/*! 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 4 (a) What is the momentum of a ga... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) What is the momentum of a garbage truck that is \(1.20 \times 10^{4} \mathrm{kg}\) and is moving at \(10.0 \mathrm{m} / \mathrm{s} ?\) (b) At what speed would an 8.00 -kg trash can have the same momentum as the truck?

Short Answer

Expert verified
The momentum of the garbage truck is \(1.20 \times 10^5 \mathrm{kg \cdot m/s}\), and the speed of the trash can to have the same momentum is \(1.50 \times 10^3 \mathrm{m/s}\).

Step by step solution

01

Calculate the Momentum of the Garbage Truck

The momentum, denoted as \(p\), of an object can be calculated using the formula \(p = m \times v\), where \(m\) is the mass of the object and \(v\) is its velocity. For the garbage truck, we have the mass \(m = 1.20 \times 10^{4} \mathrm{kg}\) and the velocity \(v = 10.0 \mathrm{m/s}\). We calculate the momentum by multiplying the two values.
02

Use the Momentum of the Truck to Find the Speed of the Trash Can

Knowing that the momentum of the trash can must equal the momentum of the garbage truck and using the formula for momentum, we can find the required speed of the trash can. Let the speed of the trash can be \(v_{\text{can}}\). The mass of the trash can is given as \(m_{\text{can}} = 8.00 \mathrm{kg}\). We set up the equation \(p_{\text{truck}} = p_{\text{can}}\), where \(p_{\text{truck}}\) is the momentum we found in Step 1. The equation becomes \(1.20 \times 10^{4} \mathrm{kg} \times 10.0 \mathrm{m/s} = 8.00 \mathrm{kg} \times v_{\text{can}}\). We then solve for \(v_{\text{can}}\).
03

Calculate the Speed of the Trash Can

By dividing both sides of the equation by the mass of the trash can, we isolate \(v_{\text{can}}\) and find its value, which represents the speed required for the trash can to have the same momentum as the garbage truck.

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 91Ó°ÊÓ!

Key Concepts

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

Linear Momentum
The idea of linear momentum is crucial when studying the motion of objects. It's defined as the product of an object's mass (\(m\)) and its velocity (\(v\)), represented by the formula \(p = m \times v\). In the context of the exercise, a garbage truck with a mass of \(1.20 \times 10^{4} \mathrm{kg}\) moving at a velocity of \(10.0 \mathrm{m/s}\) has a substantial amount of momentum because both factors contributing to momentum—mass and velocity—are significant.

The momentum of a moving body represents its capacity to impact other bodies and the force it would take to stop that body. A heavier or faster-moving object will carry more momentum, making it more influential upon collision. This ties into the broader implications of momentum in understanding vehicle crashes and in designing safety features.
Conservation of Momentum
The principle of the conservation of momentum states that in a closed system, with no external forces acting upon it, the total momentum remains constant over time. This is a fundamental law of physics that applies universally to all types of collisions and interactions.

In the context of the textbook exercise, after calculating the momentum of the garbage truck, we can predict the outcome of a hypothetical collision with the trash can by ensuring the momentum before and after the event is conserved, provided no external forces interfere. This is why a lighter trash can must move at a much higher velocity to match the momentum of the much heavier garbage truck; it's a direct consequence of the conservation of momentum. The concept assists us in understanding physical interactions and predicting the results of collisions in a diverse array of scenarios, spanning from everyday occurrences to complex astrophysical events.
Velocity
Velocity, a fundamental concept in physics, is a vector quantity that not only concerns the speed of an object but also its direction of travel. It's directly linked to momentum, as seen in the formula \(p = m \times v\). In the exercise, we use velocity to determine how fast the trash can must be moving to have the same momentum as the garbage truck.

The concept of velocity allows us to analyze motion in a more comprehensive manner, going beyond mere speed to include the path of travel. This enables precise predictions about the movement of objects and the effects of forces applied to them. Velocity can have vast implications, from tracking the orbits of celestial bodies to the design of transportation systems.

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 ball with an initial velocity of \(10 \mathrm{m} / \mathrm{s}\) moves at an angle \(60^{\circ}\) above the \(+x\) -direction. The ball hits a vertical wall and bounces off so that it is moving \(60^{\circ}\) above the \(-x\) -direction with the same speed. What is the impulse delivered by the wall?

Two cars collide at an icy intersection and stick together afterward. The first car has a mass of \(1200 \mathrm{kg}\) and is approaching at \(8.00 \mathrm{m} / \mathrm{s}\) due south. The second car has a mass of \(850 \mathrm{kg}\) and is approaching at \(17.0 \mathrm{m} / \mathrm{s}\) due west. (a) Calculate the final velocity (magnitude and direction) of the cars. (b) How much kinetic energy is lost in the collision? (This energy goes into deformation of the cars.) Note that because both cars have an initial velocity, you cannot use the equations for conservation of momentum along the \(x\) -axis and \(y\) -axis; instead, you must look for other simplifying aspects.

Calculate the voltage applied to a \(2.00 \mu \mathrm{F}\) capacitor when it holds \(3.10 \mu \mathrm{C}\) of charge.

Military rifles have a mechanism for reducing the recoil forces of the gun on the person firing it. An internal part recoils over a relatively large distance and is stopped by damping mechanisms in the gun. The larger distance reduces the average force needed to stop the internal part. (a) Calculate the recoil velocity of a \(1.00 \mathrm{kg}\) plunger that directly interacts with a 0.0200 -kg bullet fired at 600 m/s from the gun. (b) If this part is stopped over a distance of \(20.0 \mathrm{cm},\) what average force is exerted upon it by the gun? (c) Compare this to the force exerted on the gun if the bullet is accelerated to its velocity in \(10.0 \mathrm{ms}\) (milliseconds).

(a) What is the potential between two points situated \(10 \mathrm{~cm}\) and \(20 \mathrm{~cm}\) from a \(3.0 \mu \mathrm{C}\) point charge? (b) To what location should the point at \(20 \mathrm{~cm}\) be moved to increase this potential difference by a factor of two?

See all solutions

Recommended explanations on Physics Textbooks

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.