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Chapter 7: Problem 1

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

Short Answer

Expert verified
The time interval for \( \overrightarrow{F}_2 \) is 9.6 ms.

Step by step solution

01

Understand the Impulse Formula

Impulse, denoted by \( J \), is calculated by the formula \( J = F \times \Delta t \), where \( F \) is the average force and \( \Delta t \) is the time interval during which the force acts. Since both forces produce the same impulse, we have \( J_1 = J_2 \).
02

Set Up the Equation for Impulse

For force \( \overrightarrow{F}_1 \) with a magnitude \( 3F_2 \) and time interval \( \Delta t_1 = 3.2 \) ms, the impulse is \( J_1 = 3F_2 \times 3.2 \). Since \( J_1 = J_2 \), we can write: \( 3F_2 \times 3.2 = F_2 \times \Delta t_2 \).
03

Solve for \( \Delta t_2 \)

Cancel \( F_2 \) from both sides of the equation: \( 3 \times 3.2 = \Delta t_2 \). This simplifies to \( \Delta t_2 = 9.6 \) ms.

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Key Concepts

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

Average Force
When discussing forces in physics, "average force" refers to the consistent force applied over a specific time period. In real-world scenarios, forces often fluctuate. Thus, by calculating the average force, we get a simplified representation of how much force affects an object over time.
  • The formula for calculating impulse is important here, as impulse equals force times time.
  • So, if you know the impulse, you can understand how the average force acts over that period.
In our exercise, average force \( \overrightarrow{\mathrm{F}}_1 \) is three times larger than \( \overrightarrow{\mathrm{F}}_2 \). This means \( \overrightarrow{\mathrm{F}}_1 \) delivers the same impulse in less time compared to \( \overrightarrow{\mathrm{F}}_2 \). By comparing the magnitudes, we can determine the effect different forces have over their respective periods.
Time Interval
The time interval, denoted as \( \Delta t \), is the duration over which the force is applied. Understanding this concept is crucial because it influences how much work or change the force can produce. In our exercise, the difference in time intervals is key to maintaining equal impulses despite differing force magnitudes. For \( \overrightarrow{\mathrm{F}}_1 \), given as 3.2 ms, the higher force needs less time to achieve the same impulse.
  • Forces applied over shorter intervals with larger magnitudes result in quicker changes.
  • Conversely, smaller forces need longer time intervals to produce the same effect.
The exercise asks us to solve for the time interval of the second force, \( \overrightarrow{\mathrm{F}}_2 \). By understanding how time intervals work with varying forces, we determine \( \Delta t_2 = 9.6 \) ms.
Impulse Formula
The impulse formula is \( J = F \times \Delta t \), where impulse (\( J \)) is the result of force (\( F \)) applied over a time interval (\( \Delta t \)). Impulse reflects the change in momentum resulting from applied forces. In the context of the exercise, we see how manipulating the impulse formula helps solve for unknowns. Since both forces produce the same impulse, the relationship between force and time becomes clear:
  • An increase in force requires a shorter time to achieve the same impulse.
  • A decrease in force means a longer time is needed.
This exercise illustrates one of the fundamental principles of physics: balancing force magnitude and application time to understand real-world dynamics.

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Most popular questions from this chapter

The lead female character in the movie Diamonds Are Forever is standing at the edge of an offshore oil rig. As she fires a gun, she is driven back over the edge and into the sea. Suppose the mass of a bullet is \(0.010 \mathrm{~kg}\) and its velocity is \(+720 \mathrm{~m} / \mathrm{s}\). Her mass (including the gun) is \(51 \mathrm{~kg}\). (a) What recoil velocity does she acquire in response to a single shot from a stationary position, assuming that no external force keeps her in place? (b) Under the same assumption, what would be her recoil velocity if, instead, she shoots a blank cartridge that ejects a mass of \(5.0 \times 10^{-4} \mathrm{~kg}\) at a velocity of \(+720 \mathrm{~m} / \mathrm{s} ?\)

In a performance test two cars take the same time to accelerate from rest up to the same speed. Car A has a mass of \(1400 \mathrm{~kg}\), and car \(\mathrm{B}\) has a mass of \(1900 \mathrm{~kg}\). During the test, which car (a) has the greater change in momentum, (b) experiences the greater impulse, and (c) is acted upon by the greater net average force? In each case, give your reasoning.

Each of these problems consists of Concept Questions followed by a related quantitative Problem. The Concept Questions involve little or no mathematics. They focus on the concepts with which the problems deal. Recognizing the concepts is the essential initial step in any problem-solving technique. Concept Questions In a performance test two cars take the same time to accelerate from rest up to the same speed. Car A has a mass of \(1400 \mathrm{~kg},\) and car \(\mathrm{B}\) has a mass of \(1900 \mathrm{~kg} .\) During the test, which car (a) has the greater change in momentum, (b) experiences the greater impulse, and (c) is acted upon by the greater net average force? In each case, give your reasoning. Problem Each car takes 9.0 s to accelerate from rest to \(27 \mathrm{~m} / \mathrm{s}\). Find the net average force that acts on each car during the test. Verify that your answers are consistent with your answers to the Concept Questions.

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 Figure \(7-11,\) 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) John has a larger mass than Barbara has. He is standing on the \(x\) axis at \(x_{\mathrm{J}}=+9.0 \mathrm{~m}\), while she is standing on the \(x\) axis at \(x_{\mathrm{B}}=+2.0 \mathrm{~m}\). Is their centerof-mass point closer to the \(9.0\) -m point or the \(2.0\) -m point? (b) They switch positions. Is their center-of-mass point now closer to the \(9.0\) -m point or the \(2.0\) -m point? (c) In which direction, toward or away from the origin, does their center of mass move as a result of the switch?
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