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Chapter 2: Problem 81

Certain rifles can fire a bullet with a speed of 965 \(\mathrm{m} / \mathrm{s}\) just as it leaves the muzzle (this speed is called the muzzle velocity). If the muzzle is 70.0 \(\mathrm{cm}\) long and if the bullet is accelerated uniformly from rest within it, (a) what is the acceleration (in \(g^{\prime}\) s) of the bullet in the muzzle, and (b) for how long (in ms) is it in the muzzle? (c) If, when this rifle is fired vertically, the bullet reaches a maximum height \(H,\) what would be the maximum height (in terms of \(H\) ) for a new rifle that produced half the muzzle velocity of this one?

Short Answer

Expert verified
(a) The acceleration is approximately 67812.92 \(g\). (b) The bullet is in the muzzle for about 1.451 ms. (c) The new maximum height is \((1/4)H\).

Step by step solution

01

Convert Units

Convert the length of the muzzle from centimeters to meters for consistency in units. Since 1 cm = 0.01 m, the muzzle length is 70.0 cm = 0.70 m.
02

Calculate Acceleration

Use the kinematic equation to find the acceleration:\[ v^2 = u^2 + 2as \]where \( v = 965 \, \text{m/s} \) is the final velocity, \( u = 0 \, \text{m/s} \) is the initial velocity, \( a \) is the acceleration, and \( s = 0.70 \, \text{m} \) is the distance. Rearrange to solve for \( a \):\[ a = \frac{v^2 - u^2}{2s} = \frac{965^2 - 0^2}{2 \times 0.70} \]\[ a = \frac{931225}{1.4} \approx 665160.71 \, \text{m/s}^2 \]
03

Convert Acceleration to g's

The acceleration due to gravity, \( g \), is approximately \( 9.81 \, \text{m/s}^2 \). To convert the acceleration found to units of \( g \), divide by \( g \):\[ \frac{665160.71}{9.81} \approx 67812.92 \, g \]
04

Calculate Time in Muzzle

Use the kinematic equation to find the time \( t \):\[ v = u + at \]Rearrange to solve for \( t \):\[ t = \frac{v - u}{a} = \frac{965 - 0}{665160.71} \approx 0.001451 \, \text{s} \]Convert seconds to milliseconds: \( 0.001451 \, \text{s} = 1.451 \, \text{ms} \)
05

Determine Maximum Height for New Rifle

For the new rifle, the muzzle velocity is half:\[ v_{\text{new}} = \frac{965}{2} = 482.5 \, \text{m/s} \]Using energy conservation, the height \( H \) achieved by a projectile is proportional to the square of its initial velocity:\[ H_{\text{new}} = \left( \frac{v_{\text{new}}}{v} \right)^2 H = \left( \frac{482.5}{965} \right)^2 H = \frac{1}{4}H \]

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Key Concepts

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

Muzzle Velocity
Muzzle velocity refers to the speed at which a projectile, such as a bullet, leaves the muzzle of a firearm. It is a critical factor in determining the range and accuracy of the projectile.
Muzzle velocity can be affected by several factors:
  • The type of gunpowder or propellant used.
  • The length of the gun barrel.
  • The mass of the projectile.
Generally, a higher muzzle velocity results in flatter trajectories and longer ranges. In the given exercise, a rifle with a muzzle velocity of 965 meters per second accelerates the bullet uniformly along a 70 cm barrel.
Acceleration
Acceleration is the rate at which an object changes its velocity. In kinematics, it is a vector quantity, meaning it has both magnitude and direction.
For the bullet in the exercise, the calculation of acceleration involves using the kinematic equation:
\[ a = \frac{v^2 - u^2}{2s} \]
  • v is the final velocity (965 m/s),
  • u is the initial velocity (0 m/s),
  • s is the distance over which the acceleration occurs (0.70 m).
Here, the bullet's acceleration was calculated to be approximately 665160.71 m/s虏. To put that into perspective relative to Earth's gravity, we express it in terms of g's, where 1 g \( \approx 9.81 \, \text{m/s}^2 \). This gives an acceleration of around 67812.92 g's, which is immensely larger than everyday accelerations experienced on Earth.
Projectile Motion
Projectile motion describes the trajectory of an object that is thrown or propelled near the surface of the Earth. A key factor in projectile motion is the initial velocity, specifically the muzzle velocity in the case of firearms.
Two components define the projectile's path:
  • Horizontal motion, which remains constant barring air resistance.
  • Vertical motion, which is affected by gravity pulling the projectile downward.
For this exercise, when the rifle fires vertically, the bullet reaches a maximum height where its vertical velocity component is zero. The higher the initial velocity, the higher the projectile can go. This concept is used when calculating the height reached by the bullet when fired from different rifles.
Kinematic Equations
Kinematic equations are formulas that relate the motion of an object to time when acceleration is constant. These equations allow us to compute unknown values such as displacement, initial velocity, final velocity, acceleration, and time.
There are several principal kinematic equations, including:
  • \( v = u + at \) - relates velocity with time and acceleration.
  • \( s = ut + \frac{1}{2}at^2 \) - relates displacement with initial velocity, time, and acceleration.
  • \( v^2 = u^2 + 2as \) - relates velocities and acceleration with displacement, used in the exercise.
These equations are vital tools in physics and engineering, helping predict future motions of projectiles.
In the step-by-step solution, they provided a systematic approach to find the time the bullet stays in the muzzle and the potential height if fired vertically.

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Most popular questions from this chapter

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