/*! 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 68 A railroad handcar is moving alo... [FREE SOLUTION] | 91Ó°ÊÓ

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A railroad handcar is moving along straight, frictionless tracks with negligible air resistance. In the following cases, the car initially has a total mass (car and contents) of \(200 \mathrm{~kg}\) and is traveling east with a velocity of magnitude \(5.00 \mathrm{~m} / \mathrm{s}\). Find the final velocity of the car in cach casc, assuming that the handcar docs not leave the tracks. (a) A \(25.0 \mathrm{~kg}\) mass is thrown sideways out of the car with a velocity of magnitude \(2.00 \mathrm{~m} / \mathrm{s}\) relative to the car's initial velocity, (b) \(\mathrm{A} 25.0 \mathrm{~kg}\) mass is thrown backward out of the car with a velocity of \(5.00 \mathrm{~m} / \mathrm{s}\) relative to the initial motion of the car. (c) A \(25.0 \mathrm{~kg}\) mass is thrown into the car with a velocity of \(6.00 \mathrm{~m} / \mathrm{s}\) relative to the ground and opposite in direction to the initial velocity of the car.

Short Answer

Expert verified
The final velocities for each case are - a) \(5.00 m/s\), b) \(6.43 m/s\) and c) \(3.78 m/s\).

Step by step solution

01

Identify initial and final momentum for each case

The initial momentum (\(P_{initial}\)) of the system is the product of the initial mass and the initial velocity and the final momentum (\(P_{final}\)) is the product of the final mass and the final velocity. The law of conservation of momentum tells us these quantities are equal. 'a' involves throwing mass sideways out of the car, 'b' is throwing mass backward, and 'c' involves throwing mass into the car. We will calculate the final momentum for each case individually.
02

Solve for the final velocity in case (a)

In case 'a', a mass of 25 kg is thrown sideways. Even though it changes the system mass, it does not affect the system momentum because the velocity is perpendicular to the direction of motion. Thus, the final velocity (\(v_{f_a}\)) equals the initial velocity, 5.00 m/s.
03

Solve for the final velocity in case (b)

In case 'b', initial momentum is \(200 kg * 5.00 m/s = 1000 kg*m/s\). When the 25.0 kg mass is thrown backwards, the final system mass decreases to 175 kg. Also, the thrown mass adds to the system momentum ('-' sign is because it's thrown backward) an amount of \(25.0 kg * 5.00 m/s = 125 kg*m/s\). Thus, the system has to maintain a momentum of \(1000 kg*m/s + 125 kg*m/s = 1125 kg*m/s\). Therefore, the final velocity (\(v_{f_b}\)) is given by \(1125 kg*m/s / 175 kg = 6.43 m/s\).
04

Solve for the final velocity in case (c)

In case 'c', when the 25.0 kg mass is thrown into the car, the final system mass increases to 225 kg. Because the mass is thrown opposite to the motion, it decreases the system's momentum by an amount of \(25.0 kg * 6.00 m/s = 150 kg*m/s\). The system now has \(1000 kg*m/s - 150 kg*m/s = 850 kg*m/s\). Therefore, the final velocity (\(v_{f_c}\)) is \(850 kg*m/s / 225 kg = 3.78 m/s\).

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Key Concepts

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

Momentum Calculation
Momentum is a fundamental concept in physics, closely related to the movement of objects. It is the product of an object's mass and its velocity, and it helps us understand how objects interact with each other when they collide or exert forces upon one another. The formula to calculate the linear momentum (\(p\) of an object is very straightforward: \[ p = m \times v \] where \(m\) represents the mass of the object and \(v\) its velocity.

For example, if a train with a mass of 200 kg is traveling eastward at a speed of 5 m/s, the momentum can be calculated by multiplying the two: \[ p_{\text{train}} = 200\,\text{kg} \times 5\,\text{m/s} = 1000\,\text{kg}\cdot\text{m/s} \] Understanding the calculation of momentum is crucial for solving problems involving collisions and the movement of objects, as such principles govern many physical phenomena we encounter every day.
Inelastic Collision
An inelastic collision is a type of collision where the colliding objects do not bounce back from each other completely, and instead, some of the kinetic energy is converted into other forms of energy, such as heat or sound. This energy conversion typically results in objects being deformed or stuck together.

However, even in inelastic collisions, the law of conservation of momentum still applies. The total momentum before the collision must equal the total momentum after the collision, assuming no external forces are acting on the system.

For example, in a case where an object is thrown from a moving car, if we analyze the situation before and after the object is thrown, the sum of momenta will remain constant. External factors such as air resistance or friction are typically neglected in textbook problems to simplify the calculations, and to focus on the fundamental concepts of the collision itself.
Momentum Conservation in Physics
Momentum conservation is one of the most profound principles in physics. The law states that in a closed system, where no external forces are present, the total momentum remains constant throughout any process, such as a collision.

When solving a problem regarding momentum conservation, it is essential to consider both the direction and the magnitude of each object's momentum. Adding vector quantities requires attention to these details, as momentum is a vector quantity with both size and direction.

In the equation \[ P_{\text{initial}} = P_{\text{final}} \] \(P_{\text{initial}}\) is the total momentum before the event, and \(P_{\text{final}}\) is the total momentum after the event. To fully understand momentum conservation, students should practice breaking down problems into initial and final states, calculating the momentum for each and ensuring that, without external forces, the momentum totals of the two states are equal. This understanding reinforces the fundamental symmetry of nature and helps in predicting the outcomes of interactions in isolated systems.

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Most popular questions from this chapter

A \(15.0 \mathrm{~kg}\) block is attached to a very light horizontal spring of force constant \(500.0 \mathrm{~N} / \mathrm{m}\) and is resting on a frictionless horizontal table (Fig. \(\mathrm{LX.44}\) ). Suddenly it is struck by a \(3.00 \mathrm{~kg}\) stone traveling horizontally at \(8.00 \mathrm{~m} / \mathrm{s}\) to the right, whereupon the stone rebounds at \(2.00 \mathrm{~m} / \mathrm{s}\) horizontally to the left. Find the maximum distance that the block will compress the spring after the collision.

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Two large blocks of wood are sliding toward cach other on the frictionless surface of a frozen pond. Block \(A\) has mass \(4.00 \mathrm{~kg}\) and is initially sliding cast at \(2.00 \mathrm{~m} / \mathrm{s}\). Block \(B\) has mass \(6.00 \mathrm{~kg}\) and is initially sliding west at \(2.50 \mathrm{~m} / \mathrm{s}\). The blocks collide head-on. Aner the collision block \(B\) is sliding east at \(0.50 \mathrm{~m} / \mathrm{s}\). What is the decrease in the total kinctic encrgy of the two blocks as a result of the collision?

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