Problem 61 - College Physics

Question

A rifle bullet with mass \(8.00 \mathrm{~g}\) strikes and embeds itself in a block with a mass of \(0.992 \mathrm{~kg}\) that rests on a frictionless, horizontal surface and is attached to a coil spring. (See Figure \(8.50 .\) ) The impact compresses the spring \(15.0 \mathrm{~cm}\). Calibration of the spring shows that a force of \(0.750 \mathrm{~N}\) is required to compress the spring \(0.250 \mathrm{~cm} .\) (a) Find the magnitude of the block's velocity just after impact. (b) What was the initial speed of the bullet?

Step-by-Step Solution

Answer

(a) The block's velocity is approximately 2.6 m/s. (b) The initial speed of the bullet is 325 m/s.

Step 1: Understanding the Spring Constant

First, we'll determine the spring constant (\( k \)) using the relationship between force and displacement: \( F = kx \). Given that a force of \( 0.750 \mathrm{~N} \) compresses the spring by \( 0.250 \mathrm{~cm} \) or \( 0.0025 \mathrm{~m} \), we find \( k \) as follows:\[ k = \frac{F}{x} = \frac{0.750}{0.0025} = 300 \ \mathrm{N/m}. \]

Step 2: Calculate the Potential Energy in the Spring

The potential energy stored in the spring at maximum compression is given by the equation for elastic potential energy: \( U = \frac{1}{2}kx^2 \). With \( x = 0.15 \mathrm{~m} \), we find:\[ U = \frac{1}{2} \cdot 300 \cdot (0.15)^2 = 3.375 \ \mathrm{J}. \]

Step 3: Determine Velocity Using Conservation of Energy

Assuming no energy loss, the kinetic energy just after the bullet embeds in the block is converted to spring potential energy. Use the kinetic energy formula: \( KE = \frac{1}{2} mv^2 \). Set the kinetic energy equal to the potential energy:\[ \frac{1}{2} \times (0.992 + 0.008) \times v^2 = 3.375. \]Solve for \( v \):\[ 0.5 \times 1 \times v^2 = 3.375. \]\[ v^2 = 6.75. \]\[ v = \sqrt{6.75} \approx 2.6 \ \mathrm{m/s}. \]

Step 4: Apply Conservation of Momentum for Bullet's Initial Speed

Use conservation of momentum to find the bullet's initial speed. Before impact, only the bullet has momentum, so:\[ m_b \cdot v_b = (m_b + m_B) \cdot v. \]Where \( m_b = 0.008 \) kg and \( m_B = 0.992 \) kg, and \( v = 2.6 \) m/s as found earlier:\[ 0.008 \cdot v_b = 1 \cdot 2.6. \]Solve for \( v_b \):\[ v_b = \frac{2.6}{0.008} = 325 \ \mathrm{m/s}. \]

Key Concepts

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. It is given by the equation \[ KE = \frac{1}{2} mv^2, \] where \( m \) is the mass of the object and \( v \) is its velocity. This equation tells us that kinetic energy is directly proportional to the mass of the object and the square of its velocity. Thus, even a small increase in velocity results in a large increase in kinetic energy.

In the problem, kinetic energy concepts are used to determine the block's speed after the bullet impacts it. We need to equate the kinetic energy right after the impact to the potential energy stored in the compressed spring. Doing this step helps in calculating the velocity, which is crucial for further steps in determining the bullet's initial speed.

Spring Constant

The spring constant, denoted by \( k \), is a measure of the stiffness of a spring. It tells us how much force is required to compress or extend a spring by a certain distance. The formula linking force and displacement for springs is given as:\[ F = kx, \]where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement.

In the exercise, we use the force and displacement provided to find the spring constant: \( F = 0.750 \, \text{N} \) and \( x = 0.0025 \, \text{m} \). We calculate \( k \) as:

• The given data lead us to a spring constant of 300 N/m.

This calculated spring constant is then utilized to find the elastic potential energy stored in the spring when it is compressed by the bullet-block system.

Elastic Potential Energy

Elastic potential energy is the energy stored in a spring when it is compressed or extended. It is expressed as:\[ U = \frac{1}{2}kx^2, \]where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position. This concept is crucial in our exercise as it helps to determine the energy stored in the spring after the bullet embeds in the block.

By plugging in \( k = 300 \ \, \text{N/m} \) and \( x = 0.15 \ \, \text{m} \), we find the stored elastic potential energy to be 3.375 Joules. Understanding this energy transformation helps in applying conservation of energy principles to solve for velocity, as energy is neither lost nor gained but transformed from one form to another.

Rifle Bullet Impact

When a rifle bullet impacts an object, it transfers its momentum to the object. This transfer can result in the embedding of the bullet into the object, as seen in this problem. The impact compresses the spring attached to the object based on the momentum transfer.

Initially, the bullet possesses all the motion and energy, which it imparts to the block upon impact. The embedded bullet-block system then moves together, compressing the spring and storing energy in the form of elastic potential energy.

• This component of the exercise demonstrates how applying momentum and energy conservation principles can solve real-world impact problems.

Physics Problem Solving

Solving physics problems often involves breaking down complex scenarios into simpler conceptual steps. In this exercise, we apply both conservation of energy and conservation of momentum. These are fundamental principles in physics that provide a framework for solving intricate problems.

First, we calculate how energy is converted between forms using the kinetic energy and elastic potential energy equations. Then, we use the conservation of momentum to find initial parameters, such as the bullet's initial speed.

• By following a step-by-step approach, students can develop a systematic problem-solving strategy.
• This strategy is applicable across various physics problems, bolstering their comprehension and application skills.

Remember, breaking problems down and careful analysis are key to mastering physics.