/*! 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 \(150-\mathrm{kg}\) 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 m/s.

Step by step solution

01

Understand the System

In this problem, we are dealing with a conservation of momentum scenario. Before the collision, the mine car and the chunk of coal are moving separately. After the coal comes to rest in the car, they move together as a single system.
02

Identify Known Values

Identify the quantities given in the problem. Mass of the mine car, \( m_1 = 440\, \text{kg} \); Speed of the mine car, \( v_1 = 0.50\, \text{m/s} \); Mass of the coal, \( m_2 = 150\, \text{kg} \); Speed of the coal, \( v_2 = 0.80\, \text{m/s} \).
03

Apply the Conservation of Momentum

According to the conservation of momentum, the total momentum before the collision equals the total momentum after the collision. Before the collision, the momentum of the system is \( m_1v_1 + m_2v_2 \). After the collision, the momentum is \((m_1 + m_2)v\), where \(v\) is the final velocity of the car-coal system.
04

Set Up the Equation

The equation for the conservation of momentum is:\[m_1v_1 + m_2v_2 = (m_1 + m_2)v\]Substitute the known values:\[440 \times 0.50 + 150 \times 0.80 = (440 + 150)v\]
05

Solve for the Final Velocity

Calculate the individual momenta:\[220 + 120 = 590v\]Then simplify and solve for \(v\):\[340 = 590v\]\[v = \frac{340}{590} \approx 0.576\, \text{m/s}\]
06

Interpret the Result

The final speed of the car-coal system is approximately \(0.576\, \text{m/s}\). This means that after the collision, when both the car and the coal move together, the entire system travels at this final speed.

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.

collision
In the realm of physics, a collision is an event where two or more objects come into contact, exchanging energy and momentum.
The exercise above describes a typical inelastic collision. Before the collision occurs, the mine car and the coal move independently. Once they collide, they stick together, forming a single system.
This situation is essential because it helps illustrate crucial principles such as conservation of momentum. In natural and artificial settings, understanding collisions enables us to predict the resulting motion of objects, from car crashes to merging lanes in traffic.
The key takeaway here is that in a perfectly inelastic collision like this one, the two objects become one, moving together with a new velocity. This change is predictable using knowledge of the initial movements and mass of both objects.
final velocity
The concept of final velocity refers to the speed at which objects move after an event or interaction—in this case, a collision.
In our exercise, when the coal lands in the mine car, the final velocity is the speed of the combined car-coal system.
Understanding final velocity is vital for determining the outcome of interactions. When objects come together, their final speed depends on their initial speeds and their masses.
To find this velocity, you apply the conservation of momentum theory, resulting in the speed both masses share after the interaction. This principle is not just academic; it has real-world applications, including vehicle safety, sports mechanics, and even sending spacecraft to explore other planets. The final velocity offers an understanding of kinetic energy distribution post-collision, reflecting both the conservation and transformation laws of energy.
momentum calculation
Momentum is the product of an object's mass and its velocity, expressed typically as \( p = mv \). It describes how much motion an object possesses and its tendency to keep moving once in motion.
To calculate momentum in our exercise, consider each object separately: the momentum of the mine car and the coal before they merge.
The total system's momentum before they combine is found by adding these two amounts together:
  • The mine car: \( 440 imes 0.50\) = 220 kg·m/s
  • The coal: \( 150 imes 0.80\) = 120 kg·m/s
Thus, the total initial momentum is 340 kg·m/s.
After the collision, the car and coal move as a single entity. Their combined momentum remains constant due to the principle of momentum conservation. The equation \( m_1v_1 + m_2v_2 = (m_1 + m_2)v \) guides us to find the final velocity.
Understanding this equation allows us to solve for the final velocity simply by maintaining momentum equality before and after the interaction. This method is indispensable in physics for analyzing the dynamics of collisions.

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 basketball \((m=0.60 \mathrm{kg})\) is dropped from rest. Just before striking the floor, the ball has a momentum whose magnitude is \(3.1 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} .\) At what height was the basketball dropped?

Two ice skaters have masses \(m_{1}\) and \(m_{2}\) and are initially stationary. Their skates are identical. They push against one another, as in Interactive Figure \(7.9,\) and move in opposite directions with different speeds. While they are pushing against each other, any kinetic frictional forces acting on their skates can be ignored. However, once the skaters separate, kinetic frictional forces eventually bring them to a halt. As they glide to a halt, the magnitudes of their accelerations are equal, and skater 1 glides twice as far as skater \(2 .\) What is the ratio \(m_{1} / m_{2}\) of their masses?

A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The incident water stream has a velocity of \(+16.0 \mathrm{m} / \mathrm{s},\) while the exiting water stream has a velocity of \(-16.0 \mathrm{m} / \mathrm{s} .\) The mass of water per second that strikes the blade is \(30.0 \mathrm{kg} / \mathrm{s}\). Find the magnitude of the average force exerted on the water by the blade.

A girl is skipping stones across a lake. One of the stones accidentally ricochets off a toy boat that is initially at rest in the water (see the drawing). The \(0.072-\mathrm{kg}\) stone strikes the boat at a velocity of \(13 \mathrm{m} / \mathrm{s}, 15^{\circ}\) below due east, and ricochets off at a velocity of \(11 \mathrm{m} / \mathrm{s}, 12^{\circ}\) above due east. After being struck by the stone, the boat's velocity is \(2.1 \mathrm{m} / \mathrm{s},\) due east. What is the mass of the boat? Assume the water offers no resistance to the boat's motion.

A golf ball strikes a hard, smooth floor at an angle of \(30.0^{\circ}\) and, as the drawing shows, rebounds at the same angle. The mass of the ball is \(0.047 \mathrm{kg},\) and its speed is \(45 \mathrm{m} / \mathrm{s}\) just before and after striking the floor. What is the magnitude of the impulse applied to the golf ball by the floor? (Hint: Note that only the vertical component of the ball's momentum changes during impact with the floor, and ignore the weight of the ball.)

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.