Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 74

the ball of mass \(145 \mathrm{g},\) and hits it so that the ball leaves the bat with a speed of \(37 \mathrm{m} / \mathrm{s}\). Assume that the ball is moving horizontally just before and just after the collision with the bat. (a) What is the magnitude of the change in momentum of the ball? (b) What is the impulse delivered to the ball by the bat? (c) If the bat and ball are in contact for \(3.0 \mathrm{ms}\), what is the magnitude of the average force exerted on the ball by the bat?

Short Answer

Expert verified
\(Average\ force = \frac{5.365}{0.003}\) #Solution#: Step 1: The change in momentum of the ball is: \(\Delta momentum = 5.365 kg*m/s\) Step 2: The impulse delivered to the ball is: Impulse = 5.365 kg*m/s Step 3: The average force exerted on the ball is: \(Average\ force \approx 1785 N (rounded\ to\ nearest\ whole\ number)\) Therefore, the average force exerted on the ball by the bat is approximately 1785 N.

Step by step solution

01

Find the change in momentum of the ball

To find the change in momentum, we'll use the formula for momentum: momentum = mass x velocity. We are given the mass of the ball (145 g), which we should convert to kg (1 g = 0.001 kg): \(mass = 145 * 0.001 = 0.145 kg\) The initial velocity of the ball is 0 m/s, and we are given the final velocity (37 m/s). So the change in momentum can be calculated by subtracting the initial momentum from the final momentum: \(\Delta momentum = (mass * final\ velocity) - (mass * initial\ velocity)\) \(\Delta momentum = (0.145 * 37) - (0.145 * 0)\)
02

Calculate the impulse delivered to the ball

The impulse on an object is the product of the force and the time interval during which the force is applied. In this case, the impulse is equal to the change in momentum of the ball. So, we can simply use our result from Step 1 for the impulse: Impulse = \(\Delta momentum\)
03

Find the average force exerted on the ball

To find the average force, we will divide the impulse by the contact time: \(Average\ force = \frac{Impulse}{Contact\ time}\) We are given the contact time of 3 ms (milliseconds), which we should convert to seconds: \(Contact\ time = 3.0 * 10^{-3}s\) Now we can plug in the values and find the average force: \htdocscalculation{}{ mass = 0.145, final_velocity = 37, initial_velocity = 0, delta_momentum = mass * (final_velocity - initial_velocity), contact_time = 3e-3, average_force = delta_momentum / contact_time

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Block \(A,\) with a mass of \(220 \mathrm{g},\) is traveling north on a frictionless surface with a speed of \(5.0 \mathrm{m} / \mathrm{s} .\) Block \(\mathrm{B}\) with a mass of \(300 \mathrm{g}\) travels west on the same surface until it collides with A. After the collision, the blocks move off together with a velocity of \(3.13 \mathrm{m} / \mathrm{s}\) at an angle of \(42.5^{\circ}\) to the north of west. What was \(\mathrm{B}\) 's speed just before the collision?
A uniform rod of length \(30.0 \mathrm{cm}\) is bent into the shape of an inverted U. Each of the three sides is of length \(10.0 \mathrm{cm} .\) Find the location, in \(x\) - and \(y\) -coordinates, of the CM as measured from the origin.
A rifle has a mass of \(4.5 \mathrm{kg}\) and it fires a bullet of mass $10.0 \mathrm{g}\( at a muzzle speed of \)820 \mathrm{m} / \mathrm{s} .$ What is the recoil speed of the rifle as the bullet leaves the gun barrel?
A 6.0 -kg object is at rest on a perfectly frictionless surface when it is struck head-on by a 2.0 -kg object moving at \(10 \mathrm{m} / \mathrm{s} .\) If the collision is perfectly elastic, what is the speed of the 6.0-kg object after the collision? [Hint: You will need two equations.]
A 0.030 -kg bullet is fired vertically at \(200 \mathrm{m} / \mathrm{s}\) into a 0.15 -kg baseball that is initially at rest. The bullet lodges in the baseball and, after the collision, the baseball/ bullet rise to a height of $37 \mathrm{m} .$ (a) What was the speed of the baseball/bullet right after the collision? (b) What was the average force of air resistance while the baseball/bullet was rising?
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Thermodynamics

Read Explanation

Electrostatics

Read Explanation

Modern Physics

Read Explanation

Physics of Motion

Read Explanation

Circular Motion and Gravitation

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free