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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

Expert verified
The initial speed of the bullet was approximately 265.1 m/s.

Step by step solution

01

Identify Given Information

We have the bullet mass, \(m = 5.00\) g (or 0.005 kg), and the block mass, \(M = 1.20\) kg. The coefficient of friction, \(\mu = 0.20\). The distance the block slides, \(d = 0.310\) m. We need to find the initial speed of the bullet.
02

Determine the Total Mass after Collision

Since the bullet becomes embedded in the block, we add their masses to find the total mass after collision: \[ M_{\text{total}} = m + M = 0.005\, \text{kg} + 1.20\, \text{kg} = 1.205\, \text{kg}. \]
03

Use Work-Energy Principle

The work done by friction to stop the combined mass is equal to the initial kinetic energy of the combined mass. So, using the equation for work done by friction: \[ W_{\text{friction}} = f_k \times d \] Where \( f_k = \mu \cdot M_{\text{total}} \cdot g \), with \( g = 9.81\, \text{m/s}^2 \).
04

Calculate Frictional Force

Plug the given values into the equation: \[ f_k = 0.20 \cdot 1.205\, \text{kg} \cdot 9.81\, \text{m/s}^2 = 2.3601\, \text{N}. \]
05

Calculate Work Done by Friction

Using \( W_{\text{friction}} = f_k \times d \): \[ W_{\text{friction}} = 2.3601\, \text{N} \times 0.310\, \text{m} = 0.731631\, \text{J}. \]
06

Set Up the Kinetic Energy Equation

Since the block and bullet stop, the initial kinetic energy equals the work done by friction.\[ \frac{1}{2} M_{\text{total}} v^2 = W_{\text{friction}}. \] \[ \frac{1}{2} \times 1.205\, \text{kg} \times v^2 = 0.731631\, \text{J}. \]
07

Solve for Initial Velocity v

Rearrange the kinetic energy equation to solve for \( v \): \[ v^2 = \frac{2 \times 0.731631}{1.205} \] \[ v^2 = 1.21538 \] \[ v = \sqrt{1.21538} \] \[ v \approx 1.10\, \text{m/s}. \]
08

Use Momentum Conservation for Initial Bullet Speed

Use conservation of momentum before and after the collision: \[ m \cdot v_{\text{bullet}} = M_{\text{total}} \cdot v \] \[ 0.005\, \text{kg} \cdot v_{\text{bullet}} = 1.205 \cdot 1.10 \] \[ v_{\text{bullet}} = \frac{1.205 \cdot 1.10}{0.005} \] \[ v_{\text{bullet}} = 265.1\, \text{m/s}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Work-Energy Principle
The work-energy principle is a cornerstone in solving many physics problems. It tells us that the work done on an object is equal to its change in kinetic energy. When a bullet embeds itself into a block, the system gains an initial kinetic energy from the bullet's motion. After the collision, the block and bullet together slide to a stop. The work done by the force of kinetic friction is what reduces this system's kinetic energy to zero.
  • 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: \[ \frac{1}{2} M_{\text{total}} v^2 = W_{\text{friction}} \]
Understanding how the work done by friction relates to kinetic energy gives insight into how energy is conserved and transformed during the slide.
Momentum Conservation
Momentum conservation is crucial when dealing with collisions, especially in our scenario where a bullet collides with a block. Momentum, the product of mass and velocity, is conserved in a closed system when no external forces act on the system.
  • Before the bullet enters the block, only the bullet has momentum:
\[ \text{momentum}_{\text{initial}} = m \cdot v_{\text{bullet}} \]
  • After collision, the bullet-block system moves together, sharing the same momentum:
\[ \text{momentum}_{\text{final}} = M_{\text{total}} \cdot v \]
  • Since momentum is conserved:
\[ m \cdot v_{\text{bullet}} = M_{\text{total}} \cdot v \]By following the momentum conservation principle, we link the initial speed of the bullet to the speed after it embeds in the block, maintaining the balance of motion in the absence of external forces during collision.
Kinetic Friction
Kinetic friction plays a significant role when objects move across surfaces. It is the force that opposes the motion between two sliding surfaces. The coefficient of kinetic friction, \( \mu \), is a unitless value that quantifies the degree of opposition to the sliding motion.
  • Kinetic friction force is given by:\[ f_k = \mu \cdot M_{\text{total}} \cdot g \]
Where \( M_{\text{total}} \) is the combined mass of the bullet and block, and \( g \) is the acceleration due to gravity, approximately \( 9.81 \text{ m/s}^2 \).
  • This frictional force lowers the kinetic energy, crucial in bringing the block to a stop after the collision.
Understanding kinetic friction helps us calculate how quickly an object comes to rest after sliding over a surface.
Collision Dynamics
Collision dynamics involves understanding the behavior of two colliding bodies. In our scenario, the bullet's impact with the wooden block is a perfect example of an inelastic collision, where kinetic energy is not conserved, but momentum is.
  • The bullet's initial kinetic energy is transferred to the bullet-block system.
  • Post-collision, the system moves with a common velocity.
This type of collision, where objects stick together post-impact, reduces some kinetic energy to other forms, such as thermal energy due to deformation and friction. To understand collision dynamics, focus on how the kinetic energy and momentum shift between the colliding entities and how factors like mass and velocity influence the post-collision motion of both objects.

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