/*! 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 29 Suppose a gangster sprays Superm... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose a gangster sprays Superman's chest with \(3 \mathrm{~g}\) bullets at the rate of 100 bullets \(/ \mathrm{min}\), 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?

Short Answer

Expert verified
The average force is approximately 5 N.

Step by step solution

01

Identify Given Information

The problem provides several key pieces of information: the mass of each bullet is \(3 \text{ g} = 0.003 \text{ kg}\), the rate of bullets is \(100 \text{ bullets per minute}\), the speed of each bullet is \(500 \text{ m/s}\), and the bullets rebound with no change in speed.
02

Calculate Change in Momentum for One Bullet

When a bullet rebounds, its velocity changes direction. The initial momentum is \(mv\) and the final momentum is \(-mv\) since the bullet rebounds back. The change in momentum for one bullet is therefore \( \Delta p = mv - (-mv) = 2mv \). Thus, for one bullet, \( \Delta p = 2 \times 0.003 \times 500 = 3 \text{ kg} \cdot \text{m/s} \).
03

Calculate Rate of Change of Momentum

The rate of change of momentum is equal to the force applied, which can be found by multiplying the change in momentum per bullet by the number of bullets per second. Convert the rate of bullets from minutes to seconds: \(100 \text{ bullets/minute} = \frac{100}{60} \text{ bullets/second} \approx 1.67 \text{ bullets/second} \). The force \( F \) is then \( F = 1.67 \times 3 \approx 5 \text{ N} \).
04

Final Step: Compute the Average Force

Since the rate of change of momentum equals force, the average force exerted on Superman's chest is approximately \(5 \text{ N}\).

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Key Concepts

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

Momentum
Momentum is a fundamental concept in physics that describes the motion of objects. It is a vector quantity, meaning it has both magnitude and direction. The momentum of an object is the product of its mass and velocity, given by the formula: \[ p = mv \] where
  • \( p \) is the momentum,
  • \( m \) is the mass of the object, and
  • \( v \) is the velocity of the object.
The importance of momentum lies in its conservation in isolated systems. When bullets hit Superman, they change direction and therefore their momentum. The change in momentum, denoted as \( \Delta p \), can be used to determine the force Superman experiences. Each bullet's velocity changes from going forward to backward, creating a change in momentum that's twice the initial momentum value. Understanding this helps us calculate resulting forces using principles that are consistent with real-world experiences.
Force calculation
Force is another critical concept in physics that can often be connected back to momentum changes. Newton's second law of motion directly relates the net force acting on an object to the rate of change of its momentum. Essentially, force can be calculated using the formula: \[ F = \frac{\Delta p}{\Delta t} \] where
  • \( F \) is the force,
  • \( \Delta p \) is the change in momentum, and
  • \( \Delta t \) is the change in time.
In the given scenario, the bullets' change in momentum is derived from their initial forward movement and subsequent rebound. By multiplying the momentum change of a single bullet by the frequency of bullets impacting per second, we find the total force exerted. This process involves translating the momentum change due to the rebounding bullets into a measurable force exerted over time.
Newton's laws
Newton's laws of motion provide the foundation for understanding dynamics in physical systems. In this context, the first and third laws are particularly relevant. The third law states that for every action, there is an equal and opposite reaction. This is observed when bullets, upon impacting Superman's chest, exert force and bounce back, which is consistent with Newton's action-reaction principle. The second law, which links force and momentum, is at the core of calculating the force experienced by Superman. It describes how the rate of change in momentum (or the acceleration of mass) influences the force exerted. By converting the bullets' momentum change into force, while considering the rate of bullet impacts, we've adhered to the second law. Newton's laws simplify the examination of complex interactions by enabling us to predict how forces will influence the motions of objects.

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Most popular questions from this chapter

Two particles \(P\) and \(Q\) are released from rest \(1.0 \mathrm{~m}\) apart. \(P\) has a mass of \(0.10 \mathrm{~kg}\), and \(Q\) a mass of \(0.30 \mathrm{~kg} . P\) and \(Q\) attract each other with a constant force of \(1.0 \times 10^{-2} \mathrm{~N}\). No external forces act on the system. (a) What is the speed of the center of mass of \(P\) and \(Q\) when the separation is \(0.50 \mathrm{~m} ?(\mathrm{~b})\) At what distance from \(P\) 's original position do the particles collide?

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An object is tracked by a radar station and determined to have a position vector given by \(\vec{r}=(3500-160 t) \hat{\mathrm{i}}+2700 \hat{\mathrm{j}}+300 \hat{\mathrm{k}}\), with \(\vec{r}\) in meters and \(t\) in seconds. The radar station's \(x\) axis points east, its \(y\) axis north, and its \(z\) axis vertically up. If the object is a \(250 \mathrm{~kg}\) meteorological missile, what are (a) its linear momentum, (b) its direction of motion, and (c) the net force on it?

In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land hard on the floor. Suppose the victim falls by \(0.50 \mathrm{~m}\), the mass that moves downward is \(70 \mathrm{~kg}\), and the collision on the floor lasts \(0.082\) s. What are the magnitudes of the (a) impulse and (b) average force acting on the victim from the floor during the collision?

A ball having a mass of \(150 \mathrm{~g}\) strikes a wall with a speed of \(5.2\) \(\mathrm{m} / \mathrm{s}\) and rebounds with only \(50 \%\) of its initial kinetic energy. (a) What is the speed of the ball immediately after rebounding? (b) What is the magnitude of the impulse on the wall from the ball? (c) If the ball is in contact with the wall for \(7.6 \mathrm{~ms}\), what is the magnitude of the average force on the ball from the wall during this time interval?

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