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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?

Short Answer

Expert verified
The force \( \overrightarrow{\mathbf{F}}_{2} \) acts for 9.6 ms.

Step by step solution

01

Understand the Impulse-Momentum Theorem

The impulse-momentum theorem states that impulse is equal to the change in momentum of an object, and it can also be expressed as the product of average force \( \overrightarrow{\mathbf{F}} \) and time \( t \): \( I = F \times t \). Here, \( I \) represents impulse, \( F \) is the force, and \( t \) is the time over which force acts.
02

Analyze the Relationship Between Forces and Impulses

We know that the impulse produced by force \( \overrightarrow{\mathbf{F}}_{1} \) is equal to that produced by force \( \overrightarrow{\mathbf{F}}_{2} \). Thus, \( F_1 \times t_1 = F_2 \times t_2 \). Given \( F_1 = 3F_2 \), the equation becomes \( (3F_2) \times t_1 = F_2 \times t_2 \).
03

Solve the Equation for \( t_2 \)

Substitute the time \( t_1 = 3.2 \ ms = 3.2 \times 10^{-3} \ s \) and solve for \( t_2 \). The equation simplifies to \( 3F_2 \times (3.2 \times 10^{-3}) = F_2 \times t_2 \). The \( F_2 \) terms cancel out, giving \( 3 \times (3.2 \times 10^{-3}) = t_2 \).
04

Calculate the Time Interval \( t_2 \)

Perform the multiplication: \( 3 \times 3.2 \times 10^{-3} = 9.6 \times 10^{-3} \). Therefore, the time interval for force \( \overrightarrow{\mathbf{F}}_{2} \) is \( 9.6 \ ms \).

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Key Concepts

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

Understanding Average Force
In physics, an average force refers to the constant force that would produce the same change in an object's motion as a variable force does over a specific time period. It is expressed through the equation of impulse \[ I = \overrightarrow{\mathbf{F}} \times t \] where \( I \) is the impulse, \( \overrightarrow{\mathbf{F}} \) is the average force, and \( t \) is the time over which the force is applied.

When determining average force, it is crucial to understand its relationship to changes in velocity and mass. For the problem at hand, knowing that one force is three times the other offers a direct path to calculating the necessary time intervals. Simplifying the relationship between forces allows solving for unknowns like the time interval \( t_2 \). This relationship is a key tool in analyzing and solving motion-related problems.

The ability to compute average force accurately helps solve real-world challenges, such as determining the force exerted by athletes during sports activities or calculating the impact forces in vehicle collisions.
The Role of Time Interval in Motion
The time interval \( t \) represents the duration over which an average force acts on an object. In the impulse-momentum theorem, this factor plays a significant role as it directly influences the total impulse produced by a force. The equation \[ I = \overrightarrow{\mathbf{F}} \times t \] clearly demonstrates how time interval affects the magnitude of impulse.

In problem-solving, understanding the time interval assists in analyzing how longer or shorter application of force impacts an object's momentum. For example, in our exercise, knowing that force \( \overrightarrow{\mathbf{F}}_{1} \) acts over a time of \( 3.2 \, \text{ms} \) enables us to determine the equivalent time interval \( t_2 \) for force \( \overrightarrow{\mathbf{F}}_{2} \) using the impulse equivalence.

Ultimately, the intuitive grasp of how time interval extends or reduces the effects of force is crucial. It also plays a role in engineering, where designing systems that withstand forces over time depends on precise timing calculations.
Essential Problem Solving in Physics
Problem solving in physics often involves breaking down complicated scenarios into manageable steps. The methodology includes the use of fundamental theorems, like the impulse-momentum theorem, to link known quantities and solve for unknowns. Here’s how the process typically unfolds:
  • Identify what information is given and what is required.
  • Write down relevant formulas and remember their relationships.
  • Substitute known values into these formulas to uncover unknowns.
  • Perform algebraic manipulations to isolate the desired variable.

The given problem elegantly exemplifies this process. By understanding the impact of average forces and time intervals, we solve for \( t_2 \). With force \( \overrightarrow{\mathbf{F}}_{1} \) being three times \( \overrightarrow{\mathbf{F}}_{2} \) and yielding the same impulse, solving the equation for different time intervals becomes straightforward using mathematical expressions.

Practicing this methodical way of thinking equips students to handle not only textbook problems but also real-world situations. Such skills are foundational in developing analytical thinking and practical application of physics concepts.

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Most popular questions from this chapter

A 46-kg skater is standing still in front of a wall. By pushing against the wall she propels herself backward with a velocity of 21.2 m/s. Her hands are in contact with the wall for 0.80 s. Ignore friction and wind resistance. Find the magnitude and direction of the average force she exerts on the wall (which has the same magnitude as, but opposite direction to, the force that the wall applies to her).

A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total mass of the wagon, rider, and rock is \(95.0 \mathrm{kg}\). The mass of the rock is \(0.300 \mathrm{kg}\). Initially the wagon is rolling forward at a speed of \(0.500 \mathrm{m} / \mathrm{s} .\) Then the person throws the rock with a speed of \(16.0 \mathrm{m} / \mathrm{s}\) Both speeds are relative to the ground. Find the speed of the wagon after the rock is thrown directly forward in one case and directly backward in another.

A golf ball strikes a hard, smooth floor at an angle of \(30.0^{\circ}\) and, as the drawing shows, rebounds at the same angle. The mass of the ball is \(0.047 \mathrm{kg},\) and its speed is \(45 \mathrm{m} / \mathrm{s}\) just before and after striking the floor. What is the magnitude of the impulse applied to the golf ball by the floor? (Hint: Note that only the vertical component of the ball's momentum changes during impact with the fl oor, and ignore the weight of the ball.)

The drawing shows a human figure in a sitting position. For purposes of this problem, there are three parts to the figure,and the center of mass of each one is shown in the drawing. These parts are: (1) the torso, neck, and head (total mass \(=41 \mathrm{kg}\) ) with a center of mass located on the \(y\) axis at a point \(0.39 \mathrm{m}\) above the origin, (2) the upper legs (mass \(=17 \mathrm{kg}\) ) with a center of mass located on the \(x\) axis at a point \(0.17 \mathrm{m}\) to the right of the origin, and (3) the lower legs and feet (total mass \(=9.9 \mathrm{kg}\) ) with a center of mass located \(0.43 \mathrm{m}\) to the right of and \(0.26 \mathrm{m}\) below the origin. Find the \(x\) and \(y\) coordinates of the center of mass of the human figure. Note that the mass of the arms and hands (approximately 12\% of the whole-body mass) has been ignored to simplify the drawing.

A 4.00-g bullet is moving horizontally with a velocity of \(+355 \mathrm{m} / \mathrm{s},\) where the \(+\) sign indicates that it is moving to the right (see part \(a\) of the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part \(b\). Note that both blocks are moving after the collision with the bullet. The mass of the first block is \(1150 \mathrm{g}\), and its velocity is \(+0.550 \mathrm{m} / \mathrm{s}\) after the bullet passes through it. The mass of the second block is \(1530 \mathrm{g}\). (a) What is the velocity of the second block after the bullet embeds itself? (b) Find the ratio of the total kinetic energy after the collisions to that before the collisions.

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