Question

A rifle bullet with mass 8.00 g and initial horizontal velocity 280 m/s strikes and embeds itself in a block with mass 0.992 kg that rests on a frictionless surface and is attached to one end of an ideal spring. The other end of the spring is attached to the wall. The impact compresses the spring a maximum distance of 15.0 cm. After the impact, the block moves in SHM. Calculate the period of this motion.

Short answer

The period of the motion is approximately 0.420 seconds.

Step-by-step solution

Convert Mass Units

The mass of the bullet is given in grams and needs to be converted to kilograms for consistent units. The bullet's mass is \(8.00 \, \text{g} = 0.0080 \, \text{kg}\).

Apply the Conservation of Momentum

Just before the impact, the momentum of the system is the momentum of the bullet, as the block is initially at rest. The conservation of momentum gives us: \[ m_b v_b = (m_b + M) v_f \]where\( m_b = 0.0080 \, \text{kg} \) (mass of bullet),\( v_b = 280 \, \text{m/s} \) (velocity of bullet),\( M = 0.992 \, \text{kg} \) (mass of block),\( v_f \) is the final velocity after collision. Solve for \( v_f \):\[ v_f = \frac{m_b v_b}{m_b + M} = \frac{0.0080 \times 280}{0.0080 + 0.992} \, \text{m/s} \]

Calculate Final Velocity After Collision

Substitute the values into the equation from Step 2 to calculate the final velocity:\[ v_f = \frac{0.0080 \times 280}{1.0000} = 2.24 \, \text{m/s} \].Thus, the combined velocity of the bullet and block after the collision is 2.24 m/s.

Use Energy Conservation to Find Spring Constant

The kinetic energy after collision is transformed into elastic potential energy at maximum compression of the spring. Set the initial kinetic energy equal to the elastic potential energy:\[ \frac{1}{2}(m_b + M)v_f^2 = \frac{1}{2}k x^2 \]where \( x = 0.15 \, \text{m} \) (compression of spring) and \( k \) is the spring constant.\[ k = \frac{(m_b + M) v_f^2}{x^2} = \frac{1.0000 \times 2.24^2}{0.15^2} \].

Substitute and Calculate Spring Constant

Substitute \( v_f = 2.24 \, \text{m/s} \), mass \( m_b + M = 1.0000 \, \text{kg} \), \( x = 0.15 \, \text{m} \): \[ k = \frac{1.0000 \times 2.24^2}{0.0225} \approx 223.78 \, \text{N/m} \].Thus, the spring constant \( k \) is approximately 223.78 N/m.

Calculate the Period of Simple Harmonic Motion

The formula for the period \( T \) of SHM is:\[ T = 2\pi \sqrt{\frac{m_b + M}{k}} \].Substitute \( m_b + M = 1.0000 \, \text{kg} \) and \( k = 223.78 \, \text{N/m} \):\[ T = 2\pi \sqrt{\frac{1.0000}{223.78}} \approx 0.420 \, \text{s} \].

Final Result

The period of the simple harmonic motion is approximately 0.420 seconds.

Key concepts

Conservation of Momentum

Conservation of momentum is a core principle in physics that describes how the total momentum of a closed system remains constant unless acted on by an external force. In the given problem, initially, the only moving object is the bullet. Thus, the entire system's momentum is conserved just before and immediately after the bullet embeds into the block.

To understand this better:

• Before impact: Only the bullet is moving, so the total momentum is the momentum of the bullet, calculated as mass (\( m_b = 0.0080 \, \text{kg} \)) times velocity (\( v_b = 280 \, \text{m/s} \)).
• After impact: The bullet sticks to the block, moving together. So, their combined mass is \( m_b + M = 1.000 \, \text{kg} \), and their shared velocity is \( v_f \).

Applying the conservation of momentum principle:\[m_b v_b = (m_b + M) v_f \]This equation allows the calculation of the final velocity (\( v_f \)). This principle ensures the total momentum of the system stays the same before and after the collision.

Spring Constant

The spring constant, denoted by \( k \), is a measure of a spring's stiffness. In the given exercise, after the bullet becomes embedded in the block, the kinetic energy of the combined system is converted into potential energy when the spring is compressed. To find the spring constant, we equate the initial kinetic energy of the moving block and bullet with the potential energy stored in the spring.

Here's how:

• Kinetic energy just after collision is given by: \( \frac{1}{2}(m_b + M)v_f^2 \)
• At maximum spring compression, the kinetic energy becomes potential energy: \( \frac{1}{2}kx^2 \)
• By equating these two energies: \( \frac{1}{2}(m_b + M)v_f^2 = \frac{1}{2}kx^2 \)

Solving for \( k \), the following formula provides its value:\[k = \frac{(m_b + M) v_f^2}{x^2} = \frac{1.000 \times 2.24^2}{0.15^2} \approx 223.78 \, \text{N/m}\]This result tells us how rigid or supportive the spring is during compression, crucial for calculating other aspects of the system like the period of motion.

Kinetic and Potential Energy Transformation

The transformation between kinetic and potential energy is a vital concept in understanding simple harmonic motion (SHM). Initially, the bullet's kinetic energy is transferred to the block, setting the system in motion. This kinetic energy is gradually converted into potential energy at the peak spring compression, allowing us to analyze the SHM.

Think of it this way:

• After the bullet fuses with the block, they share a kinetic energy governed by the equation \( \frac{1}{2}(m_b + M)v_f^2 \).
• As the system compresses the spring to its maximum, this energy converts to potential: \( \frac{1}{2}kx^2 \).
• Once released, the potential energy changes back to kinetic energy, enabling the back-and-forth motion of SHM.

Understanding the exchange of energy forms between kinetic and potential during SHM enables predictions of the system’s dynamics, exemplified by the calculation of the motion's period using:\[T = 2\pi \sqrt{\frac{m_b + M}{k}} \approx 0.420 \, \text{s}\]An insight into these energy transformations enhances our grasp of the rhythmic oscillation characteristic of systems in SHM.