Question

A rifle bullet with mass 8.00 g strikes and embeds itself in a block with mass 0.992 kg that rests on a frictionless, horizontal surface and is attached to a coil spring \((\textbf{Fig. P8.79})\). The impact compresses the spring 15.0 cm. Calibration of the spring shows that a force of 0.750 N is required to compress the spring 0.250 cm. (a) Find the magnitude of the block's velocity just after impact. (b) What was the initial speed of the bullet?

Short answer

(a) 2.605 m/s; (b) 325.625 m/s.

Step-by-step solution

Determine the spring constant

The spring constant \( k \) can be determined using Hooke's Law, which states \( F = kx \). Given that a force of 0.750 N compresses the spring by 0.250 cm (or 0.0025 m), we set up the equation \( 0.750 = k \times 0.0025 \). Solving for \( k \), we find \( k = \frac{0.750}{0.0025} = 300 \, \text{N/m} \).

Calculate the potential energy stored in the spring

The potential energy \( U \) stored in a compressed spring is given by \( U = \frac{1}{2}kx^2 \). We are told the spring is compressed 15.0 cm (or 0.15 m), so substituting in, we have \( U = \frac{1}{2} \times 300 \times (0.15)^2 = 3.375 \, \text{J} \).

Find the block's velocity post-impact using energy conservation

Since the surface is frictionless, the kinetic energy of the block right after impact equals the potential energy stored in the spring. Using the equation \( \frac{1}{2}mv^2 = 3.375 \), with the block's mass \( m = 0.992 \, \text{kg} \), we solve for \( v \):\[ v = \sqrt{\frac{2 \times 3.375}{0.992}} \approx 2.605 \, \text{m/s}. \]

Determine the initial speed of the bullet using momentum conservation

Applying conservation of momentum for the bullet and the block: \( m_b v_b = (m_b + m_{block}) v_{post-impact} \), where \( m_b = 0.008 \, \text{kg} \) (mass of the bullet), \( v_b \) is the initial bullet speed, and \( v_{post-impact} = 2.605 \, \text{m/s} \). Set up the equation: \[ 0.008 \times v_b = (0.008 + 0.992) \times 2.605. \]Solving for \( v_b \),\[ v_b = \frac{1.000 \times 2.605}{0.008} \approx 325.625 \, \text{m/s}. \]

Key concepts

Spring Constant

The spring constant, denoted as \( k \), was determined using Hooke's Law. This law is a key principle for understanding how springs behave, saying that the force required to compress or extend a spring is linearly proportional to the distance the spring is stretched or compressed. This relationship
can be captured with the formula:

• \( F = kx \)

Here:

• \( F \) is the force applied to the spring,
• \( k \) is the spring constant, and
• \( x \) is the distance by which the spring is compressed or extended.

Given the problem specifics that a force of 0.750 N compresses the spring by 0.250 cm, we substitute these values into Hooke's Law. After calculating, we find that the spring constant \( k \) turns out to be 300 N/m. It's crucial because it tells us how stiff the spring is: greater the spring constant, the stiffer the spring.

Kinetic Energy

Kinetic energy refers to the energy that an object possesses due to its motion. It plays a significant role in the scenario of the bullet impacting the block. When the bullet hits the block, it transfers its kinetic energy to the block, causing the block to move. This movement changes the system's kinetic energy into potential energy stored in the spring.
The kinetic energy \( KE \) can be defined with the formula:

• \( KE = \frac{1}{2}mv^2 \)

where:

• \( m \) is the mass of the object, and
• \( v \) is the velocity of the object.

The problem demonstrates the conservation of energy as the kinetic energy of the bullet-block system is fully converted into potential energy stored in the compressed spring. Understanding this conversion helps us find the initial velocity of the bullet.

Hooke's Law

Hooke's Law is fundamental when dealing with springs and elastic objects. It establishes the linear relationship between the force applied to a material and the material's extension or compression. Specifically,
this law is expressed as:

• \( F = kx \)

where:

• \( F \) is the applied force,
• \( k \) is the spring constant, and
• \( x \) is the displacement caused by the applied force.

In the exercise, Hooke's Law allows us to find the spring constant from the known force and compression measurements. Understanding this law is essential to grasp how springs and elastic forces behave. It also lays the groundwork for analyzing potential energy stored in springs.

Potential Energy

Potential energy in a spring is the stored energy due to its compression or elongation. In this context, it arises from compressing the spring upon impact with the bullet-block system. The
potential energy stored in the spring can be calculated using:

• \( U = \frac{1}{2}kx^2 \)

where:

• \( U \) is the potential energy,
• \( k \) is the spring constant, and
• \( x \) is the compression length of the spring.

The scenario specifies that the spring compresses by 0.15 m, resulting in a stored energy of 3.375 J. This potential energy reflects the system's capacity to perform work as the spring returns to its natural length, converting energy back into kinetic form.

Frictionless Surface

In physics problems, a frictionless surface is an idealization, meaning no frictional forces resist motion. It ensures that energy conservation between kinetic and potential forms is not disturbed by external dissipating forces.
In this scenario, it simplifies calculations since the only forces to consider are those from the collision and the spring. A frictionless environment allows for a perfect exchange of energy between kinetic energy of motion and the potential energy stored in the spring.
This concept makes it easier to calculate the outcomes of motion without dealing with energy lost as heat or sound. In the exercise, the absence of friction implies that the entire kinetic energy of the bullet-block system converts into potential energy stored in the spring, simplifying our analysis significantly.