/*! 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 86 A 5.00-g bullet is shot \(throug... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 86

A 5.00-g bullet is shot \(through\) a 1.00-kg wood block suspended on a string 2.00 m long. The center of mass of the block rises a distance of 0.38 cm. Find the speed of the bullet as it emerges from the block if its initial speed is 450 m/s.

Short Answer

Expert verified
The bullet exits the block at 395.4 m/s.

Step by step solution

01

Understand the Conservation of Energy

When the bullet goes through the block, the block gains some kinetic energy which is then converted into potential energy as it rises. The potential energy gained by the block corresponds to the change in height, which is given as 0.38 cm or 0.0038 m. We'll first calculate the potential energy gained by the block.
02

Calculate Potential Energy of the Block

The potential energy gained by the block as it rises can be calculated using the formula: \[ PE = mgh \]where \(m = 1 \) kg (mass of the block), \(g = 9.81 \) m/se (acceleration due to gravity), and \(h = 0.0038 \) m (height). This gives:\[PE = 1 \times 9.81 \times 0.0038 = 0.037278\, \text{J}\]
03

Use Conservation of Momentum for Bullet and Block Collision

Before the collision, only the bullet has kinetic energy as it moves towards the block with an initial speed. As the bullet exits the block, the speed of the bullet changes, and the block gains some velocity. We use the principle of conservation of momentum:\[ m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \]where \(m_1 = 0.005\) kg, \(v_1 = 450\) m/s, \(m_2 = 1\) kg, and \(v_2 = 0\) m/s (since the block is initially at rest). \(v_1'\) is the final velocity of the bullet and \(v_2'\) is the velocity of the block after the bullet passes through.
04

Calculate Kinetic Energy of the Block After Collision

The kinetic energy of the block right after the bullet exits it is equal to the potential energy gained because energy is conserved. Therefore, the kinetic energy \(KE = 0.037278\) J is:\[ KE = \frac{1}{2} m_2 v_2'^2 \]Solving for \(v_2'\), we have:\[ v_2'^2 = \frac{2 \times 0.037278}{1} = 0.074556 \]\[ v_2' = \sqrt{0.074556} = 0.273\, \text{m/s} \]
05

Apply Conservation of Momentum to Find Bullet's Final Speed

Using the momentum conservation equation:\[ 0.005 \times 450 = 0.005 \times v_1' + 1 \times 0.273 \]\[ 2.25 = 0.005v_1' + 0.273 \]Solving for \(v_1'\),\[ 0.005v_1' = 2.25 - 0.273 \]\[ 0.005v_1' = 1.977 \]\[ v_1' = \frac{1.977}{0.005} = 395.4\, \text{m/s} \]
06

Conclusion

Therefore, the speed of the bullet as it emerges from the block is 395.4 m/s.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Kinetic Energy
Kinetic energy is a form of energy associated with objects in motion. It quantifies the energy an object possesses due to its movement, calculated using the formula: \[ KE = \frac{1}{2} m v^2 \]where \( m \) is the mass of the object and \( v \) is its velocity.
For example, when our bullet with a mass of 5 grams moves at an initial speed of 450 m/s, it has kinetic energy. When it collides with the block, some of this kinetic energy is transferred, illustrating the concept of energy conservation.
Kinetic energy plays a crucial role during collisions, as seen when the bullet transfers energy to the block, causing the block to rise. This rise is a testament to kinetic energy converting into potential energy, which we will discuss further below.
  • Energy due to motion
  • Calculated as \( \frac{1}{2} m v^2 \)
  • Transfers between objects during collisions
Potential Energy
Potential energy is the energy stored in an object due to its position or configuration. In the context of our exercise, the block gains potential energy as it rises when the bullet transfers some of its kinetic energy. This potential energy can be calculated using the formula:\[ PE = mgh \]where \( m \) is the mass of the block, \( g \) is the acceleration due to gravity, and \( h \) is the height the block has risen.
By rising 0.38 cm or 0.0038 m, the block stores energy in the form of gravitational potential energy, which fundamentally comes from the initial kinetic energy of the bullet.
Potential energy often contrasts with kinetic energy because it鈥檚 related to an object's position rather than its motion. In our scenario, even though the block is momentarily at rest at its peak height, the energy is still present in the form of potential.
  • Stored energy due to position
  • Calculated as \( mgh \)
  • Converted from kinetic energy
Conservation of Energy
The principle of conservation of energy is pivotal in physics. It states that energy in a closed system is constant, transforming from one form to another but never lost. In our exercise, this concept is beautifully illustrated as the system shifts back and forth between kinetic and potential energy.
Initially, the bullet possesses kinetic energy by virtue of its high speed. As it moves through the block, some kinetic energy is transferred to the block, causing it to move and subsequently rise.
As the block rises, kinetic energy is transformed into potential energy, visible in its increased height. This conversion underscores the conservation of energy, where the total energy before and after the interaction remains the same, albeit transformed from kinetic to potential form.
  • Energy cannot be created or destroyed
  • Transforms between kinetic and potential
  • Total energy remains constant in a closed system

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

On a frictionless, horizontal air table, puck \(A\) (with mass 0.250 kg) is moving toward puck \(B\) (with mass 0.350 kg), which is initially at rest. After the collision, puck A has a velocity of 0.120 m/s to the left, and puck \(B\) has a velocity of 0.650 m/s to the right. (a) What was the speed of puck \(A\) before the collision? (b) Calculate the change in the total kinetic energy of the system that occurs during the collision.

A machine part consists of a thin, uniform 4.00-kg bar that is 1.50 m long, hinged perpendicular to a similar vertical bar of mass 3.00 kg and length 1.80 m. The longer bar has a small but dense 2.00-kg ball at one end (\(\textbf{Fig. E8.55}\)). By what distance will the center of mass of this part move horizontally and vertically if the vertical bar is pivoted counterclockwise through 90\(^\circ\) to make the entire part horizontal?

Obviously, we can make rockets to go very fast, but what is a reasonable top speed? Assume that a rocket is fired from rest at a space station in deep space, where gravity is negligible. (a) If the rocket ejects gas at a relative speed of 2000 m/s and you want the rocket's speed eventually to be 1.00\(\times\) 10\(^{-3}c\), where \(c\) is the speed of light in vacuum, what fraction of the initial mass of the rocket and fuel is \(not\) fuel? (b) What is this fraction if the final speed is to be 3000 m/s?

A 0.150-kg glider is moving to the right with a speed of 0.80 m/s on a frictionless, horizontal air track. The glider has a head-on collision with a 0.300-kg glider that is moving to the left with a speed of 2.20 m/s. Find the final velocity (magnitude and direction) of each glider if the collision is elastic.

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage. In one such accident, a 1750-kg car traveling to the right at 1.50 m/s collides with a 1450-kg car going to the left at 1.10 m/s. Measurements show that the heavier car's speed just after the collision was 0.250 m/s in its original direction. Ignore any road friction during the collision. (a) What was the speed of the lighter car just after the collision? (b) Calculate the change in the combined kinetic energy of the two-car system during this collision.
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Torque and Rotational Motion

Read Explanation

Modern Physics

Read Explanation

Translational Dynamics

Read Explanation

Force

Read Explanation

Electromagnetism

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.