/*! 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 block 1 of mass \(m_{1}\) slides... [FREE SOLUTION] | 91Ó°ÊÓ

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block 1 of mass \(m_{1}\) slides along an \(x\) axis on a frictionless floor with a speed of \(v_{1 j}=4.00 \mathrm{~m} / \mathrm{s}\). Then it undergoes a one- dimensional elastic collision with stationary block 2 of mass \(m_{2}=0.500 m_{1}\). Next, block 2 undergoes a one-dimensional elastic collision with stationary block 3 of mass \(m_{3}=0.500 \mathrm{~m}_{2}\). (a) What then is the speed of block 3 ? Are (b) the speed, (c) the kinetic energy, and (d) the momentum of block 3 greater than, less than, or the same as the initial values for block \(1 ?\)

Short Answer

Expert verified
(a) The speed of block 3 is less than 4 m/s. (b) Speed is less, (c) kinetic energy is less, (d) momentum is less.

Step by step solution

01

Understand Elastic Collision

In an elastic collision, both momentum and kinetic energy are conserved. This means that for each collision, we will use these two principles to find the final velocities of the blocks involved.
02

Set Up Equations for Block 1 and 2 Collision

For the collision between block 1 and block 2, we have conservation of momentum: \[ m_1v_{1j} = m_1v_{1f} + m_2v_{2f} \] where \( v_{1f} \) and \( v_{2f} \) are the final velocities of block 1 and 2 respectively. Using the given that \( m_2 = 0.5m_1 \) and initial velocity of block 2 being 0, set up the equations accordingly.
03

Apply Conservation of Kinetic Energy

Apply the conservation of kinetic energy for the collision between block 1 and block 2: \[ \frac{1}{2}m_1v_{1j}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2 \]. Solve these two equations from Steps 2 and 3 to find \( v_{1f} \) and \( v_{2f} \).
04

Calculate Velocities for Collision with Block 3

Now block 2 with velocity \( v_{2f} \) collides with block 3 which is initially stationary. Use conservation of momentum: \[ m_2v_{2f} = m_2v_{2f}' + m_3v_{3f} \]. Here \( v_{2f}' \) is the new velocity of block 2 and \( v_{3f} \) is the velocity of block 3 after the collision. With \( m_3 = 0.5m_2 \), simplify and solve for \( v_{3f} \).
05

Apply Conservation of Kinetic Energy Again

For the collision between block 2 and block 3, use the conservation of kinetic energy: \[ \frac{1}{2}m_2v_{2f}^2 = \frac{1}{2}m_2v_{2f}'^2 + \frac{1}{2}m_3v_{3f}^2 \]. Using the values from Step 4, solve for \( v_{3f} \).
06

Compare Speeds

Now, compare \( v_{3f} \) with the initial speed of block 1, \( v_{1j} = 4.00 \text{ m/s} \). You'll find that \( v_{3f} \) is less than \( v_{1j} \).
07

Compare Kinetic Energy

The kinetic energy of block 3 is \( \frac{1}{2}m_3v_{3f}^2 \). Calculate it and compare it to the initial kinetic energy of block 1, \( \frac{1}{2}m_1v_{1j}^2 \), to determine if it is less, equal, or greater.
08

Compare Momentum

Evaluate the momentum of block 3, \( m_3v_{3f} \), compare it with the initial momentum of block 1, \( m_1v_{1j} \), and determine if it is less, equal, or greater.

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Key Concepts

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

Conservation of Momentum
Momentum is a measure of motion for an object, defined as the product of its mass and velocity. In physical interactions, especially in collisions, the principle of conservation of momentum is paramount. This principle states that the total momentum of a system of objects is constant if no external forces are acting on it.
  • Think of momentum as the amount of "push" something has because of its movement.
  • When two objects collide, the idea is that whatever "push" one loses, the other gains.
  • This sharing ensures that nothing is lost overall, even though individual velocities might change.
For the particular collision scenario in our exercise, we observe two collisions. Each collision respects this law, ensuring that, before and after the collision, the sum of the momentums of the interacting bodies remains unchanged.
The equation we use in collisions is often written as: \[ m_1 v_{1} + m_2 v_{2} = m_1 v'_{1} + m_2 v'_{2} \] Where:- \( m_1 \) and \( m_2 \) are the masses of two colliding objects.- \( v_1 \) and \( v_2 \) are their velocities before collision.- \( v'_1 \) and \( v'_2 \) are their velocities after collision.
This equation allows us to find the velocities of the objects after a collision, assuming no outside forces intervene.
Conservation of Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. It is a crucial concept in understanding how energy is transferred or transformed during collisions. In elastic collisions, a special type of collision, not only momentum but also kinetic energy is conserved.
  • When objects collide elastically, they bounce off each other without shredding any energy as heat.
  • This means the total kinetic energy before the collision equals the total kinetic energy after.
  • This conservation is why elastic collisions are considered "perfect" in terms of energy efficiency.
The formula used for kinetic energy is: \[ KE = \frac{1}{2} mv^2 \] In the context of our exercise, for each pair of collisions, the total kinetic energy is expressed as:\[ \frac{1}{2}m_1v_{1}^2 + \frac{1}{2}m_2v_{2}^2 = \frac{1}{2}m_1v'_{1}^2 + \frac{1}{2}m_2v'_{2}^2 \]
  • This equation reflects how kinetic energy remains unchanged during the elastic collisions between blocks.
  • Understanding this principle helps predict post-collision velocities.
One-dimensional Collisions
Collisions can occur in one, two, or three dimensions. One-dimensional collisions simplify things as everything happens along a single line. This means all motion is linear, and calculations are generally straightforward.
  • Imagine cars on a straight road—one-dimensional movement describes their interactions somewhat simply.
  • Colliding objects only move forward or backward, which simplifies our equations.
For one-dimensional collisions like in our exercise, the conservation laws become easier to apply.
Here are some key points about one-dimensional collisions:
  • Equations become clear since there are no angles or vector components involved.
  • All measurements and equations pertain to a straight line of action, reducing complexity.
  • Velocities are either positive or negative, easily showing direction along the line.

This focus on simplicity allows learners to grasp the core principles of momentum and energy conservation without the added layers of multidimensional calculations.

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Most popular questions from this chapter

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