Chapter 8: Problem 42
A 5.00-g bullet is fired horizontally into a 1.20-kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.310 m along the surface before stopping. What was the initial speed of the bullet?
Short Answer
Step by step solution
Identify Given Information
Determine the Total Mass after Collision
Use Work-Energy Principle
Calculate Frictional Force
Calculate Work Done by Friction
Set Up the Kinetic Energy Equation
Solve for Initial Velocity v
Use Momentum Conservation for Initial Bullet Speed
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Work-Energy Principle
- Work done by friction is calculated using: \[ W_{\text{friction}} = f_k \times d \]
- Force of kinetic friction, \( f_k \), is the product of the coefficient of friction (\( \mu \)), the gravitational force (\( g \cdot M_{\text{total}} \)), and the stopping distance \( d \).
- The principle equation forms: \[ \frac{1}{2} M_{\text{total}} v^2 = W_{\text{friction}} \]
Momentum Conservation
- Before the bullet enters the block, only the bullet has momentum:
- After collision, the bullet-block system moves together, sharing the same momentum:
- Since momentum is conserved:
Kinetic Friction
- Kinetic friction force is given by:\[ f_k = \mu \cdot M_{\text{total}} \cdot g \]
- This frictional force lowers the kinetic energy, crucial in bringing the block to a stop after the collision.
Collision Dynamics
- The bullet's initial kinetic energy is transferred to the bullet-block system.
- Post-collision, the system moves with a common velocity.