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91Ó°ÊÓ

A 75 kg man rides on a \(39 \mathrm{~kg}\) cart moving at a velocity of \(2.3 \mathrm{~m} / \mathrm{s}\). He jumps off with zero horizontal velocity relative to the ground. What is the resulting change in the cart's velocity, including sign?

Short Answer

Expert verified
The cart's velocity increases by 4.423 m/s.

Step by step solution

01

Understanding Initial Conditions

We start by understanding that when the man is on the cart, the initial total momentum of the system is the sum of the momenta of the man and the cart, both moving at 2.3 m/s. The total initial momentum is given by the formula \( p_{initial} = (m_{man} + m_{cart}) \times v_{initial} \), where \( m_{man} = 75 \, \text{kg} \), \( m_{cart} = 39 \, \text{kg} \), and \( v_{initial} = 2.3 \, \text{m/s} \).
02

Calculating Initial Momentum

Calculate the initial momentum of the entire system using\[ p_{initial} = (75 \text{ kg} + 39 \text{ kg}) \times 2.3 \text{ m/s} = 114 \text{ kg} \times 2.3 \text{ m/s} = 262.2 \text{ kg} \cdot \text{m/s}. \]
03

Understanding Final Conditions

After the man jumps off with zero horizontal velocity relative to the ground, the man has zero final momentum, and only the cart contributes to the system's final momentum.
04

Calculating Man's Momentum After Jumping Off

The man's momentum after jumping is zero because his velocity relative to the ground is zero. Mathematically, this means \( p_{man_{final}} = 75 \text{ kg} \times 0 \text{ m/s} = 0 \text{ kg} \cdot \text{m/s} \).
05

Applying Conservation of Momentum

According to the conservation of momentum, the initial momentum of the system must be equal to the final momentum. Therefore, \( p_{cart_{final}} = p_{initial} - p_{man_{final}} = 262.2 \text{ kg} \cdot \text{m/s}.\)
06

Calculating Cart's Final Velocity

The cart’s final velocity \( v_{cart_{final}} \) can be calculated using \( p_{cart_{final}} = m_{cart} \cdot v_{cart_{final}} \). Substituting the numbers, we have \( 262.2 \text{ kg} \cdot \text{m/s} = 39 \text{ kg} \cdot v_{cart_{final}} \). Solving for \( v_{cart_{final}} \), we get \( v_{cart_{final}} = \frac{262.2}{39} \approx 6.723 \text{ m/s}.\)
07

Calculating the Change in Cart's Velocity

The change in the cart's velocity \( \Delta v_{cart} \) is the difference between its final and initial velocities: \( \Delta v_{cart} = v_{cart_{final}} - v_{initial} = 6.723 \text{ m/s} - 2.3 \text{ m/s} = 4.423 \text{ m/s}.\)

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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 quantity of motion an object possesses. It is calculated as the product of an object’s mass and its velocity. Mathematically, it is expressed as:
\[ p = m \times v \]where:
  • \( p \) is momentum, measured in kilogram meters per second (\( ext{kg} imes ext{m/s} \))
  • \( m \) is mass, measured in kilograms (\( ext{kg} \))
  • \( v \) is velocity, measured in meters per second (\( ext{m/s} \))
The concept of momentum is crucial in understanding how objects move and interact. According to the law of conservation of momentum, the total momentum of an isolated system remains constant if no external forces act on it. This principle helps predict how objects will behave after they collide or interact in other ways.
When the man jumps off the cart, the total momentum before and after this event must remain the same, allowing us to calculate the resulting change in velocity.
Velocity
Velocity is one of the primary determinants in calculating momentum and is essential in understanding motion. Unlike speed, which is only a measure of how fast something is moving, velocity is directional. This means that it takes into account both the speed of an object and the direction of its movement.
The formula for velocity is:
\[ v = \frac{d}{t} \]where:
  • \( v \) is velocity, measured in meters per second (\( ext{m/s} \))
  • \( d \) is the distance traveled, measured in meters (\( ext{m} \))
  • \( t \) is the time taken, measured in seconds (\( ext{s} \))
In our exercise, the man's initial velocity on the cart was 2.3 m/s. When he jumps off, his horizontal velocity becomes zero, which impacts the system's momentum. The calculation of the cart's velocity change shows how crucial velocity is when predicting an object's motion post-interaction.
Physics Problem Solving
Solving physics problems, like the one in the exercise, requires a structured approach to ensure all aspects of a question are addressed. Here’s a simplified method to solve such problems:
  • **Understand the Problem**: Identify all known quantities (masses, velocities) and unknowns (final velocities, changes in momentum). Diagrammatically representing the system may help clarify interactions.
  • **Apply Relevant Physics Principles**: Use principles, such as conservation of momentum, to set up mathematical equations. This involves keeping track of all initial and final values to ensure the system's properties are appropriately compared.
  • **Solve the Equations**: Carefully solve the relevant equations. Pay attention to sign conventions; for instance, a negative velocity may indicate movement in the opposite direction.
  • **Check Your Work**: Once a solution is obtained, verifying if it aligns logically with physical laws and all given conditions is vital. Review initial steps if errors are suspected.
This systematic approach helps build understanding and develops habits for tackling increasingly complex problems in physics.

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Most popular questions from this chapter

The last stage of a rocket, which is traveling at a speed of 7600 \(\mathrm{m} / \mathrm{s}\), consists of two parts that are clamped together: a rocket case with a mass of \(290.0 \mathrm{~kg}\) and a payload capsule with a mass of \(150.0\) \(\mathrm{kg}\). When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of \(910.0 \mathrm{~m} / \mathrm{s}\). What are the speeds of (a) the rocket case and (b) the payload after they have separated? Assume that all velocities are along the same line. Find the total kinetic energy of the two parts (c) before and (d) after they separate. (e) Account for the difference.

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