Problem 53 - College Physics

Question

A rifle weighs \(50.0 \mathrm{~N}\) and its barrel is \(0.750 \mathrm{~m}\) long. It shoots a 25.0-g bullet, which leaves the barrel at a speed (muzzle velocity) of \(300 \mathrm{~m} / \mathrm{s}\) after being uniformly accelerated. What is the magnitude of the force exerted on the rifle by the bullet?

Step-by-Step Solution

Answer

The force exerted on the rifle by the bullet is 1500 N.

Step 1: Identify the Given Values

- Weight of the rifle: \(50.0 \text{ N}\).- Length of the barrel: \(0.750 \text{ m}\).- Mass of the bullet: \(25.0 \text{ g} = 0.025 \text{ kg}\).- Muzzle velocity of the bullet: \(300 \text{ m/s}\).

Step 2: Calculate Bullet's Acceleration

Use the formula for uniformly accelerated motion: \[ v^2 = u^2 + 2as \]where \( v = 300 \text{ m/s} \), \( u = 0 \text{ m/s} \), and \( s = 0.750 \text{ m} \). Re-arrange to find acceleration \( a \):\[ a = \frac{v^2 - u^2}{2s} = \frac{300^2 - 0^2}{2 \times 0.750} = \frac{90000}{1.5} = 60000 \text{ m/s}^2 \].

Step 3: Calculate the Force Exerted by the Bullet

Using Newton's second law, \( F = ma \), where \( m = 0.025 \text{ kg} \) and \( a = 60000 \text{ m/s}^2 \):\[ F = 0.025 \times 60000 = 1500 \text{ N} \].

Key Concepts

Uniformly Accelerated Motion

When we talk about uniformly accelerated motion, we are describing a situation where an object's velocity changes at a constant rate. This type of motion is described by a set of equations, one of the most common being the equation \( v^2 = u^2 + 2as \), wherein:

• \( v \) represents the final velocity, in this case, the bullet's muzzle velocity of 300 m/s.
• \( u \), the initial velocity, is 0 m/s because we assume the bullet starts at rest.
• \( a \) is the acceleration which we need to find.
• \( s \) is the distance covered during acceleration, which is the length of the rifle's barrel, 0.750 m.

This equation allows us to find the acceleration by rearranging it: \( a = \frac{v^2 - u^2}{2s} \). For the bullet, this computes to an astounding 60000 m/s², which indicates how quickly the bullet's speed increases over the short barrel length.

Muzzle Velocity

Muzzle velocity is a term used to describe the speed of a projectile, in this case, a bullet, as it exits the barrel of a firearm. It is a crucial factor because it affects the bullet's range and impact. In the given exercise, the bullet has a muzzle velocity of 300 m/s, a considerable speed which is achieved in a very short time. This velocity results from the uniformly accelerated motion within the rifle barrel.
High muzzle velocity is often desired in firearms as it contributes to a flatter trajectory and greater accuracy. To achieve such high speeds, the bullet must experience high acceleration, which is facilitated by the strong force exerted as gases expand in the rifle barrel upon firing. Understanding muzzle velocity is essential for calculating other factors like force and energy involved in the shooting process.

Force Calculation

To find the force exerted on the rifle by the bullet, we utilize Newton's Second Law, which states that \( F = ma \), where \( F \) is the force applied, \( m \) is the mass of the object, and \( a \) is its acceleration. In this exercise, the mass of the bullet is 0.025 kg, converted from grams, and its acceleration is the previously calculated 60000 m/s².
By multiplying the mass and the acceleration, \( F = 0.025 \times 60000 \), we find that the force is 1500 N. This is a substantial force, which explains the recoil often felt when firing a firearm. Calculating the force helps illustrate the principles of physics involved in shooting and how these principles ensure that the bullet achieves the desired velocity and trajectory.