/*! 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 40 A mine car (mass \(=440 \mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A mine car (mass \(=440 \mathrm{kg}\) ) rolls at a speed of \(0.50 \mathrm{m} / \mathrm{s}\) on a horizontal track, as the drawing shows. A 150-kg chunk of coal has a speed of \(0.80 \mathrm{m} / \mathrm{s}\) when it leaves the chute. Determine the speed of the car-coal system after the coal has come to rest in the car.

Short Answer

Expert verified
The final speed of the car-coal system is approximately \(0.576\text{ m/s}\).

Step by step solution

01

Understand the Conservation of Momentum

The conservation of momentum formula states that the total momentum before an event is equal to the total momentum after an event, provided no external forces act on it. The formula is \( m_1v_1 + m_2v_2 = (m_1 + m_2)v_f \), where \(v_f\) is the final speed of the system.
02

Identify Initial Momentums

Identify the initial momentums of the mine car and the coal chunk separately. The mine car's initial momentum is given as \(440 \text{ kg} \times 0.50 \text{ m/s} = 220 \text{ kg m/s}\). The coal's initial momentum is \(150 \text{ kg} \times 0.80 \text{ m/s} = 120 \text{ kg m/s}\).
03

Calculate Total Initial Momentum

Add the initial momentum of the mine car to that of the coal to find the total initial momentum. So, \(220 \text{ kg m/s} + 120 \text{ kg m/s} = 340 \text{ kg m/s}\).
04

Apply Conservation of Momentum

Using the formula \( m_1v_1 + m_2v_2 = (m_1 + m_2)v_f \), substitute the total initial momentum and the masses of the mine car and coal into the equation. So: \[340 = (440 + 150)v_f\]
05

Solve for Final Velocity \(v_f\)

Simplify and solve the equation for \(v_f\). Calculate \(v_f = \frac{340}{440 + 150} = \frac{340}{590} \approx 0.576 \text{ m/s}\).

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.

Momentum
Momentum is a fundamental concept in physics that describes the motion of an object. It reflects how difficult it is to stop an object when it's moving. The momentum of an object is calculated by multiplying its mass by its velocity.
Understanding momentum helps explain the strength and direction of motion. If you have a heavy and fast-moving object, it has a large momentum. This means it requires more effort to bring it to a stop compared to a lighter or slower object.
  • The equation for momentum is given by: \[ p = m imes v \] where \( p \) is the momentum, \( m \) is the mass, and \( v \) is the velocity.
  • This relation means that an increase in either mass or velocity will result in a proportional increase in momentum.
Grasping momentum is essential for understanding the behavior of moving objects, especially when they interact with each other.
Initial Momentum
Initial momentum is the momentum that objects have before they interact. It sets the stage for analyzing how systems will behave after an event, such as a collision or combination. This exercise uses the initial momenta of both the mine car and the coal chunk.
To understand how the total momentum before the interaction is calculated, you need to assess both objects individually:
  • The initial momentum of the mine car can be calculated as: \[ 440 \text{ kg} \times 0.50 \text{ m/s} = 220 \text{ kg m/s} \]
  • The initial momentum of the coal is: \[ 150 \text{ kg} \times 0.80 \text{ m/s} = 120 \text{ kg m/s} \]
By summing these, the total initial momentum is determined, which is crucial for finding the final speed after these two objects interact.
Final Speed
Final speed refers to how fast an object or system is moving after an event, such as colliding or combining with another object. In the example, finding the final speed is about determining the new velocity of the mine car plus coal system.
The conservation of momentum is the principle used here. It dictates that the total initial momentum should equal the total final momentum if no external force acts on the system. Thus, using the initial momentum calculated:
  • Total initial momentum: \[ 340 \text{ kg m/s} \]
  • Mass of combined system (mine car + coal): \[ 440 \text{ kg} + 150 \text{ kg} = 590 \text{ kg} \]
To find the final speed \( v_f \), apply the formula:\[ v_f = \frac{\text{Total initial momentum}}{\text{Total mass}} \]Substituting the values we have:\[ v_f = \frac{340}{590} \approx 0.576 \text{ m/s} \] This shows how the speed changes after the coal comes to rest in the mine car, highlighting the transformation due to momentum conservation.

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

An astronaut in his space suit and with a propulsion unit (empty of its gas propellant) strapped to his back has a mass of \(146 \mathrm{kg}\). The astronaut begins a space walk at rest, with a completely filled propulsion unit. During the space walk, the unit ejects some gas with a velocity of \(+32 \mathrm{m} / \mathrm{s}\). As a result, the astronaut recoils with a velocity of \(-0.39 \mathrm{m} / \mathrm{s}\). After the gas is ejected, the mass of the astronaut (now wearing a partially \(y\) empty propulsion unit) is 165 kg. What percentage of the gas was ejected from the completely filled propulsion unit?

A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. The masses of the ball and block are, respectively, \(1.60 \mathrm{kg}\) and \(2.40 \mathrm{kg},\) and the length of the wire is 1.20 m. Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision.

Multiple-Concept Example 7 presents a model for solving problems such as this one. A \(1055-\mathrm{kg}\) van, stopped at a traffic light, is hit directly in the rear by a 715 -kg car traveling with a velocity of \(+2.25 \mathrm{m} / \mathrm{s}\). Assume that the transmission of the van is in neutral, the brakes are not being applied, and the collision is elastic. What are the final velocities of (a) the car and (b) the van?

A model rocket is constructed with a motor that can provide a total impulse of \(29.0 \mathrm{N} \cdot \mathrm{s}\) The mass of the rocket is 0.175 kg. What is the speed that this rocket achieves when launched from rest? Neglect the effects of gravity and air resistance.

A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total mass of the wagon, rider, and rock is \(95.0 \mathrm{kg}\). The mass of the rock is \(0.300 \mathrm{kg}\). Initially the wagon is rolling forward at a speed of \(0.500 \mathrm{m} / \mathrm{s} .\) Then the person throws the rock with a speed of \(16.0 \mathrm{m} / \mathrm{s}\) Both speeds are relative to the ground. Find the speed of the wagon after the rock is thrown directly forward in one case and directly backward in another.

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.