Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 25

A freight car, open at the top, weighing \(9.75\) metric tons, is coasting along a level track with negligible friction at \(1.36 \mathrm{~m} / \mathrm{s}\) when it begins to rain hard. The raindrops fall vertically with respect to the ground. What is the speed of the car when it has collected \(0.50\) metric tons of rain? What assumptions, if any, must you make to get your answer?

Short Answer

Expert verified
The final speed of the freight car after collecting 0.50 metric tons of rain will be \(1.30 \, \mathrm{m/s}\). The calculations assume that raindrops fall vertically with respect to the ground, momentum is conserved and the friction in the system is negligible.

Step by step solution

01

Understand the Problem

We need to find the speed of the freight car after it has collected a certain weight of water by the rain. The rain falls vertically and system experiences no external force so total momentum should be conserved. An important point to keep in mind is that the weights given in the problem need to converted to masses by dividing with acceleration due to gravity. In SI units, acceleration due to gravity (g) is approximately \(9.8 \, \mathrm{m/s^2}\).
02

Apply the Principle of Conservation of Momentum

According to the conservation of momentum, the initial momentum of the system equal to the final momentum. The initial momentum is only the momentum of the car (mass_car * velocity_car). The final momentum when the rain has been collected is the mass of car plus the collected rain mass, both moving with the same final speed (velocity_final). Setting these two equal to each other: mass_car * velocity_car = (mass_car + mass_rain) * velocity_final. Solve this equation to get the final velocity.
03

Perform the Calculations

First convert the weights to mass by dividing with g: Mass of the car, mass_car = 9.75 metric tons / g = 9.75 * 10^3 kg / 9.8 m/s^2 = 995 kg (approximately). Mass of the rain, mass_rain = 0.50 metric tons / g = 0.50 * 10^3 kg / 9.8 m/s^2 = 51 kg (approximately). Substitute these values and the known initial velocity into the equation from Step 2:995 kg * 1.36 m/s = (995 kg + 51 kg) * v_final. Simplifying and solving the equation gives v_final = 1.30 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.

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

Show that the ratio of the distances \(x_{1}\) and \(x_{2}\) of two particles from their center of mass is the inverse ratio of their masses; that is, \(x_{1} / x_{2}=m_{2} / m_{1}\).

A shell is fired from a gun with a muzzle velocity of \(466 \mathrm{~m} / \mathrm{s}\), at an angle of \(57.4^{\circ}\) with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming level terrain?

A rocket at rest in space, where there is virtually no gravity, has a mass of \(2.55 \times 10^{5} \mathrm{~kg}\), of which \(1.81 \times 10^{5} \mathrm{~kg}\) is fuel. The engine consumes fuel at the rate of \(480 \mathrm{~kg} / \mathrm{s}\), and the exhaust speed is \(3.27 \mathrm{~km} / \mathrm{s}\). The engine is fired for \(250 \mathrm{~s}\). (a) Find the thrust of the rocket engine. \((b)\) What is the mass of the rocket after the engine burn? \((c)\) What is the final speed attained?

How far is the center of mass of the Earth-Moon system from the center of the Earth? (From Appendix C, obtain the masses of the Earth and Moon and the distance between the centers of the Earth and Moon. It is interesting to compare the answer to the Earth's radius.)

An \(84.4\) -kg man is standing at the rear of a \(425-\mathrm{kg}\) iceboat that is moving at \(4.16 \mathrm{~m} / \mathrm{s}\) across ice that may be considered to be frictionless. He decides to walk to the front of the \(18.2-\mathrm{m}\) -long boat and does so at a speed of \(2.08 \mathrm{~m} / \mathrm{s}\) with respect to the boat. How far does the boat move across the ice while he is walking?
See all solutions

Recommended explanations on Physics Textbooks

Electricity and Magnetism

Read Explanation

Solid State Physics

Read Explanation

Quantum Physics

Read Explanation

Physics of Motion

Read Explanation

Fluids

Read Explanation

Further Mechanics and Thermal 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