ}) \)

Understanding these components aids in the analysis of motion and force directions, crucial for determining the overall effect on the ball's trajectory.
Velocity Components
When dealing with motion, breaking down velocity into components simplifies the problem. Velocity can often be described in terms of direction and speed:

- **Horizontal and Vertical Components**: Initial velocities are given directly, but final velocities at an angle require decomposition using trigonometry.
- **Baseball Context**: Initially, the ball moves horizontally at 50 m/s. After being hit, the velocity splits into two: the horizontal component \( v_{f,x} \) and the vertical component \( v_{f,y} \). This division helps understand the impact of the collision on the ball's path.
- **Calculations**: The change in each component is vital for finding acceleration and subsequently, the forces.

Recognizing velocity components is essential in predicting and analyzing the motion, crucial in both theoretical understanding and practical applications.
Impulse and Momentum
Impulse and momentum are related concepts that explain the motion resulting from a force applied over time.

- **Momentum**: It's the product of an object's mass and velocity. In our baseball problem, momentum initially is simply its mass times initial velocity.
- **Impulse**: It's the change in momentum, reflecting the force applied over a time interval. The impulse is calculated by finding the change in velocity components, which in turn help determine the force with Newton's Second Law.
- **Importance**: For the baseball, the impulse explains how the bat's force influences the velocity and direction of the ball after impact.

Having a strong grasp of these concepts allows for deeper insights into how forces change motion and how objects react to collisions.
Forces in Two Dimensions
In physics, examining forces in two dimensions means analyzing forces that have both horizontal and vertical components.

- **Multidimensional Analysis**: Unlike simple cases with only one dimension, real-world problems often involve forces acting in various directions.
- **In Practice**: The baseball's force problem requires understanding how both the horizontal and vertical forces interact. By treating these components individually, we simplify calculations while gaining a complete picture of the ball's motion after being struck.
- **Balancing Forces**: Ultimately, understanding these components allows us to better predict and manipulate real life scenarios, such as aiming a ball towards a specific direction.

Comprehending force combinations and their directional components is crucial in physics, aiding in the full understanding of objects in motion.

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Most popular questions from this chapter

A fireworks rocket is fired vertically upward. At its maximum height of \(80.0 \mathrm{m},\) it explodes and breaks into two pieces: one with mass 1.40 \(\mathrm{kg}\) and the other with mass 0.28 \(\mathrm{kg} .\) In the explosion, 860 \(\mathrm{J}\) of chemical energy is converted to kinetic energy of the two fragments. (a) What is the speed of each fragment just after the explosion? (b) It is observed that the two fragments hit the ground at the same time. What is the distance between the points on the ground where they land? Assume that the ground is level and air resistance can be ignored.

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One 110 -kg football lineman is running to the right at 2.75 \(\mathrm{m} / \mathrm{s}\) while another 125 -kg lineman is running directly toward him at 2.60 \(\mathrm{m} / \mathrm{s}\) . What are (a) the magnitude and direction of the net momentum of these two athletes, and (b) their total kinetic energy?

In a shipping company distribution center, an open cart of mass 50.0 \(\mathrm{kg}\) is rolling to the left at a speed of 5.00 \(\mathrm{m} / \mathrm{s}\) (Fig. P8.95). You can ignore friction between the cart and the floor. A 15.0 -kg package slides down a chute that is inclined at \(37^{\circ}\) from the horizontal and leaves the end of the chute with a speed of 3.00 \(\mathrm{m} / \mathrm{s}\) . The package lands in the cart and they roll off together. If the lower end roll off together. If the lower end of the chute is a vertical distance of 4.00 \(\mathrm{m}\) above the bottom of the cart, what are (a) the speed of the package just before it lands in the cart and (b) the final speed of the cart?

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