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(II) A 12-kg hammer strikes a nail at a velocity of 7.5 m/s and comes to rest in a time interval of 8.0 ms. \((a)\) What is the impulse given to the nail? \((b)\) What is the average force acting on the nail?

Short Answer

Expert verified
Impulse is -90 Ns and the average force is -11250 N.

Step by step solution

01

Understanding Impulse

Impulse is defined as the change in momentum of an object when a force is applied over a time interval. It can be calculated using the formula \( J = ext{Δ}p = m(v_f - v_i) \) where \( m \) is the mass, \( v_f \) is the final velocity, and \( v_i \) is the initial velocity. Given the hammer comes to rest, \( v_f \) is 0.
02

Calculating Initial Momentum Change

First, we identify the initial velocity as \( v_i = 7.5 \, ext{m/s} \) and the mass \( m = 12 \, ext{kg} \). The final velocity \( v_f = 0 \, ext{m/s} \) since it comes to rest. Thus, the change in momentum \( ext{Δ}p \) is \( 12 imes (0 - 7.5) = -90 \, ext{kg} \, ext{m/s} \). This is also the impulse \( J \) provided to the nail.
03

Determining the Impulse

The impulse \( J \), which equals the change in momentum, is \( -90 \, ext{Ns} \). The negative sign indicates a reduction in velocity towards rest.
04

Understanding Average Force

To find the average force, we use the relationship between impulse and force: \( J = F_{ ext{avg}} imes ext{Δ}t \), where \( F_{ ext{avg}} \) is the average force and \( ext{Δ}t \) is the time interval during which the force acts.
05

Calculating Average Force

Given \( ext{Δ}t = 8.0 \, ext{ms} = 8.0 imes 10^{-3} \, ext{s} \) and impulse \( J = -90 \, ext{Ns} \), we find \( F_{ ext{avg}} = rac{-90}{8.0 imes 10^{-3}} = -11250 \, ext{N} \). The negative sign indicates the force's direction is opposite to the initial motion direction.

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Key Concepts

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

Impulse
Impulse is an important concept in physics that can help us understand how forces interact with objects over short periods. When you think about impulse, think about it as the influence that changes an object's momentum.
It happens when a force acts on an object over a certain time. The formula to calculate impulse is given by\[ J = \Delta p = m(v_f - v_i) \]
where:
  • \( J \) is the impulse.
  • \( m \) is the mass of the object.
  • \( v_f \) is the final velocity.
  • \( v_i \) is the initial velocity.
In the exercise, the hammer initially has a velocity of 7.5 m/s and then comes to rest, meaning its final velocity is 0 m/s.
This change in velocity results in a change in momentum, providing us with the impulse given to the nail.
Keep in mind the sign of the impulse can indicate direction; in this case, a negative impulse shows that the hammer's velocity is reduced till it stops.
Average Force
Average force is the constant force that would produce the same change in an object's motion during the same time interval as the actual variable force does.
Knowing the impulse, we can relate it to average force through the formula:\[ J = F_{\text{avg}} \times \Delta t \]
where:
  • \( J \) is the impulse.
  • \( F_{\text{avg}} \) is the average force.
  • \( \Delta t \) is the time interval over which the force acts.
The given problem provides a time interval of 8.0 ms, or 8.0 \( \times \) 10^{-3} seconds.
Using the impulse value of -90 Ns, we can solve for the average force:\[ F_{\text{avg}} = \frac{-90}{8.0 \times 10^{-3}} = -11250 \text{ N} \]
The negative sign tells us that the force's direction is opposite to the direction the hammer was initially moving.
Momentum Change
Understanding momentum change is crucial for analyzing collisions and other interactions in physics. Momentum itself is the product of an object's mass and its velocity:\[ p = m \times v \]
The larger the momentum, the harder it is to stop or change the object's direction.
In this exercise, the hammer experiences a momentum change due to a force acting over time.
Here's how it breaks down:
  • Initial momentum: The hammer starts with a momentum \( 12 \times 7.5 = 90 \text{ kg m/s} \).
  • Final momentum: As the hammer comes to rest, its momentum becomes 0 \text{ kg m/s}.
  • Change in momentum: This equals \(-90 \text{ kg m/s}\), the same as the impulse.
Momentum change gives us a clear indication of how much motion has been altered due to a given force, which is why it's directly connected to impulse.

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Most popular questions from this chapter

(III) A 725-kg two-stage rocket is traveling at a speed of \(6.60 \times 10^3 m/s\) away from Earth when a predesigned explosion separates the rocket into two sections of equal mass that then move with a speed of \(2.80 \times 10^3 m/s\) relative to each other along the original line of motion. \((a)\) What is the speed and direction of each section (relative to Earth) after the explosion? \((b)\) How much energy was supplied by the explosion? [\(Hint\):What is the change in kinetic energy as a result of the explosion?]

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