Chapter 14: Problem 83
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
Step by step solution
Convert Mass Units
Apply the Conservation of Momentum
Calculate Final Velocity After Collision
Use Energy Conservation to Find Spring Constant
Substitute and Calculate Spring Constant
Calculate the Period of Simple Harmonic Motion
Final Result
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Conservation of Momentum
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 \( v_f \).
Spring Constant
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: \( \frac{1}{2}(m_b + M)v_f^2 = \frac{1}{2}kx^2 \)
Kinetic and Potential Energy Transformation
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.