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Chapter 7: Problem 56

A baseball \((m=149 \mathrm{g})\) approaches a bat horizontally at a speed of \(40.2 \mathrm{m} / \mathrm{s}(90 \mathrm{mi} / \mathrm{h})\) and is hit straight back at a speed of \(45.6 \mathrm{m} / \mathrm{s}\) \((102 \mathrm{mi} / \mathrm{h}) .\) If the ball is in contact with the bat for a time of \(1.10 \mathrm{ms},\) what is the average force exerted on the ball by the bat? Neglect the weight of the ball, since it is so much less than the force of the bat. Choose the direction of the incoming ball as the positive direction.

Short Answer

Expert verified
The average force on the ball is approximately -11622 N.

Step by step solution

01

Understand the Problem

We need to calculate the average force exerted on a baseball by a bat. The problem provides the mass of the baseball, its initial and final velocities, and the time of contact with the bat. We will assume the only force acting is from the bat, and we'll use Newton's second law of motion, which relates force, mass, and acceleration.
02

Identify Given Values and Assumptions

The mass of the ball, \( m \), is 149 g. Let's convert this to kilograms: \( m = 0.149 \, \text{kg} \). The initial velocity of the ball, \( v_i \), is 40.2 m/s, and the final velocity, \( v_f \), is -45.6 m/s (negative because direction is opposite). The contact time, \( \Delta t \), is \( 1.10 \times 10^{-3} \, \text{s} \). The weight of the ball is neglected.
03

Calculate Change in Velocity

Since the initial and final velocities are given in m/s, we calculate the change in velocity: \[ \Delta v = v_f - v_i = -45.6 \, \text{m/s} - 40.2 \, \text{m/s} = -85.8 \, \text{m/s} \].
04

Determine the Acceleration

Using the change in velocity and the time of contact, calculate acceleration, \( a \), using the formula:\[ a = \frac{\Delta v}{\Delta t} = \frac{-85.8 \, \text{m/s}}{1.10 \times 10^{-3} \, \text{s}} = -78000 \, \text{m/s}^2 \].
05

Calculate the Average Force

Now, using Newton's second law \( F = ma \), we find the average force. Substitute \( m = 0.149 \, \text{kg} \) and \( a = -78000 \, \text{m/s}^2 \):\[ F = 0.149 \, \text{kg} \times (-78000 \, \text{m/s}^2) = -11622 \, \text{N} \].The negative sign indicates that the direction of force is opposite to the initial motion direction, consistent with the ball being hit back.

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Key Concepts

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

average force
When we talk about average force, we are referring to the constant force acting on an object over a given time period. In the case of the baseball and the bat, we're examining how much force the bat exerts on the baseball over the brief time they are in contact with each other. The calculation of average force involves Newton's Second Law of Motion, which states that force is the product of mass and acceleration. To put it simply, if you know how fast an object's velocity changes over time, and you know the object's mass, you can find the average force acting on it with the formula:
  • \[ F = ma \]
  • Here, "m" is the mass and "a" is the acceleration.
In our baseball scenario, once the contact time and acceleration are known, they are used to find the average force by seeing how quickly the ball goes from moving towards the bat to away from it. The calculation shows not just the magnitude of the force, but also its direction. This is important for understanding the interactions in any sports or physics context.
momentum change
Momentum is a key concept in physics, representing the product of an object's mass and its velocity. The change in momentum gives insight into how the object's motion is altered over time, particularly when acted upon by an external force. For our baseball problem, we calculate the change in momentum to understand the effects of the bat's force on the ball.
  • The initial momentum is calculated as \( p_i = m imes v_i \).
  • The final momentum is \( p_f = m imes v_f \).

Using these, the change in momentum can be found with:
  • \[ ext{Change in momentum} = p_f - p_i \]
  • This can also be expressed as \( m imes (v_f - v_i) \).
This change in momentum is fundamental in determining the amount of force applied, since force is the rate of change of momentum over time. The baseball's negative change in velocity leads to a sizable shift in momentum, correspondingly indicating the substantial force exerted by the bat.
velocity calculation
Velocity is the measure of an object's speed and the direction it is moving. Calculating the velocity change of an object can help determine how forces are impacting it. In the baseball step-by-step solution, understanding velocity change is integral to finding out how the ball's interaction with the bat alters its state of motion.
When the bat hits the baseball:
  • The initial velocity \( v_i \) is 40.2 m/s.
  • The bat imparts a force that reverses its motion, making the final velocity \( v_f = -45.6 \) m/s (the negative sign indicates a change in direction).
  • The change in velocity can be represented as:
\[ \Delta v = v_f - v_i \]The result is a total velocity change of -85.8 m/s, illustrating just how swiftly the bat alters the ball's trajectory. This change in velocity is what allows us to calculate acceleration, which is then used to determine the average force. Understanding these steps helps clarify the pivotal role velocity plays within the laws of motion, especially in high-speed sports scenarios.

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Most popular questions from this chapter

Adolf and Ed are wearing harnesses and are hanging at rest from the ceiling by means of ropes attached to them. Face to face, they push off against one another. Adolf has a mass of \(120 \mathrm{kg},\) and Ed has a mass of \(78 \mathrm{kg} .\) Following the push, Adolf swings upward to a height of \(0.65 \mathrm{m}\) above his starting point. To what height above his own starting point does Ed rise?

Multiple-Concept Example 7 deals with some of the concepts that are used to solve this problem. A cue ball (mass \(=0.165 \mathrm{kg}\) ) is at rest on a frictionless pool table. The ball is hit dead center by a pool stick, which applies an impulse of \(+1.50 \mathrm{N} \cdot \mathrm{s}\) to the ball. The ball then slides along the table and makes an elastic head-on collision with a second ball of equal mass that is initially at rest. Find the velocity of the second ball just after it is struck.

An electron collides elastically with a stationary hydrogen atom. The mass of the hydrogen atom is 1837 times that of the electron. Assume that all motion, before and after the collision, occurs along the same straight line. What is the ratio of the kinetic energy of the hydrogen atom after the collision to that of the electron before the collision? \(?\)

Two particles are moving along the \(x\) axis. Particle 1 has a mass \(m_{1}\) and a velocity \(v_{1}=+4.6 \mathrm{m} / \mathrm{s} .\) Particle 2 has a mass \(m_{2}\) and a velocity \(v_{2}=-6.1 \mathrm{m} / \mathrm{s} .\) The velocity of the center of mass of these two particles is zero. In other words, the center of mass of the particles remains stationary, even though each particle is moving. Find the ratio \(m_{1} / m_{2}\) of the masses of the particles.

Two ice skaters have masses \(m_{1}\) and \(m_{2}\) and are initially stationary. Their skates are identical. They push against one another, as in Figure \(7.9,\) and move in opposite directions with different speeds. While they are pushing against each other, any kinetic frictional forces acting on their skates can be ignored. However, once the skaters separate, kinetic frictional forces eventually bring them to a halt. As they glide to a halt, the magnitudes of their accelerations are equal, and skater 1 glides twice as far as skater 2 . What is the ratio \(m_{1} / m_{2}\) of their masses?
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