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Chapter 7: Problem 28

After sliding down a snow-covered hill on an inner tube, Ashley is coasting across a level snowfield at a constant velocity of \(+2.7 \mathrm{m} / \mathrm{s} .\) Miranda runs after her at a velocity of \(+4.5 \mathrm{m} / \mathrm{s}\) and hops on the inner tube. How fast do the two slide across the snow together on the inner tube? Ashley's mass is \(71 \mathrm{kg}\) and Miranda's is \(58 \mathrm{kg} .\) Ignore the mass of the inner tube and any friction between the inner tube and the snow.

Short Answer

Expert verified
They slide together at approximately 3.51 m/s.

Step by step solution

01

Identify the Conservation Principle

To solve this problem, we will use the principle of conservation of linear momentum. Since there are no external forces acting on the system (ignoring friction), the total momentum of the system before Miranda jumps onto the tube is equal to the total momentum after she jumps on.
02

Calculate Initial Momentum

First, calculate the initial momentum of Ashley and Miranda separately.*Ashley:* The initial momentum of Ashley is given by: \[ p_{Ashley} = m_{Ashley} \cdot v_{Ashley} = 71 \, \text{kg} \cdot 2.7 \, \text{m/s} \]\[ p_{Ashley} = 191.7 \, \text{kg m/s} \]*Miranda:* The initial momentum of Miranda is given by: \[ p_{Miranda} = m_{Miranda} \cdot v_{Miranda} = 58 \, \text{kg} \cdot 4.5 \, \text{m/s} \]\[ p_{Miranda} = 261 \, \text{kg m/s} \]
03

Calculate Total Initial Momentum

The total initial momentum of the system is the sum of the momenta of Ashley and Miranda before the interaction (before Miranda jumps on the tube).\[ p_{initial} = p_{Ashley} + p_{Miranda} = 191.7 \, \text{kg m/s} + 261 \, \text{kg m/s} = 452.7 \, \text{kg m/s} \]
04

Set Up the Equation for Final Momentum

After Miranda hops onto the tube, their masses combine, and they slide with a common velocity \( v_f \). According to the conservation of momentum,\[ p_{initial} = p_{final} \]\[ 452.7 \, \text{kg m/s} = (m_{Ashley} + m_{Miranda}) \cdot v_f \]\[ 452.7 \, \text{kg m/s} = (71 \, \text{kg} + 58 \, \text{kg}) \cdot v_f \]
05

Solve for Final Velocity

Now solve the equation for \( v_f \) to find their final velocity. \[ 452.7 \, \text{kg m/s} = 129 \, \text{kg} \cdot v_f \]\[ v_f = \frac{452.7 \, \text{kg m/s}}{129 \, \text{kg}} \approx 3.51 \, \text{m/s} \]

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Key Concepts

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

Initial Momentum
The concept of **initial momentum** plays a crucial role in problems involving the conservation of linear momentum. To start, let's define momentum. Momentum is a measure of the motion of an object and is calculated by multiplying the object's mass by its velocity. It helps us understand how hard it would be to stop that object. In this exercise, both Ashley and Miranda contribute to the system's initial momentum before Miranda hops onto the inner tube.
  • Ashley's Momentum: Based on her mass and velocity, we can calculate Ashley's initial momentum using the formula: \( p_{Ashley} = m_{Ashley} \cdot v_{Ashley} \). Substituting her mass (71 kg) and velocity (2.7 m/s) gives \( p_{Ashley} = 191.7 \, \text{kg m/s} \). This tells us how much momentum Ashley has while coasting.

  • Miranda's Momentum: Similarly, Miranda's initial momentum can be computed. Using her mass (58 kg) and her faster running speed of 4.5 m/s, we find \( p_{Miranda} = 261 \, \text{kg m/s} \). This represents Miranda's momentum as she approaches the inner tube.
The total initial momentum is the sum of these two individual momenta, forming the foundation for calculating later changes in the system.
Final Velocity
The **final velocity** describes how fast Ashley and Miranda slide together on the inner tube after Miranda jumps on. It relies heavily on the concept of conservation of linear momentum. In cases where no external forces like friction significantly alter the system, the total momentum before and after an interaction stays the same. Here, this principle allows us to find the combined velocity after everything settles.
The equation we set is:
  • \( p_{initial} = p_{final} \)
  • \( 452.7 \, \text{kg m/s} = (m_{Ashley} + m_{Miranda}) \cdot v_f \)
Substituting the combined mass of both Ashley and Miranda (129 kg) into the equation and rearranging gives us the final velocity:
  • \( v_f = \frac{452.7 \, \text{kg m/s}}{129 \, \text{kg}} \approx 3.51 \, \text{m/s} \)
This result shows how quickly they move together, maintaining the momentum balance set up before Miranda's hop.
Momentum Calculation
When tackling physics problems that involve motion, understanding the process of **momentum calculation** is fundamental. In this exercise, our goal was to find the velocity after an interaction by using momentum calculations. Here鈥檚 how we systematically approached it:
  • Step 1: Determine Individual Momenta
    Calculate each person's momentum using their mass and velocity, providing a piece of the total initial picture. This part of the calculation breaks down the effect of each person's motion.

  • Step 2: Sum of Initial Momentum
    Add these values to determine the system's total initial momentum. This step reveals the starting state for combined motion calculations.

  • Step 3: Apply Conservation of Momentum
    Using the conservation principle, equate the initial total momentum to the final total momentum post-interaction. This allows for solving unknowns like the final velocity, ensuring the total quantity of motion remains unchanged.
This sequence is not only used in this exercise but can be applied to any problem involving momentum where interactions happen without external influences. Understanding these steps helps students unlock many physics scenarios with similar dynamics.

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Most popular questions from this chapter

A cannon of mass \(5.80 \times 10^{3} \mathrm{kg}\) is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires an \(85.0-\mathrm{kg}\) shell horizontally with an initial velocity of \(+551 \mathrm{m} / \mathrm{s}\). Suppose the cannon is then unbolted from the earth, and no external force hinders its recoil. What would be the velocity of an identical shell fired by this loose cannon? (Hint: In both cases assume that the burning gunpowder imparts the same kinetic energy to the system.)

Multiple-Concept Example 7 deals with some of the concepts that are used to solve this problem. A cue ball (mass \(=0.165 \mathrm{kg}\) ) is at rest on a frictionless pool table. The ball is hit dead center by a pool stick, which applies an impulse of \(+1.50 \mathrm{N} \cdot \mathrm{s}\) to the ball. The ball then slides along the table and makes an elastic head-on collision with a second ball of equal mass that is initially at rest. Find the velocity of the second ball just after it is struck.

One average force \(\overrightarrow{\mathbf{F}}_{1}\) has a magnitude that is three times as large as that of another average force \(\overrightarrow{\mathbf{F}}_{2} .\) Both forces produce the same impulse. The average force \(\overrightarrow{\mathbf{F}}_{1}\) acts for a time interval of \(3.2 \mathrm{ms}\). For what time interval does the average force \(\overrightarrow{\mathbf{F}}_{2}\) act?

The carbon monoxide molecule (CO) consists of a carbon atom and an oxygen atom separated by a distance of \(1.13 \times 10^{-10} \mathrm{m}\). The mass \(m_{\mathrm{c}}\) of the carbon atom is 0.750 times the mass \(m_{\mathrm{o}}\) of the oxygen atom, or \(m_{\mathrm{C}}=0.750 \mathrm{m}_{\mathrm{o}}\). Determine the location of the center of mass of this molecule relative to the carbon atom.

A projectile (mass \(=0.20 \mathrm{kg}\) ) is fired at and embeds itself in a stationary target (mass \(=2.50 \mathrm{kg}\) ). With what percentage of the projectile's incident kinetic energy does the target (with the projectile in it) fly off after being struck?
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