/*! 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 14 It is well known that bullets an... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

It is well known that bullets and other missiles fired at Superman simply bounce off his chest (Fig.7-22). Suppose that a gangster sprays Superman's chest with \(3 \mathrm{~g}\) bullets at the rate of 100 bullets/min, and the speed of each bullet is \(500 \mathrm{~m} / \mathrm{s}\). Suppose too that the bullets rebound straight back with no change in speed. What is the magnitude of the average force on Superman's chest from the stream of bullets?

Short Answer

Expert verified
The magnitude of the average force on Superman's chest is 5 N.

Step by step solution

01

Understand the Problem

Determine the initial and final momentum, the rate of change of momentum, and how to use these to find the force applied to Superman's chest.
02

Calculate the Momentum of One Bullet

Use the formula for momentum: \[ p = mv \] where \(m = 3 \text{ g} = 0.003 \text{ kg}\) and \(v = 500 \text{ m/s}\). Therefore, \[ p = 0.003 \times 500 = 1.5 \text{ kgâ‹…m/s} \]
03

Determine the Change in Momentum for One Bullet

Since the bullet rebounds with no change in speed, it has an initial momentum of \(1.5 \text{ kgâ‹…m/s}\) and a final momentum of \(-1.5 \text{ kgâ‹…m/s}\). The change in momentum \( \triangle p \) is calculated as: \[ \triangle p = p_{final} - p_{initial} = -1.5 - 1.5 = -3 \text{ kgâ‹…m/s} \]
04

Calculate the Total Change in Momentum per Minute

Given the rate of 100 bullets per minute, the total change in momentum per minute is: \[ \triangle p_{total} = 100 \times (-3) = -300 \text{ kgâ‹…m/s} \]
05

Convert Time to Seconds and Calculate Average Force

1 minute is 60 seconds. The average force, \( F_{avg} \), can be calculated using the formula: \[ F_{avg} = \frac{ \triangle p_{total} }{ \triangle t } = \frac{ -300 }{ 60 } = -5 \text{ N} \] The magnitude of the force is \( 5 \text{ N} \).

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.

momentum
Momentum is a fundamental concept in physics, describing the quantity of motion an object has. Momentum is defined as the product of an object's mass and velocity, represented by the formula: \( p = mv \). Here, 'p' is momentum, 'm' is mass, and 'v' is velocity.

In the problem, each bullet has a mass of \(3 \text{ g} \) or \(0.003 \text{ kg} \) and a speed of \(500 \text{ m/s} \). Using the momentum formula, we calculated the momentum of one bullet to be: \[ p = 0.003 \times 500 = 1.5 \text{ kgâ‹…m/s} \]

Understanding momentum helps us analyze changes in motion when objects interact. In this scenario, the gangster’s bullets hit and bounce back from Superman's chest without slowing down, leading to a reversal in momentum.
force calculation
Force is what causes an object to accelerate, and it is calculated using Newton’s second law of motion, which states that force is equal to the rate of change of momentum. The formula is: \( F = \frac{ \triangle p }{ \triangle t } \) where \( \triangle p \) is the change in momentum and \( \triangle t \) is the time interval.

In the exercise, each bullet changes its momentum when it bounces back. The initial momentum is \(1.5 \text{ kgâ‹…m/s} \) and the final momentum is \(-1.5 \text{ kgâ‹…m/s} \). Therefore, the change in momentum for one bullet is: \[ \triangle p = p_{ \text{ final }} \text{ - } p_{ \text{ initial }} \text{ = } (-1.5) - 1.5 = -3 \text{ kgâ‹…m/s} \]

Given 100 bullets per minute, the total change in momentum per minute is \(-300 \text{ kg⋅m/s} \). Converting this into seconds, the average force on Superman’s chest is: \[ F_{ \text{ avg }} = \frac{ -300 }{ 60 } = -5 \text{ N} \] The magnitude of which is \(5 \text{ N} \).
projectile motion
Projectile motion describes the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. In this exercise, bullets are moving towards Superman's chest, and they follow a linear path due to the high speed and close range.

For projectile motion, key factors include initial velocity, angle of projection, and effects of gravity. Even if the bullets aren't falling freely, understanding projectile motion helps grasp how objects move when launched. Each bullet has a speed of \(500 \text{ m/s} \) and travels in a straight line towards Superman before rebounding.

Factors like air resistance and trajectory can influence projectile motion, but with bullets and a direct path, these effects are minimal. Hence, this problem simplifies to linear motion with a rebound effect, aligning with fundamental projectile motion concepts.

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

In February 1955, a paratrooper fell \(370 \mathrm{~m}\) from an airplane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was \(56 \mathrm{~m} / \mathrm{s}\) (terminal speed), that his mass (including gear) was \(85 \mathrm{~kg}\), and that the magnitude of the force on him from the snow was at the survivable limit of \(1.2 \times 10^{5} \mathrm{~N}\). What are (a) the minimum depth of snow that would have stopped him safely and (b) the magnitude of the impulse on him from the snow?

The last stage of a rocket, which is traveling at a speed of \(7600 \mathrm{~m} / \mathrm{s}\), consists of two parts that are clamped together: a rocket case with a mass of \(290.0 \mathrm{~kg}\) and a payload capsule with a mass of \(150.0 \mathrm{~kg}\). When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of \(910.0 \mathrm{~m} / \mathrm{s}\). What are the speeds of (a) the rocket case and (b) the payload after they have separated? Assume that all velocities are along the same line.

Isaac Newton studied many types of collisions and invented the definition of momentum about twenty years before he developed his three laws of motion. As a result of his observations of collision processes, he formulated the law of conservation of momentum as a statement of experimental fact. Let's assume for the sake of argument that Newton had already defined the concepts of force and momentum but had not yet formulated his laws of motion. Also assume that he had an electronic force sensor and was able to verify the impulse-momentum theorem. Explain in words how Newton could use the impulse- momentum theorem and the law of conservation of momentum to predict the existence of the third law of motion and to explain the nature of the interaction forces between two colliding objects.

A projectile body of mass \(m_{A}\) and initial \(x\) component velocity \(v_{A x}\left(t_{1}\right)=10.0 \mathrm{~m} / \mathrm{s}\) collides with an initially stationary target body of mass \(m_{B}=2.00 m_{A}\) in a one-dimensional collision. What is the velocity of \(m_{B}\) following the collision if the two masses stick together?

Car with Grain A railroad car moves at a constant speed of \(3.20 \mathrm{~m} / \mathrm{s}\) under a grain elevator. Grain drops into it at the rate of \(540 \mathrm{~kg} / \mathrm{min}\). What is the magnitude of the force needed to keep the car moving at constant speed if friction is negligible?

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.