/*! 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 31 An incoming 0.14 -kg baseball ha... [FREE SOLUTION] | 91Ó°ÊÓ

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An incoming 0.14 -kg baseball has a speed of \(45 \mathrm{~m} / \mathrm{s}\). The batter hits the ball, giving it a speed of \(60 \mathrm{~m} / \mathrm{s}\). If the contact time is \(0.040 \mathrm{~s},\) what is the average force of the bat on the ball?

Short Answer

Expert verified
The average force is 52.5 N.

Step by step solution

01

Identify known values

Firstly, gather all the values given in the problem. The mass of the baseball is \( m = 0.14 \, \text{kg} \), the initial speed \( v_i = 45 \, \text{m/s} \), the final speed \( v_f = 60 \, \text{m/s} \), and the contact time \( \Delta t = 0.040 \, \text{s} \).
02

Calculate the change in velocity

Use the initial and final velocities to find the change in velocity (\( \Delta v \)) of the baseball. \( \Delta v = v_f - v_i = 60 \, \text{m/s} - 45 \, \text{m/s} = 15 \, \text{m/s} \).
03

Determine the change in momentum

The change in momentum (\( \Delta p \)) is given by \( \Delta p = m \cdot \Delta v \). Substitute the known values into the equation: \( \Delta p = 0.14 \, \text{kg} \times 15 \, \text{m/s} = 2.1 \, \text{kg} \cdot \text{m/s} \).
04

Calculate the average force

The average force (\( F \)) on the baseball is the change in momentum divided by the contact time: \( F = \frac{\Delta p}{\Delta t} \). Substituting the known values: \( F = \frac{2.1 \, \text{kg} \cdot \text{m/s}}{0.040 \, \text{s}} = 52.5 \, \text{N} \).

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Key Concepts

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

Momentum
Momentum is a fundamental concept in physics, closely linked to both mass and velocity. It encapsulates the idea of how much "motion" an object possesses. Mathematically, momentum (\( p \)) is expressed as the product of an object's mass (\( m \)) and its velocity (\( v \)).
  • Formula: \( p = m \, \cdot \, v \)
  • Units: kgâ‹…m/s
A moving baseball is a great example to understand momentum in action. A heavier object moving at the same speed as a lighter one will have more momentum. Similarly, a faster object will have more momentum than a slower one, given equal mass. In our baseball scenario, momentum changes as the speed changes from 45 m/s to 60 m/s due to an external force (the bat).
The concept helps us quantify motion precisely, offering a base to calculate how forces will affect an object's state of movement.
Change in Velocity
Change in velocity (\( \Delta v \)) is essential for understanding how speed or direction alterations affect an object. In equations and analysis, it helps in measuring the impact of a force over time.
  • Formula: \( \Delta v = v_f - v_i \)
Here, \( v_f \) is the final velocity, and \( v_i \) is the initial velocity. For our baseball, it moves from an initial speed of 45 m/s to a final speed of 60 m/s. Thus, \( \Delta v = 60 \, \text{m/s} - 45 \, \text{m/s} = 15 \, \text{m/s} \).
The change in velocity directly influences momentum changes, and as shown, becomes crucial while calculating the average force applied during the contact time with the bat. It signifies the overall impact made on the baseball’s speed by the external force.
Impulse
Impulse is closely tied with the concepts of force and time. It describes how much momentum changes as a result of a force acting over a certain period. Impulse is crucial for understanding collisions and impact scenarios.
  • Formula: Impulse (\( J \)) = \( F \times \Delta t \)
  • Also expressed as the Change in Momentum: \( J = \Delta p \)
In the baseball example, we calculated the change in momentum to be \( 2.1 \text{ kg}\cdot\text{m/s} \). This is equivalent to the impulse involved because the change resulted from the bat's action during the 0.040 s of contact.
Hence, average force is derived using impulse over the time interval, emphasizing that both magnitude and duration of force impact an object's motion significantly. Understanding impulse helps grasp how any object's motion can be altered or controlled through force application over time.

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Most popular questions from this chapter

In a laboratory setup, two frictionless carts are placed on a horizontal surface. Cart A has a mass of \(500 \mathrm{~g}\) and cart B's mass is \(1000 \mathrm{~g}\). Between them is placed an ideal (very light) spring and they are squeezed together carefully, thereby compressing the spring by \(5.50 \mathrm{~cm} .\) Both carts are then released and \(\mathrm{B}^{\prime}\) s recoil speed is measured to be \(0.55 \mathrm{~m} / \mathrm{s}\). (a) Will cart A's speed be (1) greater than, (2) less than, or (3) the same as B's speed? Explain. (b) Determine B's recoil speed to see if your conjecture in (a) was correct. (c) Determine the spring constant of the spring.

A piece of uniform sheet metal measures \(25 \mathrm{~cm}\) by \(25 \mathrm{~cm}\). If a circular piece with a radius of \(5.0 \mathrm{~cm}\) is cut from the center of the sheet, where is the sheet's center of mass now?

Two cups are placed on a uniform board that is balanced on a cylinder ( \(\mathbf{v}\) Fig. 6.40 ). The board has a mass of \(2.00 \mathrm{~kg}\) and is \(2.00 \mathrm{~m}\) long. The mass of \(\operatorname{cup} 1\) is \(200 \mathrm{~g}\) and it is placed \(1.05 \mathrm{~m}\) to the left of the balance point. The mass of cup 2 is \(400 \mathrm{~g}\). Where should cup 2 be placed for balance (relative to the right end of the board)?

A \(170-\mathrm{g}\) hockey puck sliding on ice perpendicularly impacts a flat piece of sideboard. Its incoming momentum is \(6.10 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\). It rebounds along its incoming path after having suffered a momentum change (magnitude) of \(8.80 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\). (a) If the impact with the board took \(35.0 \mathrm{~ms}\), determine the average force (including direction) exerted by the puck on the board. (b) Determine the final momentum of the puck. (c) Was this collision elastic or inelastic? Prove your answer mathematically.

A ball of mass \(200 \mathrm{~g}\) is released from rest at a height of \(2.00 \mathrm{~m}\) above the floor and it rebounds straight up to a height of \(0.900 \mathrm{~m}\). (a) Determine the ball's change in momentum due to its contact with the floor. (b) If the contact time with the floor was \(0.0950 \mathrm{~s}\), what was the average force the floor exerted on the ball, and in what direction?

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