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Chapter 7: Problem 1

One average force \(\overline{\vec{F}}_{1}\) 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 \(\overline{\mathrm{F}}_{1}\) acts for a time interval of \(3.2 \mathrm{~ms}\). For what time interval does the average force \(\overline{\mathrm{F}}_{2}\) act?

Short Answer

Expert verified
The average force \( \overline{\mathrm{F}}_{2} \) acts for 9.6 ms.

Step by step solution

01

Understanding the Relationship

Given that the impulse produced by both forces is the same, we use the impulse formula: impulse = force 脳 time interval. Thus, for both forces, we have: \( \overline{\mathrm{F}}_1 \times \Delta t_1 = \overline{\mathrm{F}}_2 \times \Delta t_2 \).
02

Expressing Force Relationship

Since \( \overline{\mathrm{F}}_1 = 3 \overline{\mathrm{F}}_2 \), we can substitute this into our equation for impulse: \( 3 \overline{\mathrm{F}}_2 \times \Delta t_1 = \overline{\mathrm{F}}_2 \times \Delta t_2 \).
03

Solving for Time Interval of Second Force

By dividing both sides by \( \overline{\mathrm{F}}_2 \), we simplify the equation to: \( 3 \Delta t_1 = \Delta t_2 \). We know \( \Delta t_1 = 3.2 \text{ ms} \), thus: \( 3 \times 3.2 \text{ ms} = \Delta t_2 \).
04

Calculate Time Interval

Calculate \( \Delta t_2 \) by multiplying: \( 3 \times 3.2 = 9.6 \). Therefore, \( \Delta t_2 = 9.6 \text{ ms} \).

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Key Concepts

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

Average Force
Average force is a fundamental concept in physics. It's the constant force that would produce the same overall effect on an object as a variable force applied over a given time period.
To understand average force, think of it as distributing the effect of a force evenly across time. No matter how the actual force varies, the average force gives us a reliable measure for calculation.
  • An example is when a soccer player kicks a ball: the average force is what would happen if the player kicked it with a steady force throughout the entire contact.

To find the average force, you can use the formula: \[ \overline{F} = \frac{\text{Total Impulse}}{\text{Time Interval}} \]This equation helps us understand that the average force depends not only on the total impulse but also on how quickly or slowly this impulse is delivered over time.
Time Interval
A time interval refers to the duration over which a force is applied in a given situation. It is a crucial factor in calculating impulse.
In physics problems, knowing the exact time interval is key to understanding how a force impacts an object.
  • The time interval is often measured in seconds, milliseconds, or any suitable unit depending on the context.
  • In the example problem, you dealt with two different time intervals: one for each average force.

To find how long a force acts, you can rearrange the impulse formula:\[ \Delta t = \frac{\text{Impulse}}{\overline{F}} \]This makes it clear that if two different forces result in the same impulse, the time intervals must adjust accordingly. This relationship showcases how the force intensity and duration are interconnected.
Impulse Formula
The impulse formula is central in understanding how forces affect motion. Impulse is defined as the change in momentum of an object when a force is applied over time.
Mathematically, impulse is calculated by:\[ \text{Impulse} = \overline{F} \times \Delta t \]Here, \( \overline{F} \) is the average force, and \( \Delta t \) is the time interval during which the force acts.
  • Impulse measures the total impact of a force, linking both the magnitude of the force and the time period it acts.
  • It's a vector quantity, implying it captures both size and direction.

This formula is particularly useful when solving problems where forces vary or when comparing different forces, as seen in the exercise. Understanding impulse helps explain events like car crashes, where the damage depends not only on the force of the impact but also on how long the impact lasts.

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Most popular questions from this chapter

A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon throws a rock. Does the momentum of the wagon increase, decrease, or remain the same (a) when the rock is thrown directly forward and (b) when the rock is thrown directly backward? (c) In which case does the wagon have the greater speed after the rock is thrown?

Starting with an initial speed of \(5.00 \mathrm{~m} / \mathrm{s}\) at a height of \(0.300 \mathrm{~m},\) a \(1.50-\mathrm{kg}\) ball swings downward and strikes a \(4.60-\mathrm{kg}\) ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the \(1.50-\mathrm{kg}\) ball just before impact. (b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision. (c) How high does each ball swing after the collision, ignoring air resistance?

A car (mass \(=1100 \mathrm{~kg}\) ) is traveling at \(32 \mathrm{~m} / \mathrm{s}\) when it collides head-on with a sport utility vehicle (mass \(=2500 \mathrm{~kg}\) ) traveling in the opposite direction. In the collision, the two vehicles come to a halt. At what speed was the sport utility vehicle traveling?

Multiple-Concept Example 7 presents a model for solving problems such as this one. A \(1055-\mathrm{kg}\) van, stopped at a traffic light, is hit directly in the rear by a \(715-\mathrm{kg}\) car traveling with a velocity of \(+2.25 \mathrm{~m} / \mathrm{s}\). Assume that the transmission of the van is in neutral, the brakes are not being applied, and the collision is elastic. What is the final velocity of (a) the car and (b) the van?

During July 1994 the comet Shoemaker-Levy 9 smashed into Jupiter in a spectacular fashion. The comet actually consisted of 21 distinct pieces, the largest of which had a mass of approximately \(4.0 \times 10^{12} \mathrm{~kg}\) and a speed of \(6.0 \times 10^{4} \mathrm{~m} / \mathrm{s}\). Jupiter, the largest planet in the solar system, has a mass of \(1.9 \times 10^{27} \mathrm{~kg}\) and an orbital speed of \(1.3 \times 10^{4} \mathrm{~m} / \mathrm{s} .\) If this piece of the comet had hit Jupiter head-on, what would have been the change (magnitude only) in Jupiter's orbital speed (not its final speed)?
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