/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 A bullet with a mass of 4.0 g an... [FREE SOLUTION] | 91Ó°ÊÓ

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A bullet with a mass of 4.0 g and a speed of \(650 \mathrm{m} / \mathrm{s}\) is fired at a block of wood with a mass of \(0.095 \mathrm{kg}\). The block rests on a frictionless surface, and is thin enough that the bullet passes completely through it. Immediately after the bullet exits the block, the speed of the block is \(23 \mathrm{m} / \mathrm{s}\). (a) What is the speed of the bullet when it exits the block? (b) Is the final kinetic energy of this system equal to, less than, or greater than the initial kinetic energy? Explain. (c) Verify your answer to part (b) by calculating the initial and final kinetic energies of the system.

Short Answer

Expert verified
(a) The speed of the bullet after the collision is 103.75 m/s. (b) The final kinetic energy is less than the initial kinetic energy. (c) The calculations show initial KE = 845 J and final KE = 46.7 J.

Step by step solution

01

Understand the Problem

We need to find the speed of the bullet as it exits the block. Also, we need to compare the initial and final kinetic energy of the system to determine if they are equal, less, or greater.
02

Conservation of Momentum

The momentum before the collision is equal to the momentum after since there's no external force. Let the initial speed of the bullet be \(v_{bi} = 650 \text{ m/s}\), mass of the bullet be \(m_b = 4.0 \text{ g} = 0.004 \text{ kg}\), speed of the block after collision be \(v_s = 23 \text{ m/s}\), and speed of the bullet after collision be \(v_{bf}\). The mass of the block is \(m_s = 0.095 \text{ kg}\). The equation is: \(m_b \cdot v_{bi} = m_b \cdot v_{bf} + m_s \cdot v_s\).
03

Solve for Bullet's Final Speed

Substitute the known values into the momentum equation: \[0.004 \cdot 650 = 0.004 \cdot v_{bf} + 0.095 \cdot 23\]\[2.6 = 0.004v_{bf} + 2.185\]. Solving for \(v_{bf}\): \[0.004v_{bf} = 2.6 - 2.185 = 0.415\] \[v_{bf} = \frac{0.415}{0.004} = 103.75 \text{ m/s}\]. The speed of the bullet when it exits the block is \(103.75 \text{ m/s}\).
04

Compare Kinetic Energy

The kinetic energy is given by the equation \( KE = \frac{1}{2}mv^2 \). Calculate the initial KE and compare with the final KE to see if they are equal or not. Initial KE: \( KE_{initial} = \frac{1}{2} \cdot 0.004 \cdot 650^2 \). Final KE: \( KE_{final} = \frac{1}{2} \cdot 0.004 \cdot 103.75^2 + \frac{1}{2} \cdot 0.095 \cdot 23^2 \).
05

Calculate Initial Kinetic Energy

\[KE_{initial} = \frac{1}{2} \times 0.004 \times (650)^2 = 845 \, \text{J}\]
06

Calculate Final Kinetic Energy

Calculate the kinetic energy for the bullet and block after collision. Bullet: \[KE_{bullet} = \frac{1}{2} \times 0.004 \times (103.75)^2 = 21.52 \, \text{J}\] Block: \[KE_{block} = \frac{1}{2} \times 0.095 \times (23)^2 = 25.18 \, \text{J}\] Total final KE: \[KE_{final} = 21.52 + 25.18 = 46.7 \, \text{J}\]
07

Compare Initial and Final Energies

The initial kinetic energy was 845 J and the final kinetic energy is 46.7 J. The final kinetic energy is much less than the initial kinetic energy. This indicates energy was lost, likely as heat and sound, during the collision.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy of motion. It's given by the equation \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity.
In our exercise, we calculated both the initial and final kinetic energy of the bullet and wood block system. Initially, the bullet had a kinetic energy of \( 845 \, \text{J} \) resulting from its high speed and small mass.
The final kinetic energy, including both the bullet after passing through and the block, summed up to just \( 46.7 \, \text{J} \). This stark decrease indicates that not all kinetic energy was retained as mechanical energy; some transformed into other forms such as heat and sound.
Important points about kinetic energy:
  • It's impacted by both mass and velocity, increasing rapidly with higher velocity.
  • It's always positive, since it's based on squared velocity.
  • Real-life interactions often involve conversion into other energy forms, especially in collisions.
Collision Physics
Collision physics focuses on how objects behave during interactions. These interactions can result in momentum and energy transfer.
In this exercise, a bullet collides and passes through a block, showcasing an inelastic collision, where momentum is conserved, but kinetic energy is not fully retained as mechanical energy.
We applied the conservation of momentum: \[m_b \cdot v_{bi} = m_b \cdot v_{bf} + m_s \cdot v_s\]where \( v_{bi} \) and \( v_{bf} \) are the initial and final speeds of the bullet, and \( v_s \) is the speed of the block post-collision.
Key aspects of collision physics include:
  • The difference between elastic and inelastic collisions - only the former fully conserve kinetic energy.
  • The importance of understanding forces and duration in analyzing how they affect motion and energy.
Law of Conservation of Energy
The law of conservation of energy is a fundamental principle stating that energy cannot be created or destroyed, only transformed. This principle helps us understand what happens to energy during processes like collisions.
In our example, despite kinetic energy dropping significantly (from \( 845 \, \text{J} \) initially to \( 46.7 \, \text{J} \) finally), the energy hasn't vanished. Instead, it converted into other energy forms, demonstrating energy's persistent transformation in closed systems.
Considerations for energy conservation include:
  • Conversion types include mechanical to thermal energy, sound, or even deformation in collision contexts.
  • In perfectly elastic collisions, kinetic energy would remain conserved; however, inelastic collisions like our example show translational kinetic energy becoming less apparent.
Understanding energy transformations sheds light on real-world physics and energy management.

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