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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 each case, assuming that the handcar does 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. \((b) A 25.0-k g\) 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
(a) 5.00 m/s, (b) 5.71 m/s, (c) 3.78 m/s.

Step by step solution

01

Understand Conservation of Momentum

The law of conservation of momentum states that if no external forces act on a system, the total momentum of the system remains constant. For a system that involves a car and a mass being thrown, apply conservation of momentum to find the final velocity of the car.
02

Initial Momentum Calculation

Calculate the initial momentum of the system (handcar plus contents). Since the total mass is 200 kg and it is traveling east at 5.00 m/s, the initial momentum \( p_{\text{initial}} = m \cdot v = 200 \, \text{kg} \times 5.00 \, \text{m/s} = 1000 \, \text{kg} \cdot \text{m/s} \).
03

Case (a) analyze mass throw to the side

If the mass is thrown sideways, it has no horizontal component relative to the initial velocity of the car, hence it does not affect the car's velocity. Thus, the final velocity of the car remains the same as the initial velocity. \( v_f = 5.00 \, \text{m/s} \).
04

Case (b) analyze mass thrown backward

When the mass is thrown backward relative to the car, we adjust our calculation as the mass is thrown opposite to the motion of the car, changing its momentum. Set up the equation for momentum conservation: \( 1000 \, \text{kg} \cdot \text{m/s} = (m_f \cdot v_f) + (m_{\text{mass}} \cdot v_{\text{mass}}) \), where \( m_f = 175 \, \text{kg} \), \( v_{\text{mass}} = 5.00 \, \text{m/s} \). Solve for \( v_f \).
05

Compute final velocity for case (b)

Rearrange the equation to find \( v_f \): \( v_f = \frac{1000 \, \text{kg} \cdot \text{m/s} - 25 \, \text{kg} \cdot (-5.00 \, \text{m/s})}{175 \, \text{kg}} = 5.71 \, \text{m/s} \). The final velocity of the car increases to 5.71 m/s to the east.
06

Case (c) analyze mass thrown into the car

For a mass thrown into the car from the opposite direction to its motion, use the conservation of momentum: \( 1000 \, \text{kg} \cdot \text{m/s} + (25 \, \text{kg} \cdot (-6.00 \, \text{m/s})) = (225 \, \text{kg}) \cdot v_f \).
07

Compute final velocity for case (c)

Solve for \( v_f \): \( v_f = \frac{1000 \, \text{kg} \cdot \text{m/s} - 150 \, \text{kg} \cdot \text{m/s}}{225 \, \text{kg}} = 3.78 \, \text{m/s} \). Therefore, the final velocity of the car is 3.78 m/s east.

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Key Concepts

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

Physics Problem Solving
When tackling any physics problem, especially those involving complex concepts like momentum and impulse, it is important to systematically break down the scenario.
Start by carefully reading the problem and identifying the known variables, including mass, initial velocities, and any forces or impulses. Then, determine what exactly needs to be solved or understood.
  • Begin by writing down all known information and drawing a diagram if needed.
  • Think about which fundamental principles of physics (such as conservation laws) apply to the problem.
  • Translate the physical situation into mathematical equations. Often this involves inserting the known values into physical laws like speed, velocity, or momentum.
  • Keep your units consistent throughout your calculations to avoid mistakes.
  • Finally, solve the equations step-by-step, ensuring that each step logically follows from the last.
Physics problem-solving is like putting together a puzzle; every piece of information helps. By methodically analyzing what's given, you can find the missing pieces needed to complete the understanding of the problem.
Impulse and Momentum
The concepts of impulse and momentum are pivotal when discussing motion. These ideas allow us to describe how objects move and interact with forces.
Momentum is essentially a measure of an object's motion, defined as the product of an object's mass and velocity (\( p = m imes v \)). It is a vector quantity, meaning it has both magnitude and direction.
Impulse refers to a change in momentum, resulting from a force applied over a period of time (\( ext{Impulse} = ext{Force} imes ext{time} \)). This change can alter an object's velocity and direction.
  • In the absence of external forces, momentum is conserved. This principle is fundamental in analyzing collisions and separations in physics problems.
  • Impulse provides a way to handle forces that are not constant—a common scenario in real-world situations, like a ball hitting a bat.
  • For any system, if you calculate the total momentum before an event and compare it to the total momentum after, they should be equal if the system is closed and no external forces are acting.
Understanding both impulse and momentum helps in solving problems involving moving objects and how they interact with forces.
Newton's Laws of Motion
Newton's Laws of Motion are the foundation of classical mechanics, providing the principles needed to understand the motion of objects. Each of these laws offers insight into the behavior of objects and forces:
1. **Newton's First Law (Inertia)**: An object at rest stays at rest, and an object in motion stays in motion at a constant velocity unless acted upon by a net external force. This is often called the law of inertia.
- In the context of the railroad handcar, if not for changes like throwing off a mass, the handcar would continue traveling at its initial speed.2. **Newton's Second Law (F=ma)**: The force acting on an object is equal to the mass of that object multiplied by its acceleration (\( F = m imes a \)).
- This law helps us calculate changes in motion due to added or reduced forces, which are crucial for understanding how and why velocities change in momentum problems.3. **Newton's Third Law (Action and Reaction)**: For every action, there is an equal and opposite reaction.
- This explains why the car’s velocity changes when a mass is thrown out; as one mass moves in one direction, there is an equal momentum change in the car, maintaining momentum conservation.Understanding these laws allows us to predict precisely how objects will move, interact, and react to forces. They're fundamental in examining everyday phenomena and complex mechanical systems alike.

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Most popular questions from this chapter

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