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Suppose 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?

Short Answer

Expert verified
The average force on Superman's chest is 5.01 N.

Step by step solution

01

Understand the Problem

We need to find the average force exerted on Superman's chest when 100 bullets per minute, each with a mass of 3 g and moving at 500 m/s, hit and rebound off his chest.
02

Convert Units

First, convert the mass of each bullet from grams to kilograms: \(3\, \text{g} = 0.003\, \text{kg}\). Also, convert the rate of bullets from per minute to per second: \(100\, \text{bullets/min} = \frac{100}{60}\, \text{bullets/s} = \approx 1.67\, \text{bullets/s}\).
03

Calculate Impulse per Bullet

For each bullet, compute the change in momentum (impulse) as it rebounds. The change in momentum for one bullet that rebounds directly back is: \(\Delta p = m(v - (-v)) = 2mv\). So, \(\Delta p = 2 \times 0.003 \times 500\, \text{kg m/s} = 3\, \text{kg m/s}\).
04

Calculate Total Impulse per Second

Since there are approximately 1.67 bullets hitting per second, calculate the total impulse per second: Total impulse per second \(= 1.67 \times 3 = 5.01\, \text{kg m/s}\).
05

Determine Average Force

The average force exerted on Superman's chest is the total impulse per second, which is equivalent to the change in momentum per second: \(F_{\text{avg}} = 5.01\, \text{N}\).

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Key Concepts

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

Newton's Second Law
Newton's Second Law is one of the foundational principles in physics. It tells us how the motion of an object changes under the influence of different forces. This law is often expressed with the equation \(F = ma\), where \(F\) represents force, \(m\) is mass, and \(a\) is acceleration. When looking at the problem of bullets hitting Superman, we can think of Newton's Second Law as explaining how the change in motion (or the stopping and rebounding of bullets) is related to the force applied. Each bullet experiences a force when it hits and bounces back. This repetitive event creates an average force, calculated by considering the change in speed and mass of each bullet.In essence, Newton's Second Law allows us to calculate how the combined impact of many small forces (like those from multiple bullets) results in a significant average force on Superman's chest. Recognizing these changes in momentum and how they apply forcefully illustrates the power of Newton's principles in real-world scenarios.
Impulse and Momentum
Understanding impulse and momentum is crucial when tackling problems involving collisions or impacts in physics. Momentum is defined as the product of an object's mass and velocity, expressed as \(p = mv\). Impulse, on the other hand, refers to the change in momentum resulting from a force acting over time, and is given by the formula \(\Delta p = F\Delta t\). In the case of Superman and the bullets, each bullet carries a certain momentum as it moves toward and then away from his chest. The impulse each bullet experiences is due to the force exerted while it changes direction upon impact, resulting in a doubled change in momentum, \(\Delta p = 2mv\). By calculating this change for every bullet and considering the rate at which bullets hit (and rebound), we derive a continuous sequence of impulses. This helps us calculate the total impulse applied each second and thus the average force experienced by Superman's chest. The understanding of impulse and momentum not only makes solving problems more manageable but also enhances our grasp of how forces interact with moving objects.
Unit Conversion
In physics, unit conversion is essential for ensuring the accuracy of calculations. Different units measure the same quantities, such as mass, time, or speed. Converting these units to a common format allows for consistency and correctness in applying mathematical formulas. Before calculating forces or momenta, converting the bullet's mass from grams to kilograms is necessary since the standard unit in physics for mass is kilograms. Similarly, when handling rates, like bullets per minute, converting this to bullets per second aligns with the SI unit for time, seconds. Here's how critical unit conversion is: - By converting 3 grams to 0.003 kilograms, calculations become compatible with other values measured in standard units. - Changing the rate from 100 bullets per minute to approximately 1.67 bullets per second enables precise computations related to momentum and force. Unit conversion simplifies complex problems by standardizing measurements, allowing students to apply physics laws more effortlessly. It serves as a foundational skill in physics problem-solving, particularly when dealing with diverse units in real-world contexts.

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Most popular questions from this chapter

During a lunar mission, it is necessary to increase the speed of a spacecraft by \(2.2 \mathrm{~m} / \mathrm{s}\) when it is moving at \(400 \mathrm{~m} / \mathrm{s}\) relative to the Moon. The speed of the exhaust products from the rocket engine is \(1000 \mathrm{~m} / \mathrm{s}\) relative to the spacecraft. What fraction of the initial mass of the spacecraft must be burned and ejected to accomplish the speed increase?

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball moves at \(3.50 \mathrm{~m} / \mathrm{s}\) along a line making an angle of \(22.0^{\circ}\) with the cue ball's original direction of motion, and the second ball has a speed of \(2.00 \mathrm{~m} / \mathrm{s}\). Find (a) the angle between the direction of motion of the second ball and the original direction of motion of the cue ball and (b) the original speed of the cue ball. (c) Is kinetic energy (of the centers of mass, don't consider the rotation) conserved?

Two \(2.0 \mathrm{~kg}\) bodies, \(A\) and \(B\), collide. The velocities before the collision are \(\vec{v}_{A}=(15 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}\) and \(\vec{v}_{B}=(-10 \hat{\mathrm{i}}+5.0 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}\). After the collision, \(\vec{v}_{A}^{\prime}=(-5.0 \hat{1}+20 \hat{j}) \mathrm{m} / \mathrm{s} .\) What are (a) the final velocity of \(B\) and (b) the change in the total kinetic energy (including sign)?

A \(5.0 \mathrm{~kg}\) block with a speed of \(3.0 \mathrm{~m} / \mathrm{s}\) collides with a 10 kg block that has a speed of \(2.0 \mathrm{~m} / \mathrm{s}\) in the same direction. After the collision, the \(10 \mathrm{~kg}\) block travels in the original direction with a speed of \(2.5 \mathrm{~m} / \mathrm{s}\). (a) What is the velocity of the \(5.0 \mathrm{~kg}\) block immediately after the collision? (b) By how much does the total kinetic energy of the system of two blocks change because of the collision? (c) Suppose, instead, that the \(10 \mathrm{~kg}\) block ends up with a speed of \(4.0 \mathrm{~m} / \mathrm{s}\). What then is the change in the total kinetic energy? (d) Account for the result you obtained in (c).

A \(3.0 \mathrm{~kg}\) object moving at \(8.0 \mathrm{~m} / \mathrm{s}\) in the positive direction of an \(x\) axis has a one-dimensional elastic collision with an object of mass \(M\), initially at rest. After the collision the object of mass \(M\) has a velocity of \(6.0 \mathrm{~m} / \mathrm{s}\) in the positive direction of the axis. What is mass \(M ?\)

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