/*! 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 109 A fireworks rocket is fired vert... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 109

A fireworks rocket is fired vertically upward. At its maximum height of \(80.0 \mathrm{m},\) it explodes and breaks into two pieces: one with mass 1.40 \(\mathrm{kg}\) and the other with mass 0.28 \(\mathrm{kg} .\) In the explosion, 860 \(\mathrm{J}\) of chemical energy is converted to kinetic energy of the two fragments. (a) What is the speed of each fragment just after the explosion? (b) It is observed that the two fragments hit the ground at the same time. What is the distance between the points on the ground where they land? Assume that the ground is level and air resistance can be ignored.

Short Answer

Expert verified
(a) Speeds are 30 m/s and 150 m/s; (b) Distance is 484.8 m.

Step by step solution

01

Understanding the Problem

We have an explosion at the maximum height of a firework rocket. The total energy from the explosion converts into kinetic energy of the two fragments. We need to calculate their speeds and then determine the distance between their landing positions.
02

Conservation of Momentum

Before the explosion, the system's momentum is zero (the rocket was momentarily at a standstill at maximum height). Consequently, the momentum of the two fragments must be equal and opposite after the explosion. Let the speed of the 1.40 kg fragment be \( v_1 \), and the speed of the 0.28 kg fragment be \( v_2 \). By conservation of momentum: \[ 1.40 \cdot v_1 = 0.28 \cdot v_2 \].
03

Kinetic Energy Distribution

The kinetic energy from the explosion is 860 J, which is shared between the two fragments: \[ \frac{1}{2} \times 1.40 \times v_1^2 + \frac{1}{2} \times 0.28 \times v_2^2 = 860. \] We have two equations and two unknowns from Steps 2 and 3.
04

Solving the System of Equations

Substitute \( v_2 = \frac{1.40}{0.28}v_1 \) from Step 2 into Step 3's energy equation and solve for \( v_1 \). After solving the quadratic equation, we find \( v_1 = 30 \text{ m/s} \). Substitute back to find \( v_2 \): \( v_2 = 150 \text{ m/s} \).
05

Determine the Horizontal Distance Traveled

Since the fragments hit the ground simultaneously, consider the projectile motion of both fragments under gravity for the vertical descent of 80.0 m. Time to fall can be computed using: \[ t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 80}{9.81}}. \] Calculate it to find \( t \approx 4.04 \text{ s} \).
06

Calculate Landing Positions

Assume horizontal components of velocities are equal to \( v_1 \) and \( v_2 \) calculated. Distances are \( d_1 = v_1 \cdot t \) and \( d_2 = v_2 \cdot t \). Compute these: \( d_1 = 30 \times 4.04 = 121.2 \text{ m} \) and \( d_2 = 150 \times 4.04 = 606.0 \text{ m} \).
07

Finalize the Landing Distance

The difference in landing positions \( |d_2 - d_1| \) gives the distance between the fragments' landing points. This evaluates to \( 606.0 - 121.2 = 484.8 \text{ m} \).

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 is the energy an object has because of its motion. It's one of the key concepts in understanding how energy from an explosion is transformed and utilized. When the fireworks rocket explodes, a portion of the chemical energy stored within is converted into kinetic energy.

Kinetic energy (\( KE \)) is expressed mathematically as: \[ KE = \frac{1}{2}mv^2 \]where:
  • \( m \) is the mass of the object.
  • \( v \) is the velocity of the object.
In the given problem, the total kinetic energy after the explosion is 860 J. This energy is distributed between the two fragments, giving them different speeds due to their differing masses. Understanding this distribution allows us to calculate each fragment's speed using the law of conservation of momentum and the kinetic energy equation. Conserving energy in this system means that energy does not disappear; it simply changes forms from chemical to kinetic.
Projectile Motion
Projectile motion describes the motion of an object thrown or projected into the air. It is subject to only the acceleration of gravity. In the exercise, the exploded fragments of the rocket behave like projectiles when they fall back to the ground.

Projectile motion can be analyzed in horizontal and vertical components:
  • Vertically, the objects are acted upon by gravity, causing them to accelerate towards the earth. The time to fall can be calculated using the formula:\[ t = \sqrt{\frac{2h}{g}} \]where:
    • \( h \) is the height from which the object descends (80 m in this case).
    • \( g \) is the acceleration due to gravity (approximately 9.81 m/s虏).
    This gives the time it takes for both fragments to fall to the ground.
  • Horizontally, the velocity at the moment of the explosion determines how far the fragments travel, given by:\[ d = vt \]where:
    • \( v \) is the horizontal component of the velocity.
    • \( t \) is the time of flight calculated from the vertical motion.
By treating the motion of the fragments as independent in the vertical and horizontal directions, we can accurately predict where they will land on the ground.
Explosion Fragments
When the rocket explodes, it breaks into two pieces or fragments. Each fragment is subjected to the conservation of momentum, as the total momentum before and after the explosion must remain the same. This implies that the momentum gained by one fragment is equal and opposite to the other.

The law of conservation of momentum can be expressed as:\[ m_1v_1 = -m_2v_2 \]where:
  • \( m_1 \) and \( v_1 \) are the mass and speed of the first fragment.
  • \( m_2 \) and \( v_2 \) are the mass and speed of the second fragment.
This relationship helps us determine the speeds of the fragments post-explosion. This principle ensures that even though the rocket bursts apart, the system's total momentum is conserved.

Understanding the behavior of fragments also involves calculating how far each fragment travels once they hit the ground. This distance, as found from projectile motion analysis, gives us the separation between their landing points. Thus, fragmentation is key to exploring different aspects of motion and energy in this physics problem.

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 \(20.0-\mathrm{kg}\) projectile is fired at an angle of \(60.0^{\circ}\) above the horizontal with a speed of 80.0 \(\mathrm{m} / \mathrm{s} .\) At the highest point of its trajectory, the projectile explodes into two fragments with equal mass, one of which falls vertically with zero initial speed. You can ignore air resistance. (a) How far from the point of firing does the other fragment strike if the terrain is level? (b) How much energy is released during the explosion?

A 0.160 -kg hockey puck is moving on an icy, frictionless, horizontal surface. At \(t=0,\) the puck is moving to the right at 3.00 \(\mathrm{m} / \mathrm{s} .\) (a) Calculate the velocity of the puck (magnitude and direction) after a force of 25.0 \(\mathrm{N}\) directed to the right has been applied for 0.050 s. (b) If, instead, a force of 12.0 \(\mathrm{N}\) directed to the left is applied from \(t=0\) to \(t=0.050 \mathrm{s}\) , what is the final velocity of the puck?

Jonathan and Jane are sitting in a sleigh that is at rest on frictionless ice. Jonathan's weight is 800 \(\mathrm{N}\) , Jane's weight is \(600 \mathrm{N},\) and that of the sleigh is 1000 \(\mathrm{N}\) . They see a poisonous spider on the floor of the sleigh and immediately jump off. Jonathan jumps to the left with a velocity of 5.00 \(\mathrm{m} / \mathrm{s}\) at \(30.0^{\circ}\) above the horizontal (relative to the ice), and Jane jumps to the right at 7.00 \(\mathrm{m} / \mathrm{s}\) at \(36.9^{\circ}\) above the horizontal (relative to the ice). Calculate the sleigh's horizontal velocity (magnitude and direction) after they jump out.

A bat strikes a 0.145 -kg baseball. Just before impact, the ball is traveling horizontally to the right at \(50.0 \mathrm{m} / \mathrm{s},\) and it leaves the bat traveling to the left at an angle of \(30^{\circ}\) above horizontal with a speed of 65.0 \(\mathrm{m} / \mathrm{s}\) . If the ball and bat are in contact for 1.75 \(\mathrm{ms}\) find the horizontal and vertical components of the average force on the ball.

\(\mathrm{A}^{232} \mathrm{Th}\) (thorium) nucleus at rest decays to a \(^{228} \mathrm{Ra}\) (radium) nucleus with the emission of an alpha particle. The total kinetic energy of the decay fragments is \(6.54 \times 10^{-13} \mathrm{J}\) . An alpha particle has 1.76\(\%\) of the mass of a \(^{228} \mathrm{Ra}\) nucleus. Calculate the kinetic energy of (a) the recoiling 228 nucleus and (b) the alpha particle.
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Translational Dynamics

Read Explanation

Torque and Rotational Motion

Read Explanation

Magnetism

Read Explanation

Kinematics Physics

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.