/*! 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 88 Consider a one-dimensional, head... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider a one-dimensional, head-on elastic collision. One object has a mass \(m_{1}\) and an initial velocity \(v_{1} ;\) the other has a mass \(m_{2}\) and an initial velocity \(v_{2}\). Use momentum conservation and energy conservation to show that the final velocities of the two masses are $$\begin{array}{l} v_{1, f}=\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) v_{1}+\left(\frac{2 m_{2}}{m_{1}+m_{2}}\right) v_{2} \\ v_{2, f}=\left(\frac{2 m_{1}}{m_{1}+m_{2}}\right) v_{1}+\left(\frac{m_{2}-m_{1}}{m_{1}+m_{2}}\right) v_{2} \end{array}$$

Short Answer

Expert verified
The final velocities are \(v_{1,f} = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) v_1 + \left(\frac{2m_2}{m_1 + m_2}\right) v_2\) and \(v_{2,f} = \left(\frac{2m_1}{m_1 + m_2}\right) v_1 + \left(\frac{m_2 - m_1}{m_1 + m_2}\right) v_2\).

Step by step solution

01

Understand Conservation Laws

In an elastic collision, both momentum and kinetic energy are conserved. This means that the total momentum and total kinetic energy before and after the collision are equal.
02

Write Momentum Conservation Equation

The total momentum before the collision is \(m_1 v_1 + m_2 v_2\). After the collision, it becomes \(m_1 v_{1,f} + m_2 v_{2,f}\). Therefore, you can write the equation as: \[ m_1 v_1 + m_2 v_2 = m_1 v_{1,f} + m_2 v_{2,f} \]
03

Write Kinetic Energy Conservation Equation

The total kinetic energy before the collision is \( \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \). After the collision, it becomes \( \frac{1}{2} m_1 v_{1,f}^2 + \frac{1}{2} m_2 v_{2,f}^2 \). Therefore, \[ \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} m_1 v_{1,f}^2 + \frac{1}{2} m_2 v_{2,f}^2 \]
04

Solve Momentum Equation for One Variable

From the momentum conservation equation, solve for one of the final velocities, for example, \(v_{2,f}\):\[ v_{2,f} = \frac{m_1 v_1 + m_2 v_2 - m_1 v_{1,f}}{m_2} \]
05

Substitute Into Kinetic Energy Equation

Substitute the expression found for \(v_{2,f}\) from the momentum equation into the kinetic energy equation. This helps in reducing the equations and allows solving for \(v_{1,f}\).
06

Simplify and Solve for \(v_{1,f}\)

Simplify the equation obtained after substitution to solve for the expression of \(v_{1,f}\). You will find that \[ v_{1,f} = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) v_1 + \left(\frac{2m_2}{m_1 + m_2}\right) v_2 \]
07

Solve for \(v_{2,f}\)

Using the equation for \(v_{2,f}\) from Step 4, substitute \(v_{1,f}\) using its expression from Step 6. Simplify to find:\[ v_{2,f} = \left(\frac{2m_1}{m_1 + m_2}\right) v_1 + \left(\frac{m_2 - m_1}{m_1 + m_2}\right) v_2 \]

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 91Ó°ÊÓ!

Key Concepts

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

Momentum Conservation
In physics, the conservation of momentum is a fundamental concept applied to elastic collisions. Momentum itself is the product of an object's mass and velocity, and is symbolized with "p." In any closed system, the total momentum stays constant, meaning it's conserved. For this reason, momentum before a collision equals momentum after it.

Given two objects with masses and initial velocities, their total momentum before the collision can be expressed as \( m_1 v_1 + m_2 v_2 \). After the collision, the same total momentum is represented by \( m_1 v_{1,f} + m_2 v_{2,f} \). Therefore, the momentum conservation equation becomes:
  • \( m_1 v_1 + m_2 v_2 = m_1 v_{1,f} + m_2 v_{2,f} \)
This equation forms the basis for determining the final velocities of both objects post-collision. Transforming this equation is the first key step in unravelling the effects of the collision.
Kinetic Energy Conservation
In elastic collisions, kinetic energy also follows a conservation law similar to momentum. Kinetic energy depends on both mass and velocity; specifically, it equals \( \frac{1}{2}mv^2 \), where "m" is mass and "v" is velocity. During an elastic collision, total kinetic energy remains unchanged.

Before the collision occurs, the kinetic energy can be represented as \( \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \). After the collision, the kinetic energy is expressed as \( \frac{1}{2} m_1 v_{1,f}^2 + \frac{1}{2} m_2 v_{2,f}^2 \). The equation for kinetic energy conservation then is:
  • \( \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} m_1 v_{1,f}^2 + \frac{1/2} m_2 v_{2,f}^2 \)
Understanding and applying this equation allows us to delve deeper into the motions and velocities of the colliding bodies. The challenge lies in using this equation in conjunction with the momentum conservation equation to find exact post-collision velocities.
Final Velocity Equations
Determining the final velocities of the two masses after an elastic collision involves solving the equations of momentum and kinetic energy conservation simultaneously. Using the provided equations, one can derive the final velocities as follows:

For object 1, the final velocity can be expressed using:
  • \( v_{1,f} = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) v_1 + \left(\frac{2m_2}{m_1 + m_2}\right) v_2 \)
    • This equation shows that the final velocity of object 1 relies on both initial velocities and their masses. Similarly, object 2’s final velocity formula is:
      • \( v_{2,f} = \left(\frac{2m_1}{m_1 + m_2}\right) v_1 + \left(\frac{m_2 - m_1}{m_1 + m_2}\right) v_2 \)
        • These equations are derived from the interplay of the masses and the initial velocities of both bodies involved in the collision.

          It's fascinating that, despite variations in mass and speed, these simple formulas provide the solutions where conservation laws are satisfied, demonstrating how elegantly physics can explain real-world phenomena.

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 long, uniform rope with a mass of 0.135 kg per meter lies on the ground. You grab one end of the rope and lift it at the constant rate of \(1.13 \mathrm{m} / \mathrm{s}\). Calculate the upward force you must exert at the moment when the top end of the rope is \(0.525 \mathrm{m}\) above the ground.

Moderating a Neutron In a nuclear reactor, neutrons released by nuclear fission must be slowed down before they can trigger additional reactions in other nuclei. To see what sort of material is most effective in slowing (or moderating) a neutron, calculate the ratio of a neutron's final kinetic energy to its initial kinetic energy, \(K_{f} / K_{i},\) for a head-on elastic collision with each of the following stationary target particles. (Note: The mass of a neutron is \(m=1.009 \mathrm{u},\) where the atomic mass unit, \(u,\) is defined as follows: \(\left.1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg} \cdot\right)\) (a) An electron \(\left(M=5.49 \times 10^{-4} \mathrm{u}\right)\) (b) A proton \((M=1.007 \mathrm{u})\) (c) The nucleus of a lead atom \((M=207.2 \mathrm{u})\).

A player bounces a \(0.43-\mathrm{kg}\) soccer ball off her head, changing the velocity of the ball from \(\overrightarrow{\mathbf{v}}_{i}=(8.8 \mathrm{m} / \mathrm{s}) \hat{\mathrm{x}}+(-2.3 \mathrm{m} / \mathrm{s}) \hat{\mathrm{y}}\) to \(\overrightarrow{\mathbf{v}}_{f}=(5.2 \mathrm{m} / \mathrm{s}) \hat{\mathrm{x}}+(3.7 \mathrm{m} / \mathrm{s}) \hat{y} .\) If the ball is in contact with the player's head for \(6.7 \mathrm{ms},\) what are (a) the direction and (b) the magnitude of the impulse delivered to the ball?

A bullet with a mass of 4.0 g and a speed of \(650 \mathrm{m} / \mathrm{s}\) is fired at a block of wood with a mass of \(0.095 \mathrm{kg}\). The block rests on a frictionless surface, and is thin enough that the bullet passes completely through it. Immediately after the bullet exits the block, the speed of the block is \(23 \mathrm{m} / \mathrm{s}\). (a) What is the speed of the bullet when it exits the block? (b) Is the final kinetic energy of this system equal to, less than, or greater than the initial kinetic energy? Explain. (c) Verify your answer to part (b) by calculating the initial and final kinetic energies of the system.

A hoop of mass \(M\) and radius \(R\) rests on a smooth, level surface. The inside of the hoop has ridges on either side, so that it forms a track on which a ball can roll, as indicated in Figure \(9-25\). If a ball of mass \(2 M\) and radius \(r=R / 4\) is released as shown, the system rocks back and forth until it comes to rest with the ball at the bottom of the hoop. When the ball comes to rest, what is the \(x\) coordinate of its center?

See all solutions

Recommended explanations on Physics Textbooks

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.