Problem 67 - College Physics

Question

A rifle bullet with mass 8.00 g strikes and embeds itself in a block with a mass of 0.992 kg that rests on a frictionless, horizontal surface and is attached to a coil spring. (See Figure \(8.47 . )\) 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 just after impact is 2.60 m/s. (b) The initial speed of the bullet was 325 m/s.

Step 1: Find the Spring Constant

First, let's calculate the spring constant, \( k \). Using Hooke's Law, \( F = kx \), we can rearrange it to \( k = \frac{F}{x} \). Given that \( F = 0.750 \) N and \( x = 0.0025 \) m (converting 0.250 cm to meters), the spring constant can be computed as follows:\[ \begin{align*} k & = \frac{0.750 \text{ N}}{0.0025 \text{ m}} \ & = 300 \text{ N/m} \end{align*} \]

Step 2: Calculate Maximum Potential Energy in the Spring

Now, we calculate the potential energy stored in the spring at maximum compression using the formula for elastic potential energy, \( U = \frac{1}{2}kx^2 \). Here, \( x = 0.15 \) m (converted from 15.0 cm), and \( k = 300 \) N/m:\[ U = \frac{1}{2} \times 300 \text{ N/m} \times (0.15 \text{ m})^2 = 3.375 \text{ J} \]

Step 3: Use Energy Conservation to Find Velocity after Impact

At maximum compression, all kinetic energy of the block-bullet system is converted to potential energy. Thus, kinetic energy \( KE = U \). Knowing that \( KE = \frac{1}{2}mv^2 \), where \( m = 1.00 \text{ kg} \) (mass of the block + bullet), we solve for \( v \):\[ \begin{align*} \frac{1}{2} \times 1.00 \text{ kg} \times v^2 & = 3.375 \text{ J} \ v^2 & = \frac{3.375 \text{ J}}{0.5 \text{ kg}} \ v & = \sqrt{6.75} = 2.60 \text{ m/s} \end{align*} \]

Step 4: Use Conservation of Momentum for Initial Speed of Bullet

Using the principle of conservation of momentum, the initial momentum of the bullet is equal to the final momentum of the block-bullet system. If \( v_0 \) is the initial velocity of the bullet and \( v = 2.60 \) m/s:\[ \begin{align*} \text{Initial Momentum: } & m_b \times v_0, \ \text{Final Momentum: } & (m_b + m_B) \times 2.60 \text{ m/s} \0.008 \text{ kg} \times v_0 & = 1.00 \text{ kg} \times 2.60 \text{ m/s} \ v_0 & = \frac{2.60 \text{ m/s}}{0.008 \text{ kg}} = 325 \text{ m/s} \end{align*} \]

Key Concepts

Spring Constant

The spring constant, often denoted as \( k \), is a crucial factor in understanding how a spring behaves when subjected to a force. It is a measure of the stiffness of the spring. In simpler terms, it tells us how much force we need to apply to compress or extend the spring by a certain distance.

To understand the spring constant better, think of it as a measure of resistance against deformation. If you have a very stiff spring, it will have a higher spring constant because it resists changes more. Conversely, a soft spring, which is easier to stretch, will have a lower spring constant.

We calculate the spring constant using Hooke’s Law, which is stated as \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement caused in the spring. By rearranging this formula to \( k = \frac{F}{x} \), you can determine \( k \) using the known values of force and displacement.

For example, if a force of 0.750 N compresses a spring by 0.0025 m, then the spring constant \( k \) would be \( 300 \text{ N/m} \). This simple yet powerful concept helps in calculating how energy is stored in a spring, which we will cover in the following sections.

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. This type of energy depends on two main factors: the mass of the object and its velocity. The formula used to calculate kinetic energy (often abbreviated as KE) is \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is velocity.

When examining any problem involving movement, such as a bullet striking a block, kinetic energy plays a vital role. The block and bullet system have kinetic energy immediately after impact, which then transforms as the system compresses the spring. At the point where the spring is most compressed, this kinetic energy is fully converted to elastic potential energy.

Understanding kinetic energy is crucial when determining how systems behave upon impact. The calculation \( KE = 3.375 \text{ J} \), shown here, is based on turning the entire kinetic energy into elastic potential energy, aligning with the energy conservation principle. This simplification helps students learn how energy shifts within a system, forming a base understanding of dynamics in motion.

Elastic Potential Energy

Elastic potential energy is a form of potential energy stored when materials are compressed, stretched, or bent. Specifically, in the context of springs, it represents the energy stored as the spring deforms (compresses or stretches). Elastic potential energy is given by the formula \( U = \frac{1}{2}kx^2 \), where \( k \) is the spring constant and \( x \) is the deformation distance of the spring.

In the scenario outlined in the exercise, when the bullet embeds in the block and the system compresses the spring, the kinetic energy is entirely converted into elastic potential energy. This relationship helps us solve problems where energy transitions from movement (kinetic) to being stored (potential).

For example, calculating the elastic potential energy with a spring constant of \( 300 \text{ N/m} \) and a compression distance of \( 0.15 \text{ m} \) results in \( 3.375 \text{ J} \). It illustrates how energy can be stored within a spring due to compression, ready to convert back to kinetic energy once the spring releases.

Understanding this concept is essential for grasping how energy conservation and transformation dictate the behavior of physical systems, where movement slows down as energy transforms into stored potential energy.