/*! 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 64 A steel ball with mass 40.0 \(g\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 64

A steel ball with mass 40.0 \(g\) is dropped from a height of 2.00 m onto a horizontal steel slab. The ball rebounds to a height of 1.60 m. (a) Calculate the impulse delivered to the ball during impact. (b) If the ball is in contact with the slab for 2.00 ms, find the average force on the ball during impact.

Short Answer

Expert verified
(a) Impulse = 0.4744 kg鈰卪/s; (b) Average force = 237.2 N.

Step by step solution

01

Understand the Problem

We need to find the impulse delivered to the ball during its impact with the slab and the average force exerted on it. The ball falls from a height 2.00 m, rebounds to a height 1.60 m, with its mass given as 40.0 g, and the time of contact with the slab is 2.00 ms.
02

Convert Units

The mass of the ball is given as 40.0 g. We need to convert it into kilograms for our calculations: \( m = 40.0 \: g = 0.040 \: kg \).The time of contact is given as 2.00 ms, which can be converted to seconds: \( 2.00 \: ms = 0.002 \: s \).
03

Calculate Initial and Final Velocities

For the ball dropped from a height, use the energy conservation principle. The velocity just before impact, \( v_i \), can be computed as:\[ v_i = \sqrt{2gh_i} \quad \text{where } g = 9.8 \: m/s^2, \: h_i = 2.00 \: m. \]Plug in the values: \( v_i = \sqrt{2 \times 9.8 \times 2.00} \approx 6.26 \: m/s \).For the ball rebounding to height \( h_f = 1.60 \: m \), the velocity just after leaving the surface, \( v_f \), is:\[ v_f = \sqrt{2gh_f} \]Plug in the values: \( v_f = \sqrt{2 \times 9.8 \times 1.60} \approx 5.60 \: m/s \).
04

Calculate Impulse

Impulse is defined as the change in momentum: \( J = \Delta p = m(v_f - (-v_i)) \). Substitute \( m = 0.040 \: kg, \: v_f = 5.60 \: m/s, \: v_i = 6.26 \: m/s \) into:\[ J = 0.040 (5.60 + 6.26) = 0.040 \times 11.86 = 0.4744 \: kg \cdot m/s \].
05

Calculate Average Force

The average force during impact \( F_{avg} \) can be calculated using the formula: \[ F_{avg} = \frac{J}{\Delta t} \]where \( J = 0.4744 \: kg \cdot m/s \) and \( \Delta t = 0.002 \: s \):\[ F_{avg} = \frac{0.4744}{0.002} = 237.2 \: N \].

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Kinetic Energy
Kinetic Energy plays a pivotal role in understanding how objects move and interact after being dropped from a height. When the steel ball is released from a height of 2.00 meters, it is initially at rest. Its potential energy is converted into kinetic energy as it falls towards the ground. Kinetic energy
  • depends on mass and velocity
  • is given by the equation: \( KE = \frac{1}{2}mv^2 \)
  • is a scalar quantity
As the ball falls, its velocity increases due to gravity, and so does its kinetic energy. At the moment right before it strikes the slab, all its potential energy has been transformed into kinetic energy.
Kinetic energy at impact can be calculated using the velocity obtained just before the ball hits the slab, which is 6.26 m/s here. This impact velocity, along with the ball's mass, determines the maximum kinetic energy it carries upon striking the solid surface. Thus, understanding kinetic energy is crucial for analyzing the impact phenomena and the rebound height calculations.
Conservation of Energy
The principle of Conservation of Energy is essential for solving this exercise, especially in finding the velocities before and after the impact of the steel ball.Initially, when the ball is at a height of 2.00 meters and 1.60 meters respectively, energy is conserved between potential and kinetic forms.
  • Before impact, all gravitational potential energy is converted into kinetic energy.
  • Upon impact and rebound, kinetic energy transformed back into potential energy as the ball rises.
This conversion is governed by the equation:\[ mgh = \frac{1}{2}mv^2 \]where \( h \) is the height, \( m \) is mass, \( g \) is gravity, and \( v \) is velocity.
Applying this principle enabled us to calculate the initial and final velocities, 6.26 m/s and 5.60 m/s respectively, by associating the original heights with potential energy. Understanding these energy transformations is key to explain how the ball rebounces and the energy dissipates during the impact with the slab.
Average Force Calculation
Average Force Calculation is a straightforward yet vital aspect of mechanics when finding the force applied over a short timeframe. In this scenario, the average force during the ball's impact with the slab is calculated.With the time of contact given as 2 milliseconds (ms), this duration when multiplied by the rate of change of momentum gives the average force. The formula used is:\[ F_{avg} = \frac{J}{ riangle t} \]where \( J \) represents the impulse or change in momentum, and \( \triangle t \) is the time duration (0.002 seconds)Substituting in the calculated impulse of 0.4744 kg路m/s leads to:\[ F_{avg} = \frac{0.4744}{0.002} \approx 237.2 \text{ N} \]
This calculation shows the force experienced by the ball during its contact with the slab. Average force is important to understand how collisions affect objects, which in turn helps in designing materials and systems that can withstand high forces or impacts efficiently.

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

A 4.00-g bullet, traveling horizontally with a velocity of magnitude 400 m/s, is fired into a wooden block with mass 0.800 kg, initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 190 m/s. The block slides a distance of 72.0 cm along the surface from its initial position. (a) What is the coefficient of kinetic friction between block and surface? (b) What is the decrease in kinetic energy of the bullet? (c) What is the kinetic energy of the block at the instant after the bullet passes through it?

Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracting the cavity to force out the water through an opening. A 6.50-kg squid (including the water in the cavity) at rest suddenly sees a dangerous predator. (a) If the squid has 1.75 kg of water in its cavity, at what speed must it expel this water suddenly to achieve a speed of 2.50 m/s to escape the predator? Ignore any drag effects of the surrounding water. (b) How much kinetic energy does the squid create by this maneuver?

Obviously, we can make rockets to go very fast, but what is a reasonable top speed? Assume that a rocket is fired from rest at a space station in deep space, where gravity is negligible. (a) If the rocket ejects gas at a relative speed of 2000 m/s and you want the rocket's speed eventually to be 1.00\(\times\) 10\(^{-3}c\), where \(c\) is the speed of light in vacuum, what fraction of the initial mass of the rocket and fuel is \(not\) fuel? (b) What is this fraction if the final speed is to be 3000 m/s?

You are standing on a concrete slab that in turn is resting on a frozen lake. Assume there is no friction between the slab and the ice. The slab has a weight five times your weight. If you begin walking forward at 2.00 m/s relative to the ice, with what speed, relative to the ice, does the slab move?

A 0.150-kg glider is moving to the right with a speed of 0.80 m/s on a frictionless, horizontal air track. The glider has a head-on collision with a 0.300-kg glider that is moving to the left with a speed of 2.20 m/s. Find the final velocity (magnitude and direction) of each glider if the collision is elastic.
See all solutions

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Read Explanation

Turning Points in Physics

Read Explanation

Electrostatics

Read Explanation

Torque and Rotational Motion

Read Explanation

Physics of Motion

Read Explanation

Atoms and Radioactivity

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.