Problem 34 - College Physics

Question

The Beretta Model \(92 \mathrm{~S}\) (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 \(g^{\prime}\) s. (b) For how long is the bullet in the barrel?

Step-by-Step Solution

Answer

Acceleration: \(441612.2 \text{ m/s}^2\) and \(45016.8 \, g's\); Time in barrel: \(0.000759 \text{ s}\).

Step 1: Identify the known values

First, identify the given values in the problem. The length of the barrel is the distance, which is \(d = 127 \text{ mm} = 0.127 \text{ m}\). The muzzle velocity, which is the final velocity of the bullet, is \(v = 335 \text{ m/s}\). The initial velocity \(u\) is \(0 \text{ m/s}\) because the bullet starts from rest.

Step 2: Use kinematic equation to find acceleration

Since initial velocity \(u = 0\), final velocity \(v = 335 \text{ m/s}\), and displacement \(d = 0.127 \text{ m}\), use the kinematic equation:\[ v^2 = u^2 + 2ad \]Substitute the known values:\[ (335)^2 = 0^2 + 2a(0.127) \]Solving for \(a\), we get:\[ 112225 = 0.254a \]\[ a = \frac{112225}{0.254} \approx 441612.2 \text{ m/s}^2 \]

Step 3: Convert acceleration to g's

To express the acceleration in terms of \(g\), where \(g = 9.81 \text{ m/s}^2\), divide the calculated acceleration by \(g\):\[ a_{g} = \frac{441612.2}{9.81} \approx 45016.8 \, g's \]

Step 4: Use kinematic equation to find time

Using the kinematic equation that involves time:\[ v = u + at \]Solve for \(t\):\[ 335 = 0 + 441612.2 \times t \]\[ t = \frac{335}{441612.2} \approx 0.000759 \text{ seconds} \]

Step 5: Conclusion

The acceleration of the bullet in the barrel is approximately \(441612.2 \text{ m/s}^2\) or \(45016.8 \, g's\). The bullet remains in the barrel for about \(0.000759 \text{ seconds}\).

Key Concepts

Acceleration Calculation

To understand acceleration calculation in this context, let's break down the process. Acceleration is defined as the change of velocity over time. In this exercise, we're figuring out how fast the bullet speeds up inside the barrel before it exits. The problem assumes the bullet starts from zero speed, gaining velocity uniformly until it exits with a muzzle velocity of \(335 \text{ m/s}\).
Given that the barrel's length is the distance, \(127 \text{ mm}\) or \(0.127 \text{ m}\), we utilize a simple kinematic equation that links velocity, acceleration, and distance:

• \( v^2 = u^2 + 2ad \)

Where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( d \) is the distance. Plugging in \( u = 0 \), \( v = 335 \text{ m/s} \), and \( d = 0.127 \text{ m} \), we calculate acceleration \( a \). The resulting acceleration explains how fiercely the bullet gets pushed within such a short space. This factor is crucial when designing firearms for desired effects.

Muzzle Velocity

Muzzle velocity is the speed of a projectile at the moment it leaves the barrel of a gun. In our problem, this final speed is given as \(335 \text{ m/s}\). Understanding muzzle velocity is essential because it determines not only the bullet's potential impact but also its trajectory and range. A higher muzzle velocity generally means a flatter trajectory and greater range.
Here are some key factors affecting muzzle velocity:

• Barrel Length: In general, a longer barrel allows the expanding gas from the gunpowder more time to accelerate the projectile, thus higher velocity.
• Bullet Mass: A heavier bullet may have lower velocity due to greater inertia than a lighter one.
• Amount and type of gunpowder charge: More explosive force results in higher acceleration.

Muzzle velocity significantly influences other kinematic calculations, including how far and quickly a bullet travels or begins to drop due to gravity.

Kinematic Equations

Kinematic equations are tools to predict and analyze the motion of objects. They apply constant acceleration scenarios, such as the bullet's motion inside a gun barrel. The main kinematic equations include:

• \( v = u + at \)
• \( s = ut + \frac{1}{2} at^2 \)
• \( v^2 = u^2 + 2as \)

In our problem, we use the third equation: \( v^2 = u^2 + 2ad \) to find acceleration. It perfectly suits scenarios where initial velocity \( u \) is zero, and we need to connect the final velocity \( v \) with acceleration \( a \) and the distance \( d \). Another useful equation is \( v = u + at \), used to determine how long the bullet stays in the barrel by isolating time \( t \).
Kinematic principles allow us to breakdown and predict projectile motion, helping in fields ranging from engineering to ballistics.

Projectile Motion

Projectile motion describes the motion of an object thrown into space, influenced by the force of gravity. A bullet experiences this as soon as it exits the barrel, subsequently being at the mercy of gravity as it arcs toward its target. While our focus was the bullet's trajectory in the barrel, understanding projectile motion is still crucial.
Upon exiting, the bullet's path is a curve due to gravity pulling it downward from its initial horizontal velocity. Here's what factors into projectile motion:

• Angle of Projection: Dictates the range and flight path.
• Initial Speed (Muzzle Velocity): A higher speed increases range and maintains a flatter trajectory.
• Gravity: Constant downward acceleration affects vertical motion.

Predicting where a projectile lands involves both horizontal and vertical velocity components. Physics simplifies complex real-world trajectories into understandable models.

Unit Conversion

Accurate calculation often requires proper unit conversion, as shown in this problem. The barrel's length is initially given as \(127 \text{ mm}\), translating to \(0.127 \text{ m}\) for consistency with other measurements. Such conversion ensures compatibility with constant values like acceleration due to gravity \( g = 9.81 \text{ m/s}^2 \).
For conversion:

• Millimeters to Meters: Divide by 1000 there are 1000 mm in a meter.
• Converting acceleration to g's: Divide acceleration in \( \text{m/s}^2 \) by \( 9.81 \text{ m/s}^2 \).

Unit conversion plays a crucial role in calculations, ensuring precision, clarity, and comparability across various measurements in scientific and practical applications.