Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 42

Two bumper cars moving on a frictionless surface collide elastically. The first bumper car is moving to the right with a speed of \(20.4 \mathrm{~m} / \mathrm{s}\) and rear-ends the second bumper car, which is also moving to the right but with a speed of \(9.00 \mathrm{~m} / \mathrm{s} .\) What is the speed of the first bumper car after the collision? The mass of the first bumper car is \(188 \mathrm{~kg}\), and the mass of the second bumper car is \(143 \mathrm{~kg}\). Assume that the collision takes place in one dimension.

Short Answer

Expert verified
Short Answer: To find the final velocity of the first bumper car after an elastic collision, we first calculate the initial total momentum and kinetic energy of the system. Then, we apply the conservation of momentum and kinetic energy laws, resulting in two equations with two unknowns, \(v_{1f}\) and \(v_{2f}\). We solve these equations to get the final velocity of the first bumper car, \(v_{1f}\).

Step by step solution

01

Calculate initial momentum and kinetic energy

To start, we need to find the initial total momentum and kinetic energy of the system. The initial momentum (\(p_{initial}\)) and kinetic energy (\(KE_{initial}\)) can be calculated as follows: Initial momentum: \(p_{initial} = m_1v_{1i} + m_2v_{2i} = (188 \mathrm{~kg})(20.4 \mathrm{~m/s}) + (143 \mathrm{~kg})(9.00 \mathrm{~m/s})\) Initial kinetic energy: \(KE_{initial} = \frac{1}{2} m_1v_{1i}^2 + \frac{1}{2} m_2v_{2i}^2 = \frac{1}{2}(188 \mathrm{~kg})(20.4 \mathrm{~m/s})^2 + \frac{1}{2}(143 \mathrm{~kg})(9.00 \mathrm{~m/s})^2\) Calculate the values of \(p_{initial}\) and \(KE_{initial}\).
02

Apply conservation of momentum and kinetic energy

Since the collision is elastic, both momentum and kinetic energy are conserved. We write the conservation equations for momentum (\(p_{final}\)) and kinetic energy (\(KE_{final}\)) as follows: Conservation of momentum: \(p_{final} = m_1v_{1f} + m_2v_{2f}\) Conservation of kinetic energy: \(KE_{final} = \frac{1}{2} m_1v_{1f}^2 + \frac{1}{2} m_2v_{2f}^2\) Using the fact that \(p_{initial} = p_{final}\) and \(KE_{initial} = KE_{final}\), we have the following equations: \(m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}\) \(\frac{1}{2} m_1v_{1i}^2 + \frac{1}{2} m_2v_{2i}^2 = \frac{1}{2} m_1v_{1f}^2 + \frac{1}{2} m_2v_{2f}^2\)
03

Solve the equations for the final velocity of the first bumper car

Next, we need to solve the conservation equations to find \(v_{1f}\). We have two equations and two unknowns, \(v_{1f}\) and \(v_{2f}\). We can first solve for \(v_{2f}\) from the conservation of momentum equation: \(v_{2f} = \frac{m_1(v_{1i} - v_{1f}) + m_2v_{2i}}{m_2}\) Now, substitute this expression for \(v_{2f}\) into the conservation of kinetic energy equation and solve for \(v_{1f}\). After solving the equation, we will obtain the value for the final velocity of the first bumper car, \(v_{1f}\).

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

Bats are extremely adept at catching insects in midair. If a 50.0-g bat flying in one direction at \(8.00 \mathrm{~m} / \mathrm{s}\) catches a \(5.00-\mathrm{g}\) insect flying in the opposite direction at \(6.00 \mathrm{~m} / \mathrm{s}\), what is the speed of the bat immediately after catching the insect?

A bored boy shoots a soft pellet from an air gun at a piece of cheese with mass \(0.25 \mathrm{~kg}\) that sits, keeping cool for dinner guests, on a block of ice. On one particular shot, his 1.2-g pellet gets stuck in the cheese, causing it to slide \(25 \mathrm{~cm}\) before coming to a stop. According to the package the gun came in, the muzzle velocity is \(65 \mathrm{~m} / \mathrm{s}\). What is the coefficient of friction between the cheese and the ice?

Astronauts are playing baseball on the International Space Station. One astronaut with a mass of \(50.0 \mathrm{~kg}\), initially at rest, hits a baseball with a bat. The baseball was initially moving toward the astronaut at \(35.0 \mathrm{~m} / \mathrm{s},\) and after being hit, travels back in the same direction with a speed of \(45.0 \mathrm{~m} / \mathrm{s}\). The mass of a baseball is \(0.14 \mathrm{~kg}\). What is the recoil velocity of the astronaut?

In a severe storm, \(1.00 \mathrm{~cm}\) of rain falls on a flat horizontal roof in \(30.0 \mathrm{~min}\). If the area of the roof is \(100 \mathrm{~m}^{2}\) and the terminal velocity of the rain is \(5.00 \mathrm{~m} / \mathrm{s}\), what is the average force exerted on the roof by the rain during the storm?

An astronaut becomes stranded during a space walk after her jet pack malfunctions. Fortunately, there are two objects close to her that she can push to propel herself back to the International Space Station (ISS). Object A has the same mass as the astronaut, and Object \(\mathrm{B}\) is 10 times more massive. To achieve a given momentum toward the ISS by pushing one of the objects away from the ISS, which object should she push? That is, which one requires less work to produce the same impulse? Initially, the astronaut and the two objects are at rest with respect to the ISS.
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Radiation

Read Explanation

Measurements

Read Explanation

Fluids

Read Explanation

Circular Motion and Gravitation

Read Explanation

Astrophysics

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