/*! 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 52 In Fig. 9-59, a \(10 \mathrm{~g}... [FREE SOLUTION] | 91Ó°ÊÓ

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In Fig. 9-59, a \(10 \mathrm{~g}\) bullet moving directly upward at 1000 \(\mathrm{m} / \mathrm{s}\) strikes and passes through the center of mass of a \(5.0 \mathrm{~kg}\) block initially at rest. The bullet emerges from the block moving directly upward at \(400 \mathrm{~m} / \mathrm{s}\). To what maximum height does the block then rise above its initial position?

Short Answer

Expert verified
The block rises to a maximum height of 0.073 m.

Step by step solution

01

Calculate Initial Momentum

First, find the initial momentum of the system, which is just the momentum of the bullet, since the block is initially at rest. The initial momentum, \( p_{i} \), is given by the formula \( p = mv \), where \( m \) is the mass and \( v \) is the velocity. Thus, \( p_{i} = 0.01 \times 1000 \space kg \cdot m/s \).
02

Calculate Final Momentum of Bullet

Determine the final momentum of the bullet after it exits the block. Using the formula for momentum, we have \( p_{bullet,f} = 0.01 \times 400 \space kg \cdot m/s \).
03

Apply Conservation of Momentum

According to the conservation of momentum, the total momentum before the collision equals the total momentum after the collision. Set the initial momentum equal to the sum of the final momentum of the bullet and the block. \( 0.01 \times 1000 = 0.01 \times 400 + 5.0 \times v_{block,f} \). Solve for \( v_{block,f} \) to find the block's velocity after the bullet passes through.
04

Solve for Block's Velocity

Rearrange the equation from the previous step: \( 10 = 4 + 5v_{block,f} \), leading to \( 5v_{block,f} = 6 \). Therefore, \( v_{block,f} = 1.2 \space m/s \).
05

Calculate Maximum Height

Convert the block's kinetic energy at \( v_{block,f} \) to potential energy at the maximum height \( h \). Use the conservation of energy: \( \frac{1}{2} mv^2 = mgh \), where \( g = 9.8 \space m/s^2 \) and \( m \) is the mass of the block. Plug in the values: \( \frac{1}{2} \times 5 \times 1.2^2 = 5 \times 9.8 \times h \). Solve for \( h \).
06

Solve for Height

Simplify and solve: \( \frac{7.2}{2} = 49h \), which gives \( 3.6 = 49h \). Therefore, \( h = \frac{3.6}{49} = 0.073 \space m \).

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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 an object possesses due to its motion. When an object moves, it can perform work due to its velocity. The formula for kinetic energy (KE) is given by:
  • \( KE = \frac{1}{2} mv^2 \)
Here, \( m \) represents the mass of the object, and \( v \) is its velocity. This formula shows that kinetic energy increases with the square of the velocity, meaning that even a small increase in speed results in a large increase in energy.
When the bullet strikes the block in our exercise, both start with certain amounts of kinetic energy which change as the bullet exits the block. Initially, the bullet has high kinetic energy due to its high velocity. After exiting, it retains some energy, seen in its slower speed. The block gains kinetic energy from being struck, moving upwards, with its velocity derived from the energy transferred by the bullet.
Kinetic energy is thus a crucial player in understanding how objects conserve energy and interact through momentum.
Potential Energy
Potential energy is the energy an object holds due to its position or configuration. In gravitational terms, potential energy (PE) is calculated with:
  • \( PE = mgh \)
where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is the height above a reference point. Essentially, potential energy is stored energy waiting to be converted into motion or work.
In the context of the problem, potential energy becomes relevant when the block, after being struck by the bullet, ascends. As the block rises, its kinetic energy converts to potential energy. At its peak height, the block's upward velocity is zero, meaning its kinetic energy is also zero. Thus, all kinetic energy initially given to the block transitions fully to potential energy at maximum height.
Understanding this energy transformation is key in determining how high the block rises after the collision with the bullet.
Conservation of Energy
Conservation of energy is a fundamental concept in physics stating energy in a closed system remains constant. It cannot be created or destroyed, only transformed from one form to another.
  • Kinetic energy can transform into potential energy and vice versa.
  • The total amount of energy remains unchanged within an isolated system.
In the exercise, after the bullet and block interaction, the bullet's reduction in kinetic energy corresponds to the energy gained by the block. The block's upward movement converts its kinetic energy to potential energy. This reflects the conservation of energy principle during the entire process.
The block's rise to a certain height equates to the amount of kinetic energy converted to potential energy. Using the formulas for kinetic and potential energy, we calculate the block's maximum height.
Through applying these principles, the conservation of energy elegantly explains the transfer and transformation of energy during events like the collision described in the problem.

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