/*! 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 43 A batter swings at a baseball (m... [FREE SOLUTION] | 91Ó°ÊÓ

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A batter swings at a baseball (mass \(0.145 \mathrm{~kg}\) ) that is moving horizontally toward him at a speed of \(40.0 \mathrm{~m} / \mathrm{s} .\) He hits a line drive with the ball moving away from him horizontally at \(50.0 \mathrm{~m} / \mathrm{s}\) just after it leaves the bat. If the bat and ball are in contact for \(8.00 \mathrm{~ms}\), what is the average force that the bat applies to the ball?

Short Answer

Expert verified
The average force that the bat applies to the ball is \(1.63 \times 10^{3} \mathrm{~N}\).

Step by step solution

01

Calculate the change in velocity

The change in velocity can be calculated using the formula \(\Delta v = v_{\mathrm{final}} - v_{\mathrm{initial}}\), where \(v_{\mathrm{final}}\) is the final velocity and \(v_{\mathrm{initial}}\) is the initial velocity. Here, the final velocity is the velocity of the ball after it leaves the bat, which is \(50.0 \mathrm{~m/s}\), and the initial velocity is the velocity of the ball before it hits the bat, which is \(40.0 \mathrm{~m/s}\) in the opposite direction. Therefore, \(\Delta v = 50.0 \mathrm{~m/s} - (-40.0 \mathrm{~m/s}) = 90.0 \mathrm{~m/s}\).
02

Calculate the acceleration

The acceleration can be calculated using the formula \(a = \Delta v/ \Delta t\), where \(\Delta t\) is the change in time. Here, \(\Delta v = 90.0 \mathrm{~m/s}\) (from Step 1), and \(\Delta t = 8.00 \times 10^{-3} \mathrm{~s}\) (convert ms to s by multiplying by \(10^{-3}\)). Therefore, \(a = 90.0 \mathrm{~m/s} / 8.00 \times 10^{-3} \mathrm{~s} = 1.125 \times 10^{4} \mathrm{~m/s}^{2}\).
03

Apply Newton's Second Law to calculate the force

The force applied to the ball by the bat can be calculated using the formula \(F = ma\), where \(m\) is the mass of the ball and \(a\) is the acceleration of the ball. Here, \(m = 0.145 \mathrm{~kg}\) (given) and \(a = 1.125 \times 10^{4} \mathrm{~m/s}^{2}\) (from Step 2). Therefore, \(F = 0.145 \mathrm{~kg} \times 1.125 \times 10^{4} \mathrm{~m/s}^{2} = 1.63 \times 10^{3} \mathrm{~N}\).

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Key Concepts

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

Average Force
Average force is a crucial concept in understanding interactions between objects, typically studied in the context of Newton’s Second Law of motion. Newton’s Second Law states that the force applied to an object is equal to the mass of the object multiplied by its acceleration, expressed as \( F = ma \). In the scenario of a bat hitting a baseball, we are interested in the average force exerted over the brief time the bat is in contact with the ball.
If you know the change in velocity and the time duration of the interaction, you can calculate the average force. The average force is not just dependent on initial and final speed but also highly influenced by how quickly that change happens. Hence,
  • Average force links mass, velocity change, and time in the interaction.
  • Helps in understanding force applied over short or extended durations.
Calculating average force provides insight into the strength and effect of interactions in various physical activities from sports to machinery.
Velocity Change
In physics, velocity change is crucial for understanding motion dynamics. It is calculated as the difference between the final and initial velocities of an object. In our exercise, we deal with a baseball's velocity change during a bat swing.
The baseball was initially moving towards the batter and changes direction after being hit, reflecting a major change in velocity. To compute this change, use the formula \( \Delta v = v_{\text{final}} - v_{\text{initial}} \). Here:
  • \(v_{\text{final}} = 50.0\, \mathrm{m/s} \)
    \(v_{\text{initial}} = -40.0\, \mathrm{m/s} \)
  • Total change in velocity \( \Delta v = 50.0 - (-40.0) = 90.0\, \mathrm{m/s} \).
Understanding velocity change is essential as it directly influences the calculation of acceleration and force. It determines the kind of impact involved in such interactions.
Acceleration Calculation
Acceleration refers to the rate at which an object's velocity changes with time. To calculate acceleration, you need the change in velocity and the time over which this change occurs. The formula for acceleration is \( a = \frac{\Delta v}{\Delta t} \).
In the context of the bat hitting the ball, the change in velocity \( \Delta v \) was found to be 90.0 m/s with the time \( \Delta t \) being 8 milliseconds or 0.008 seconds. Thus, the acceleration calculation becomes:
\[ a = \frac{90.0\, \mathrm{m/s}}{8.00 \times 10^{-3}\, \mathrm{s}} = 1.125 \times 10^{4}\, \mathrm{m/s}^2 \]
This indicates how rapidly the ball's speed increased in the opposite direction, indicating a strong force interaction, a key consideration in playing and understanding the dynamics of sports.
Impulse and Momentum
Impulse and momentum are core concepts when discussing forces and motion. Momentum is the quantity of motion of a moving object, expressed as the product of mass and velocity \( p = mv \). Impulse is the change in momentum resulting from a force applied over a period, calculated using \( J = F \times \Delta t \).
In our example, the batter applies an impulse to the baseball, changing its momentum significantly. The impulse experienced by an object brings it to a new velocity, hence changing its momentum.
Funny enough, the concepts are interconnected since:
  • Impulse results in changes in momentum.
  • Impulse equals the average force applied multiplied by the contact time.
This relationship helps us understand not just how hard something was hit, but the effect of that hit over time. Specifically, the impulse-momentum theorem states \( J = \Delta p \), linking directly with Newton’s Second Law where force equals the change in momentum over change in time.

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Most popular questions from this chapter

A chair of mass \(12.0 \mathrm{~kg}\) is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force \(F=40.0 \mathrm{~N}\) that is directed at an angle of \(37.0^{\circ}\) below the horizontal, and the chair slides along the floor. (a) Draw a clearly labeled free-body diagram for the chair. (b) Use your diagram and Newton's laws to calculate the normal force that the floor exerts on the chair.

A box rests on a frozen pond, which serves as a frictionless horizontal surface. If a fisherman applies a horizontal force with magnitude \(48.0 \mathrm{~N}\) to the box and produces an acceleration of magnitude \(2.20 \mathrm{~m} / \mathrm{s}^{2}\), what is the mass of the box?

You walk into an elevator, step onto a scale, and push the "up" button. You recall that your normal weight is \(625 \mathrm{~N}\). Draw a free-body diagram. (a) When the elevator has an upward acceleration of magnitude \(2.50 \mathrm{~m} / \mathrm{s}^{2},\) what does the scale read? (b) If you hold a \(3.85 \mathrm{~kg}\) package by a light vertical string, what will be the tension in this string when the elevator accelerates as in part (a)?

A block of mass \(2.00 \mathrm{~kg}\) is initially at rest at \(x=0\) on a slippery horizontal surface for which there is no friction. Starting at time \(t=0,\) a horizontal force \(F_{x}(t)=\beta-\alpha t\) is applied to the block, where \(\alpha=6.00 \mathrm{~N} / \mathrm{s}\) anwd \(\beta=4.00 \mathrm{~N}\). (a) What is the largest positive value of \(x\) reached by the block? How long does it take the block to reach this point, starting from \(t=0,\) and what is the magnitude of the force when the block is at this value of \(x ?\) (b) How long from \(t=0\) does it take the block to return to \(x=0,\) and what is its speed at this point?

After an annual checkup, you leave your physician's office, where you weighed \(683 \mathrm{~N}\). You then get into an elevator that, conveniently, has a scale. Find the magnitude and direction of the elevator's acceleration if the scale reads (a) \(725 \mathrm{~N}\) and (b) \(595 \mathrm{~N}\).

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