/*! 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 35 Faster than a speeding bullet! T... [FREE SOLUTION] | 91Ó°ÊÓ

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Faster than a speeding bullet! The Beretta Model 92\(S\) (the standard-issue U.S. army pistol) has a barrel 127 \(\mathrm{mm}\) long. The bullets leave this barrel with a muzzle velocity of 335 \(\mathrm{m} / \mathrm{s}\) (a) What is the acceleration of the bullet while it is in the barrel, assuming it to be constant? Express your answer in \(\mathrm{m} / \mathrm{s}^{2}\) and in \(g^{\prime} \mathrm{s}\) . (b) For how long is the bullet in the barrel?

Short Answer

Expert verified
The bullet's acceleration is approximately 441712.6 m/s² (about 45032.2 g) and it is in the barrel for approximately 0.000758 seconds.

Step by step solution

01

Identify Known Values

We are given:- Length of the barrel, \( s = 127 \text{ mm} = 0.127 \text{ m} \)- Muzzle velocity, \( v = 335 \text{ m/s} \)- Initial velocity, \( u = 0 \text{ m/s} \) (since the bullet starts from rest).We need to find the acceleration \( a \) and the time \( t \) the bullet is in the barrel.
02

Use Kinematic Equation for Acceleration

We employ the equation \( v^2 = u^2 + 2as \).Plug the values into the equation:\[335^2 = 0 + 2a(0.127)\]\[112225 = 0.254a\]Solve for \( a \):\[a = \frac{112225}{0.254} = 441712.5984 \text{ m/s}^2\]
03

Convert Acceleration to g's

We now convert the acceleration \( a \) from \( \text{m/s}^2 \) to \( g \), where \( 1g = 9.81 \text{ m/s}^2 \). Calculate:\[a = \frac{441712.5984}{9.81} \approx 45032.2 \text{ }g\]
04

Compute Time in the Barrel

Use the kinematic equation \( v = u + at \) where\( u = 0 \), \( v = 335 \), and \( a = 441712.6 \text{ m/s}^2 \).Substitute the known values:\[335 = 0 + (441712.6)t\]Solving for \( t \):\[t = \frac{335}{441712.5984} \approx 0.000758 \text{ seconds}\]
05

Finalize and Verify Solution

We have calculated the acceleration \( a \) as \( 441712.6 \text{ m/s}^2 \) (\( \approx 45032.2 g \)) and the time \( t \approx 0.000758 \text{ seconds} \).Both parts of the problem are solved with consistent use of kinematic equations. Verify calculated values by rechecking arithmetic and ensuring proper unit conversions were applied.

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Key Concepts

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

Acceleration
Acceleration is a measure of how quickly an object changes its velocity. In the context of our exercise, the bullet experiences acceleration while traveling inside the barrel of the pistol. Since the bullet starts from rest and reaches its muzzle velocity by the time it exits the barrel, acceleration can be deemed constant during this phase.

By using the kinematic equation, we derived the bullet's acceleration as approximately 441712.6 \( ext{m/s}^2 \). This high value is because the distance is short and the velocity gain needs to be very rapid. It's also important to convert this magnitude of acceleration to "g-forces," which is a measure relative to Earth's gravity. By dividing by 9.81 \( ext{m/s}^2 \), we find that the bullet's acceleration is about 45032.2 g. This means the bullet is accelerating 45032 times stronger than natural gravity.

Understanding acceleration in projectile motion scenarios is crucial for fields like ballistics and engineering, as it informs the design of firearms and ammunition. It also illustrates physics concepts in practical, real-world situations.
Kinematic equations
Kinematic equations are essential tools in physics, especially when dealing with motion-related problems. They describe various aspects of motion, such as velocity, acceleration, time, and displacement, without considering the forces causing them.

In the context of the exercise, we used two main kinematic equations:
  • \[v^2 = u^2 + 2as\]For calculating acceleration where initial velocity \( u = 0 \) and final velocity \( v = 335 \, ext{m/s} \) helped solve for \( a \).
  • \[v = u + at\]This equation helped us solve for the time \( t \) it takes for the bullet to travel through the barrel.
The beauty of these equations is their versatility and simplicity in breaking down complex motion into manageable calculations. They allow users to compute unknown physical quantities when given certain known variables, facilitating the understanding of fundamental physics principles.
Muzzle velocity
Muzzle velocity refers to the speed of a projectile at the moment it leaves the barrel of a firearm. It is one of the most critical factors in ballistics as it directly affects the distance a projectile can travel and its subsequent trajectory.

In our example, the given muzzle velocity is 335 \( ext{m/s} \). This value was pivotal in calculating both acceleration and the time the bullet travels through the barrel using kinematic equations. A higher muzzle velocity typically results in a flatter trajectory, increasing accuracy over longer distances. It is also indicative of the energy transfer from the gunpowder to the bullet.

Understanding muzzle velocity is vital in designing not just firearms, but also related technologies in defense and sport shooting. It highlights the importance of precise calculations and controlled conditions for effective and safe projectile use.
Projectile motion
Projectile motion is a type of motion experienced by an object that is thrown or projected into the air, subject to only the acceleration of gravity. The bullet from a pistol also follows a form of projectile motion once it leaves the barrel, although the initial phase inside the barrel is dominated by linear acceleration. To analyze projectile motion, like that of the bullet, physics uses concepts such as velocity, angle of projection, and time of flight. Our initial exercise focuses on the linear portion, where the bullet steadily accelerates inside the barrel. Once out, the bullet will eventually follow a two-dimensional parabolic trajectory under the influence of gravity, assuming there's no air resistance. Understanding projectile motion helps in a wide array of scenarios beyond firearms, such as sports, engineering projects, and even launching spacecraft. It combines concepts of horizontal and vertical motion to predict an object's path, making it an important aspect of physics and motion studies.

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Most popular questions from this chapter

A test driver at Incredible Motors, Inc., is testing a new model car having a speedometer calibrated to read m/s rather than mi/h. The following series of speedometer readings was obtained during a test run: $$\begin{array}{llllllllll}{\text { Time (s) }} & {0} & {2} & {4} & {6} & {8} & {10} & {12} & {14} & {16} \\ {\text { Velocity }(\mathrm{m} / \mathrm{s})} & {0} & {0} & {2} & {5} & {10} & {15} & {20} & {22} & {22}\end{array}$$ (a) Compute the average acceleration during each 2 s interval. Is the acceleration constant? Is it constant during any part of the test run? (b) Make a velocity-time graph of the data shown, using scales of \(1 \mathrm{cm}=1\) s horizontally and \(1 \mathrm{cm}=\) 2 \(\mathrm{m} / \mathrm{s}\) vertically. Draw a smooth curve through the plotted points. By measuring the slope of your curve, find the magnitude of the instantaneous acceleration at times \(t=9 \mathrm{s}, 13 \mathrm{s}\) and 15 \(\mathrm{s}\) .

A cat drops from a shelf 4.0 ft above the floor and lands on all four feet. His legs bring him to a stop in a distance of 12 \(\mathrm{cm} .\) Calculate (a) his speed when he first touches the floor (ignore air resistance), (b) how long it takes him to stop, and (c) his acceleration (assumed constant) while he is stopping, in \(\mathrm{m} / \mathrm{s}^{2}\) and \(g^{\prime} \mathrm{s}\) .

An elite human sprinter reaches his top speed of 11.8 \(\mathrm{m} / \mathrm{s}\) at a time of 7.02 s after the starting gun. In the first \(1.40 \mathrm{s},\) however, he reaches a speed of \(8.00 \mathrm{m} / \mathrm{s},\) with a nearly constant acceleration. Calculate (a) his maximum acceleration during the starting phase and (b) his average acceleration to top speed. (c) Assuming constant acceleration for the first 1.40 s, how far does he travel during that time?

Two coconuts fall freely from rest at the same time, one from a tree twice as high as the other. (a) If the coconut from the taller tree reaches the ground with a speed \(V\) , what will be the speed (in terms of \(V\) ) of the coconut from the other tree when it reaches the ground? (b) If the coconut from the shorter tree takes time \(T\) to reach the ground, how long (in terms of \(T )\) will it take the other coconut to reach the ground?

An airplane travels 280 \(\mathrm{m}\) down the runway before taking off. Assuming that it has constant acceleration, if it starts from rest and becomes airborne in 8.00 s, how fast \((\) in \(\mathrm{m} / \mathrm{s})\) is it moving at takeoff?

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