/*! 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 50 A \(5.20 \mathrm{~g}\) bullet mo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 50

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 \(428 \mathrm{~m} / \mathrm{s}\). (a) What is the resulting speed of the block? (b) What is the speed of the bullet-block center of mass?

Short Answer

Expert verified
(a) The block's speed is approximately 1.82 m/s. (b) The center of mass speed is approximately 4.97 m/s.

Step by step solution

01

Identify Given Values

Start by identifying and listing the given values and their units:- Mass of the bullet, \( m_1 = 5.20 \, \mathrm{g} = 0.00520 \, \mathrm{kg} \)- Initial velocity of the bullet, \( v_{1i} = 672 \, \mathrm{m/s} \)- Final velocity of the bullet after collision, \( v_{1f} = 428 \, \mathrm{m/s} \)- Mass of the block, \( m_2 = 700 \, \mathrm{g} = 0.700 \, \mathrm{kg} \)- Initial velocity of the block, \( v_{2i} = 0 \, \mathrm{m/s} \) (since the block is at rest)
02

Apply Conservation of Momentum

For a collision happening in a closed system, the total momentum before the collision equals the total momentum after the collision.The equation is:\[ m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = m_1 \cdot v_{1f} + m_2 \cdot v_{2f} \]Plug in the values:\[ (0.00520 \, \mathrm{kg}) \cdot (672 \, \mathrm{m/s}) + (0.700 \, \mathrm{kg}) \cdot (0 \, \mathrm{m/s}) = (0.00520 \, \mathrm{kg}) \cdot (428 \, \mathrm{m/s}) + (0.700 \, \mathrm{kg}) \cdot v_{2f} \]
03

Solve for the Resulting Speed of the Block

Rearrange the equation from Step 2 to solve for \( v_{2f} \):\[ v_{2f} = \frac{ m_1 \cdot v_{1i} - m_1 \cdot v_{1f} }{ m_2 } \]Substitute the known values:\[ v_{2f} = \frac{(0.00520) \cdot (672) - (0.00520) \cdot (428)}{0.700} \approx 1.82 \, \mathrm{m/s} \]
04

Calculate the Speed of the Center of Mass

To find the velocity of the center of mass of the bullet-block system, the formula is:\[ v_\mathrm{cm} = \frac{m_1 \cdot v_{1i} + m_2 \cdot v_{2i}}{m_1 + m_2} \]Substitute the values:\[ v_\mathrm{cm} = \frac{(0.00520) \cdot (672) + (0.700) \cdot (0)}{0.00520 + 0.700} \approx 4.97 \, \mathrm{m/s} \]

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.

Collision
Collisions are a crucial concept in physics, especially in mechanics. They occur when two or more objects come into contact and exert forces on each other for a short duration. In physics, we often classify collisions based on the conservation of physical quantities like momentum and energy. The most common types are elastic and inelastic collisions.
  • Elastic Collisions: Both momentum and kinetic energy are conserved. No kinetic energy is lost. Idealized scenarios, like two billiard balls hitting each other, are often used to represent elastic collisions.
  • Inelastic Collisions: Only momentum is conserved. Some kinetic energy is transformed into other forms of energy, such as heat or sound. Most real-world collisions are inelastic to some extent.
In the scenario with the bullet and block, the collision is an inelastic one. The bullet retains some velocity after passing through the block, indicating that not all kinetic energy is conserved. However, momentum conservation is used to calculate the block鈥檚 resulting velocity.
Bullet and Block Problem
The bullet and block problem is a classic physics exercise used to illustrate the principles of momentum conservation during collision events. Such problems often involve a small object like a bullet impacting a larger, stationary object such as a block.
  • Parameters of the Problem: In this case, we have a bullet with an initial velocity striking a stationary block.
  • Momentum Conservation: The law of momentum conservation states that the total momentum before and after the collision must be equal, given no external forces are acting on the system. This allows us to find unknown quantities like the block's speed after the bullet emerges.
The key to solving this problem is to set up the momentum conservation equation and plug in known values to find the result. Then, rearrange the equation to solve for the desired variable, which in this case was the block's speed resulting from the momentum transferred to it by the bullet.
Center of Mass
The concept of center of mass is pivotal in understanding motion and dynamics of objects and systems. It represents a point where the whole mass of a system is considered to be concentrated for the purpose of analyzing translational motion.
  • Dynamic Analysis: In systems subjected to internal forces like collisions, the center of mass helps us understand how an entire system will move as a whole.
  • Calculating Center of Mass: The velocity of the center of mass ( v_{\mathrm{cm}} ) is determined using the formula: \[ v_{\mathrm{cm}} = \frac{m_1 \cdot v_{1i} + m_2 \cdot v_{2i}}{m_1 + m_2} \] This formula incorporates the individual masses and velocities, providing a single value to represent combined movement.
In the bullet and block problem, calculating the center of mass speed provides insight into how the combined system moves post-collision, which is crucial for dynamic assessments and further applications.

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

A \(0.15 \mathrm{~kg}\) ball hits a wall with a velocity of \((5.00 \mathrm{~m} / \mathrm{s}) \mathrm{i}+\) \((6.50 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}+(4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{k}} .\) It rebounds from the wall with a velocity of \((2.00 \mathrm{~m} / \mathrm{s}) \mathrm{i}+(3.50 \mathrm{~m} / \mathrm{s}) \mathrm{j}+(-3.20 \mathrm{~m} / \mathrm{s}) \mathrm{k} .\) What are (a) the change in the ball's momentum, (b) the impulse on the ball, and (c) the impulse on the wall?

Shows a slab with dimensions \(d_{1}=11.0 \mathrm{~cm}\), \(d_{2}=2.80 \mathrm{~cm}, \quad\) and \(\quad d_{3}=13.0 \mathrm{~cm} .\) Half the slab consists of aluminum (density \(=2.70 \mathrm{~g} / \mathrm{cm}^{3}\) ) and half consists of iron (density = \(7.85 \mathrm{~g} / \mathrm{cm}^{3}\) ). What are (a) the \(x\) coordinate, (b) the \(y\) coordinate, and (c) the \(z\) coordinate of the slab's center of mass?

block 1 of mass \(m_{1}\) slides from rest along a frictionless ramp from height \(h=2.50 \mathrm{~m}\) and then collides with stationary block \(2,\) which has mass \(m_{2}=2.00 m_{1}\). After the collision, block 2 slides into a region where the coefficient of kinetic friction \(\mu_{k}\) is 0.500 and comes to a stop in distance \(d\) within that region. What is the value of distance \(d\) if the collision is (a) elastic and (b) completely inelastic?

A rocket is moving away from the solar system at a speed of \(6.0 \times 10^{3} \mathrm{~m} / \mathrm{s} .\) It fires its engine, which ejects exhaust with a speed of \(3.0 \times 10^{3} \mathrm{~m} / \mathrm{s}\) relative to the rocket. The mass of the rocket at this time is \(4.0 \times 10^{4} \mathrm{~kg},\) and its acceleration is \(2.0 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the thrust of the engine? (b) At what rate, in kilograms per second, is exhaust ejected during the firing?

Particle \(A\) and particle \(B\) are held together with a com- pressed spring between them. When they are released, the spring pushes them apart, and they then fly off in opposite directions, free of the spring. The mass of \(A\) is 2.00 times the mass of \(B,\) and the energy stored in the spring was 60 J. Assume that the spring has negligible mass and that all its stored energy is transferred to the particles. Once that transfer is complete, what are the kinetic energics of (a) particle \(A\) and (b) particle \(B ?\)
See all solutions

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Read Explanation

Electricity

Read Explanation

Atoms and Radioactivity

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Famous Physicists

Read Explanation

Radiation

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.