/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 \(\bullet\) A 5.00 g bullet trav... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\bullet\) A 5.00 g bullet traveling horizontally at 450 \(\mathrm{m} / \mathrm{s}\) is shot through a 1.00 kg wood block suspended on a string 2.00 \(\mathrm{m}\) long. If the center of mass of the block rises a distance of 0.450 \(\mathrm{cm},\) find the speed of the bullet as it emerges from the block.

Short Answer

Expert verified
The bullet's speed as it emerges is approximately 449.98 m/s.

Step by step solution

01

Calculate Initial Kinetic Energy of Bullet

The initial kinetic energy of the bullet is given by the formula \( KE_i = \frac{1}{2} m v^2 \). Here, the mass \( m \) is 0.005 kg (converted from 5.00 g) and the initial velocity \( v \) is 450 m/s. The initial kinetic energy is:\[KE_i = \frac{1}{2} \times 0.005 \times (450)^2 = 506.25 \text{ J}\]
02

Calculate Maximum Height Potential Energy of Block

The potential energy at the maximum height of the block is given by \( PE = mgh \). Here, the mass \( m \) is 1.00 kg, \( g \) is 9.81 m/s², and \( h \) is 0.0045 m (converted from 0.450 cm). Thus,\[PE = 1.00 \cdot 9.81 \cdot 0.0045 = 0.044145 \text{ J}\]
03

Use Conservation of Energy for the Bullet-Block System

By conservation of energy, the total energy before the bullet enters the block is equal to the total energy after plus the energy used to raise the block's center of mass. Thus,\[KE_i = KE_f + PE \]where \( KE_f \) is the kinetic energy of the bullet after passing through the block. Let's denote the final velocity of the bullet by \( v_f \).
04

Calculate Remaining Kinetic Energy

Rearranging and solving for the final kinetic energy of the bullet:\[KE_f = KE_i - PE = 506.25 - 0.044145 = 506.205855 \text{ J}\]
05

Solve for Final Velocity of Bullet

Using the kinetic energy formula \( KE_f = \frac{1}{2} m v_f^2 \) to solve for the final velocity \( v_f \):\[506.205855 = \frac{1}{2} \times 0.005 \times v_f^2\]\[v_f^2 = \frac{506.205855 \times 2}{0.005} = 202482.342\]\[v_f = \sqrt{202482.342} \approx 449.98 \text{ m/s}\]
06

Conclusion: Final Result

The final speed of the bullet as it emerges from the block is approximately 449.98 m/s.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Kinetic Energy Calculation
Kinetic energy is a fundamental concept in physics that describes the energy an object possesses due to its motion. It is calculated using the formula:
  • \( KE = \frac{1}{2} mv^2 \)
where:
  • \( m \) is the mass of the object (in kilograms), and
  • \( v \) is the velocity of the object (in meters per second).
In our problem, we consider a bullet with mass 0.005 kg traveling at an initial speed of 450 m/s.
Substituting these values into the kinetic energy formula gives us:
  • \( KE_i = \frac{1}{2} \times 0.005 \times (450)^2 = 506.25 \) Joules.
This calculation shows us the enormous energy contained in the bullet due to its high speed.
Understanding kinetic energy is crucial when analyzing how energy is transformed during motion, especially in collision scenarios like this one.
Potential Energy
Potential energy represents the stored energy of an object due to its position within a force field, commonly a gravitational field. The formula for gravitational potential energy is:
  • \( PE = mgh \)
where:
  • \( m \) is the mass of the object (in kilograms),
  • \( g \) is the acceleration due to gravity (approximately 9.81 m/s² on Earth), and
  • \( h \) is the height the object is raised (in meters).
In the given scenario, when the bullet strikes the block, it causes the block to rise, increasing its potential energy.
Using the values from the problem, the mass of the block is 1.00 kg, the rise in height \( h \) is 0.0045 m (converted from 0.450 cm), and substituting into the equation gives us:
  • \( PE = 1.00 \times 9.81 \times 0.0045 = 0.044145 \) Joules.
This potential energy determines how high the block rises due to the kinetic energy transferred from the bullet.
Energy Conservation in Collisions
The law of conservation of energy is a key principle in physics which states that energy cannot be created or destroyed; it can only be transformed from one form to another.
In collisions, analyzing how energy is transferred between objects can reveal vital insights. For our bullet-block system, the conservation of energy principle helps us calculate the bullet's velocity after exiting the block.
  • The initial total energy equals the sum of all forms of energy after the event.
In this exercise, the initial kinetic energy of the bullet transforms into the kinetic energy of the bullet after the collision and the potential energy of the raised block.
  • Using the conservation equation: \( KE_i = KE_f + PE \), we find the kinetic energy remaining in the bullet after transferring some energy to the block.
This results in the final kinetic energy, \( KE_f \), from which we can determine the bullet's final speed. The steps included solving:
  • Remaining kinetic energy: \( KE_f = 506.25 - 0.044145 = 506.205855 \) Joules.
  • Using \( KE_f = \frac{1}{2} mv_f^2 \) proceeds to find the bullet’s new velocity, \( v_f \approx 449.98 \) m/s.
Thus, energy conservation provides the framework for understanding the transformation and conservation of energy in physical processes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\bullet\) A blue puck with mass 0.0400 \(\mathrm{kg}\) , sliding with a velocity of magnitude 0.200 \(\mathrm{m} / \mathrm{s}\) on a frictionless, horizontal air table, makes a perfectly elastic, head-on collision with a red puck with mass \(m,\) initially at rest. After the collision, the velocity of the blue puck is 0.050 \(\mathrm{m} / \mathrm{s}\) in the same direction as its initial velocity. Find (a) the velocity (magnitude and direction) of the red puck after the collision; and (b) the mass \(m\) of the red puck.

Rocket failure! Just as it has reached an upward speed of 5.0 \(\mathrm{m} / \mathrm{s}\) during a vertical launch, a rocket explodes into two pieces. Photographs of the explosion reveal that the lower piece, with a mass one-fourth that of the upper piece, was moving downward at 3.0 \(\mathrm{m} / \mathrm{s}\) the instant after the explosion. (a) Find the speed of the upper piece just after the explosion. (b) How high does the upper piece go above the point where the explosion occurred?

\(\bullet\) A movie stuntman (mass 80.0 \(\mathrm{kg}\) ) stands on a window ledge 5.0 \(\mathrm{m}\) above the floor (Fig. \(8.45 ) .\) Grabbing a rope attached to a chandelier, he swings down to grapple with the movie's villain (mass 70.0 \(\mathrm{kg}\) ), who is standing directly under the chandelier. (Assume that the stuntman's center of mass moves downward 5.0 \(\mathrm{m} .\) He releases the rope just as he reaches the villain.) (a) With what speed do the entwined foes start to slide across the floor? (b) If the coefficient of kinetic friction of their bodies with the floor is \(\mu_{k}=0.250,\) how far do they slide?

\(\cdot\) The speed of the fastest-pitched baseball was \(45 \mathrm{m} / \mathrm{s},\) and the ball's mass was 145 \(\mathrm{g}\) . (a) What was the magnitude of the mo- mentum of this ball, and how many joules of kinetic energy did it have? (b) How fast would a 57 gram ball have to travel to have the same amount of (i) kinetic energy, and (ii) momentum?

\(\bullet\) Detecting planets around other stars. Roughly 500 planets have so far been detected beyond our solar system. This is accomplished by looking for the effect the planet has on the star. The star is not truly stationary; instead, it and its planets orbit around the center of mass of the system. Astronomers can measure this wobble in the position of a star.(a) For a star with the mass and size of our sun and having a planet with five times the mass of Jupiter, where would the center of mass of this system be located, relative to the center of the star, if the distance from the star to the planet was the same as the distance from Jupiter to our sun? (Consult Appendix E.) (b) If the planet had earth's mass, where would the center of mass of the system be located if the planet was just as far from the star as the earth is from the sun? (c) In view of your results in parts (a) and (b), why is it much easier to detect stars having large planets rather than small ones?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.