Problem 83 - University Physics with Modern Physics

Question

A rifle bullet with mass 8.00 \(\mathrm{g}\) strikes and embeds itself in a block with mass 0.992 \(\mathrm{kg}\) that rests on a frictionless, horizontal surface and is attached to a coil spring (Fig. P8.83). 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

The block's velocity after impact is \( v \approx 5.88 \ \mathrm{m/s} \) and the initial bullet speed is \( u_b \approx 735 \ \mathrm{m/s} \).

Step 1: Understand Hooke's Law and Spring Constant

Hooke's Law states that the force required to compress or extend a spring by some distance is proportional to that distance. The formula for Hooke's Law is \( F = kx \). Here, \( k \) is the spring constant and \( x \) is the displacement of the spring from its equilibrium position. From the problem, a force of \( 0.750 \ \mathrm{N} \) compresses the spring by \( 0.250 \ \mathrm{cm} \). First, let's convert \( x \) from centimeters to meters: \( x = 0.250 \ \mathrm{cm} = 0.00250 \ \mathrm{m} \). We can find the spring constant \( k \) using \( k = \frac{F}{x} = \frac{0.750}{0.00250} \ \mathrm{N/m} \).

Step 2: Calculate the Energy Stored in the Spring

The energy stored in a compressed spring is given by the potential energy formula \( PE = \frac{1}{2}kx^2 \). Use the displacement \( x = 15.0 \ \mathrm{cm} = 0.150 \ \mathrm{m} \) and the spring constant \( k \) from Step 1. This energy is equal to the initial kinetic energy of the bullet-block system after impact. Therefore, set \( PE = \frac{1}{2}mv^2 \) to find the block's velocity just after the impact.

Step 3: Solve for the Block's Velocity After Impact

Set the potential energy of the spring equal to the kinetic energy of the block-bullet system after the impact: \( \frac{1}{2}kx^2 = \frac{1}{2}m_{total}v^2 \), where \( m_{total} = 1.000 \ \mathrm{kg} \). Simplify the equation to solve for the velocity \( v \): \( v = \sqrt{\frac{kx^2}{m_{total}}} \). Substitute in the known values to find \( v \).

Step 4: Determine the Initial Speed of the Bullet

Using the law of conservation of momentum, calculate the initial speed of the bullet. The momentum before and after the impact must be equal. Before the impact, the momentum is \( m_b u_b \). After the impact, it is \( m_{total} v \). Hence, \( m_b u_b = m_{total} v \). Solve this equation for the initial speed of the bullet \( u_b = \frac{m_{total} v}{m_b} \). Substitute \( m_b = 0.008 \ \mathrm{kg} \), \( m_{total} \), and \( v \) from Step 3 to find \( u_b \).

Key Concepts

Hooke's Law

Hooke's Law is a fundamental principle that relates the force needed to compress or extend a spring to the distance it is compressed or extended. This principle can be written as \( F = kx \), where:

• \( F \) is the force applied to the spring.
• \( k \) is the spring constant, a measure of the spring's stiffness.
• \( x \) is the change in position from the spring's natural length.

The spring constant \( k \) provides insight into how difficult it is to stretch or compress a spring. A higher value indicates a stiffer spring. In our given problem, a force of 0.750 N compresses the spring by 0.250 cm (or 0.00250 m). Using the formula, the spring constant is calculated as \( k = \frac{0.750}{0.00250} = 300 \ \mathrm{N/m} \). This shows the resistance of the spring against deformation.

Spring Constant

The spring constant \( k \) is a crucial parameter indicating the stiffness of a spring. It tells us how much force is needed per meter of displacement. In layman's terms, it's about how strongly the spring resists attempts to change its shape.
Consider the example from our problem. When a force of 0.750 N causes a tiny compression of 0.250 cm, we say the spring is relatively stiff because even a small force causes only a slight change. This is quantified by a spring constant of 300 N/m.

• A larger \( k \) means a stiffer spring that can handle more force without much deformation.
• A smaller \( k \) indicates a more flexible spring, deforming easily under force.

Understanding the spring constant is essential for predicting how a spring behaves under various loads, impacting everything from engineering designs to everyday applications like mattress springs.

Kinetic Energy

Kinetic energy represents the energy possessed by an object due to its motion. It is expressed as \( KE = \frac{1}{2} mv^2 \), where:

• \( m \) is the object's mass.
• \( v \) is the velocity.

In our scenario, after the bullet embeds itself into the block, the combined system moves with a certain velocity, having kinetic energy. Initially, this kinetic energy is equivalent to the potential energy stored in the spring when fully compressed. By equating the potential energy in the spring with the kinetic energy of the block-bullet system, we find the velocity of the system post-impact.

• This conversion illustrates the transformation from stored energy in the spring to the energy of motion.
• It emphasizes the role of dynamics in understanding the behavior of moving objects post-collision.

Kinetic energy is pivotal for calculating system behavior in collisions, driving our understanding of motion.

Potential Energy

Potential energy is the stored energy in a system, due to its position or configuration. For springs, this energy can be calculated using the formula \( PE = \frac{1}{2} k x^2 \), where:

• \( k \) is the spring constant.
• \( x \) is the displacement from equilibrium.

In this problem, the potential energy within the compressed spring becomes crucial. At its highest compression of 15.0 cm (0.150 m), it holds energy that transforms into kinetic energy, moving the bullet-block system.

• This type of energy highlights how stored mechanical energy can do work or convert to other forms.
• Understanding potential energy enables us to predict how systems store or release energy over time.

Overall, potential energy measures the potential for doing work and plays an integral role in many physical processes, ensuring conservation across energy transformations.