/*! 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 26 A \(5.20 \mathrm{~g}\) bullet mo... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 26

A \(5.20 \mathrm{~g}\) bullet moving at \(672 \mathrm{~m} / \mathrm{s}\) strikes a \(700 \mathrm{~g}\) wooden block at rest on a frictionless surface. The bullet emerges, traveling in the same direction with its speed reduced to \(428 \mathrm{~m} / \mathrm{s}\). What is the resulting speed of the block?

Short Answer

Expert verified
The resulting speed of the block is approximately 1.81 m/s.

Step by step solution

01

- Identify Known Values

First, list the given values from the problem:- Mass of bullet, \( m_1 = 5.20 \text{ g} \) converted to kg: \( 0.00520 \text{ kg} \)- Initial velocity of bullet, \( v_{1i} = 672 \text{ m/s} \)- Final velocity of bullet, \( v_{1f} = 428 \text{ m/s} \)- Mass of block, \( m_2 = 700 \text{ g} \) converted to kg: \( 0.700 \text{ kg} \)- Initial velocity of block, \( v_{2i} = 0 \text{ m/s} \)
02

- Apply the Law of Conservation of Momentum

According to the conservation of momentum, the total momentum before the collision equals the total momentum after the collision. Write the equation:\[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \]Now substitute the known values:\[ (0.00520 \text{ kg})(672 \text{ m/s}) + (0.700 \text{ kg})(0 \text{ m/s}) = (0.00520 \text{ kg})(428 \text{ m/s}) + (0.700 \text{ kg})(v_{2f}) \]
03

- Simplify the Equation

Simplify the equation to isolate \( v_{2f} \):\[ (0.00520 \times 672) = (0.00520 \times 428) + (0.700 \times v_{2f}) \]\[ 3.4944 = 2.2256 + 0.700 v_{2f} \]Subtract 2.2256 from both sides to isolate the term with \( v_{2f} \):\[ 1.2688 = 0.700 v_{2f} \]
04

- Solve for the Final Velocity of the Block

Divide both sides by 0.700 to solve for \( v_{2f} \):\[ v_{2f} = \frac{1.2688}{0.700} \]\[ v_{2f} \thickapprox 1.8126 \text{ m/s} \]

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Key Concepts

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

bullet-block collision
A bullet-block collision is an example of an inelastic collision where two objects interact briefly but do not stick together. In the given exercise, a bullet collides with a wooden block. The bullet does not become embedded in the block; instead, it passes through, reducing its speed. When solving these types of problems, it is essential to focus on the initial and final velocities of both objects.

Here, the key points are:
  • The bullet has an initial mass and velocity before the collision.
  • The block is initially at rest.
  • After the collision, the bullet has a reduced velocity.
  • The block, previously at rest, moves due to the impact.
These points help in setting up the right equations to use the law of conservation of momentum to find the unknown final speed of the block.
momentum conservation
Momentum conservation is a fundamental principle in physics that states that the total momentum of a closed system remains constant if no external forces act on it. Mathematically, it is expressed as: \( m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \)

In the context of the bullet-block collision:
  • The mass and initial velocity of the bullet represent its initial momentum.
  • Since the block is at rest initially, it has zero initial momentum.
  • After the collision, both the bullet and the block have different velocities, so their momenta change.
Using this principle ensures that we account for all changes in momentum within the system. Substituting the known values into the equation allows us to solve for the block's final velocity. This is crucial in understanding how objects exchange momentum during collisions.
frictionless surface dynamics
A frictionless surface implies that there are no external forces opposing the motion of the objects involved in the collision. This simplifies the analysis because we only need to consider the internal forces due to the collision itself.

Key aspects to keep in mind include:
  • No energy is lost to friction, so the calculations only consider the momentum exchanged during the collision.
  • The block can move freely without any resistance after the bullet passes through.
  • This idealization helps us focus purely on the interaction between the bullet and the block.
Assuming a frictionless surface is common in physics problems to isolate the variables and make it easier to apply theoretical principles without additional complications.

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Most popular questions from this chapter

Same Mass After a collision, two objects of the same mass and same initial speed are found to move away together at \(\frac{1}{2}\) their initial speed. Find the angle between the initial velocities of the objects.

Two \(2.0 \mathrm{~kg}\) bodies, \(A\) and \(B\), collide. The velocities before the collision are \(\vec{v}_{A 1}=(15 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(30 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) and \(\vec{v}_{B 1}=(-10 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(5.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). After the collision, \(\vec{v}_{A 2}=\) \((-5.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(20 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). What is the final velocity of \(B\) ?

Two vehicles \(A\) and \(B\) are traveling west and south, respectively, toward the same intersection, where they collide and lock together. Before the collision, \(A\) (total weight \(12.0 \mathrm{kN}\) ) has a speed of \(64.4 \mathrm{~km} / \mathrm{h}\), and \(B\) (total weight \(16.0 \mathrm{kN}\) ) has a speed of \(96.6 \mathrm{~km} / \mathrm{h}\). Find the (a) magnitude and (b) direction of the velocity of the (interlocked) vehicles immediately after the collision, assuming the collision is isolated.

Conservation in Subsystems Can a system whose momentum is conserved be made up of smaller systems whose individual momenta are not conserved? Explain why or why not and give an example.

A \(91 \mathrm{~kg}\) man lying on a surface of negligible friction shoves a \(68 \mathrm{~g}\) stone away from him, giving it a speed of \(4.0 \mathrm{~m} / \mathrm{s}\). What velocity does the man acquire as a result?
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