Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 56

Use the result of Problem 54 to show that in any elastic collision between two objects, the relative speed of the two is the same before and after the collision. [Hints: Look at the collision in its \(\mathrm{CM}\) frame - the reference frame in which the \(\mathrm{CM}\) is at rest. The relative speed of two objects is the same in any inertial reference frame.]

Short Answer

Expert verified
Answer: Yes, in an elastic collision between two objects, the relative speed of the objects remains the same before and after the collision.

Step by step solution

01

Define the variables and given information

Let's denote the two objects as A and B, with masses \(m_A\) and \(m_B\) respectively. Their velocities before the collision in the CM frame are \(v_{A, i}\) and \(v_{B, i}\), and after the collision, they are \(v_{A, f}\) and \(v_{B, f}\). We also define the initial and final relative speeds as \(v_{rel, i} = |v_{A, i} - v_{B, i}|\) and \(v_{rel, f} = |v_{A, f} - v_{B, f}|\).
02

Apply conservation of momentum in the CM frame

Since the total momentum is conserved, we can write the following equation for the CM frame: \(m_A v_{A, i} + m_B v_{B, i} = m_A v_{A, f} + m_B v_{B, f}\) Also, in the CM frame, the initial total momentum is zero: \(m_A v_{A, i} + m_B v_{B, i} = 0\)
03

Apply conservation of kinetic energy in the CM frame

In an elastic collision, the kinetic energy of the system is also conserved. Therefore, we can write the following equation in the CM frame: \(\frac{1}{2} m_A v_{A, i}^2 + \frac{1}{2} m_B v_{B, i}^2 = \frac{1}{2} m_A v_{A, f}^2 + \frac{1}{2} m_B v_{B, f}^2\)
04

Combine the conservation equations

Square the conservation of momentum equation and substitute it into the conservation of kinetic energy equation: \(m_A^2 v_{A, i}^2 + 2m_A m_B v_{A, i} v_{B, i} + m_B^2 v_{B, i}^2 = m_A^2 v_{A, f}^2 + 2m_A m_B v_{A, f} v_{B, f} + m_B^2 v_{B, f}^2\) Since the left side and right side of the equation represent the initial and final kinetic energy respectively, we can rewrite it as: \(2m_A m_B v_{A, i} v_{B, i} + m_A^2 v_{A, i}^2 + m_B^2 v_{B, i}^2 = 2m_A m_B v_{A, f} v_{B, f} + m_A^2 v_{A, f}^2 + m_B^2 v_{B, f}^2\) Now, we can cancel out the terms with \(m_A^2\) and \(m_B^2\) on both sides, which yields: \(2m_A m_B (v_{A, i} v_{B, i} - v_{A, f} v_{B, f}) = 0\)
05

Find the relation between initial and final relative speeds

Since the masses of the objects are not zero, we can divide both sides of the equation by \(2m_A m_B\) and obtain: \(v_{A, i} v_{B, i} = v_{A, f} v_{B, f}\) We will now write the initial and final relative speeds in terms of the velocities of A and B: \(v_{rel, i}^2 = (v_{A, i} - v_{B, i})^2\) \(v_{rel, f}^2 = (v_{A, f} - v_{B, f})^2\) Subtract these two equations: \(v_{rel, i}^2 - v_{rel, f}^2 = (v_{A, i}^2 - 2v_{A, i} v_{B, i} + v_{B, i}^2) - (v_{A, f}^2 - 2v_{A, f} v_{B, f} + v_{B, f}^2)\) From the conservation of momentum equation, we know that \(v_{A, i} v_{B, i} = v_{A, f} v_{B, f}\). Therefore: \(v_{rel, i}^2 - v_{rel, f}^2 = (v_{A, i}^2 + v_{B, i}^2) - (v_{A, f}^2 + v_{B, f}^2)\)
06

Conclusion

From the conservation of kinetic energy equation, we know that the initial and final total kinetic energy are equal. So, the difference between the initial and final relative speeds must be zero: \(v_{rel, i}^2 - v_{rel, f}^2 = 0\) This implies that: \(v_{rel, i} = v_{rel, f}\) Thus, in any elastic collision between two objects, the relative speed of the two is the same before and after the collision. This result is valid in any inertial reference frame.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 boy of mass \(60.0 \mathrm{kg}\) is rescued from a hotel fire by leaping into a firefighters' net. The window from which he leapt was \(8.0 \mathrm{m}\) above the net. The firefighters lower their arms as he lands in the net so that he is brought to a complete stop in a time of \(0.40 \mathrm{s}\). (a) What is his change in momentum during the 0.40 -s interval? (b) What is the impulse on the net due to the boy during the interval? [Hint: Do not ignore gravity.] (c) What is the average force on the net due to the boy during the interval?
In a nuclear reactor, a neutron moving at speed \(v_{\mathrm{i}}\) in the positive \(x\)-direction strikes a deuteron, which is at rest. The neutron is deflected by \(90.0^{\circ}\) and moves off with speed $v_{\mathrm{i}} / \sqrt{3}\( in the positive \)y\(-direction. Find the \)x\(- and \)y$-components of the deuteron's velocity after the collision. (The mass of the deuteron is twice the mass of the neutron.)
A flat, circular metal disk of uniform thickness has a radius of $3.0 \mathrm{cm} .\( A hole is drilled in the disk that is \)1.5 \mathrm{cm}$ in radius. The hole is tangent to one side of the disk. Where is the \(\mathrm{cm}\) of the disk now that the hole has been drilled? [Hint: The original disk (before the hole is drilled) can be thought of as having two pieces-the disk with the hole plus the smaller disk of metal drilled out. Write an equation that expresses \(x_{\mathrm{CM}}\) of the original disk in terms of the \(x_{\mathrm{CM}}\) 's of the two pieces. since the thickness is uniform, the mass of any piece is proportional to its area.]
Two identical gliders, each with elastic bumpers and mass \(0.10 \mathrm{kg},\) are on a horizontal air track. Friction is negligible. Glider 2 is stationary. Glider 1 moves toward glider 2 from the left with a speed of $0.20 \mathrm{m} / \mathrm{s} .$ They collide. After the collision, what are the velocities of glider 1 and glider \(2 ?\)
A hockey puck moving at \(0.45 \mathrm{m} / \mathrm{s}\) collides elastically with another puck that was at rest. The pucks have equal mass. The first puck is deflected \(37^{\circ}\) to the right and moves off at $0.36 \mathrm{m} / \mathrm{s} .$ Find the speed and direction of the second puck after the collision.
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Engineering Physics

Read Explanation

Kinematics Physics

Read Explanation

Fluids

Read Explanation

Fundamentals of Physics

Read Explanation

Space Physics

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.

Sign-up for free