/*! 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 60 A wagon is rolling forward on le... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total mass of the wagon, rider, and rock is \(95.0 \mathrm{kg} .\) The mass of the rock is \(0.300 \mathrm{kg} .\) Initially the wagon is rolling forward at a speed of \(0.500 \mathrm{m} / \mathrm{s}\). Then the person throws the rock with a speed of \(16.0 \mathrm{m} / \mathrm{s} .\) Both speeds are relative to the ground. Find the speed of the wagon after the rock is thrown directly forward in one case and directly backward in another.

Short Answer

Expert verified
Final speeds: 0.451 m/s forward, 0.552 m/s backward.

Step by step solution

01

Identify the given information

Before attempting to solve the problem, identify the given quantities:- Total mass of the wagon, rider, and rock, \( M = 95.0 \, \mathrm{kg} \).- Mass of the rock, \( m = 0.300 \, \mathrm{kg} \).- Initial speed of the wagon, \( v_i = 0.500 \, \mathrm{m/s} \).- Speed of the rock relative to the ground when thrown, \( v_r = 16.0 \, \mathrm{m/s} \).
02

Use the conservation of momentum principle

Since friction is negligible, the law of conservation of momentum applies. The total momentum before and after the rock is thrown remains the same:\[mV_i = (M-m)V_f + mv_r\] where \(V_f\) is the final velocity of the wagon after the rock is thrown.
03

Calculate the velocity of the wagon after throwing the rock forward

Substitute the given values into the momentum equation for the rock thrown forward:\[(95.0 \, \mathrm{kg} \times 0.500 \, \mathrm{m/s}) = (95.0 \, \mathrm{kg} - 0.300 \, \mathrm{kg})V_f + (0.300 \, \mathrm{kg} \times 16.0 \, \mathrm{m/s})\]Solving this equation:\[47.5 = 94.7V_f + 4.8\]\[V_f = \frac{42.7}{94.7} \, \mathrm{m/s} = 0.451 \, \mathrm{m/s}\]The final velocity of the wagon is \(0.451 \, \mathrm{m/s}\) when the rock is thrown forward.
04

Calculate the velocity of the wagon after throwing the rock backward

For the scenario where the rock is thrown backward, adjust the momentum equation:\[(95.0 \, \mathrm{kg} \times 0.500 \, \mathrm{m/s}) = (95.0 \, \mathrm{kg} - 0.300 \, \mathrm{kg})V_f - (0.300 \, \mathrm{kg} \times 16.0 \, \mathrm{m/s})\]Solving this equation:\[47.5 = 94.7V_f - 4.8\]\[V_f = \frac{52.3}{94.7} \, \mathrm{m/s} = 0.552 \, \mathrm{m/s}\]The final velocity of the wagon is \(0.552 \, \mathrm{m/s}\) when the rock is thrown backward.

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.

Kinematics
Kinematics is a branch of physics that deals with the motion of objects without considering the forces that cause or result from the motion. It focuses on various parameters such as displacement, velocity, and acceleration. In the exercise involving the wagon and the rock, kinematics helps us understand how the speeds change when the rock is thrown.

Here's a simple way to see it: When you throw the rock, whether forwards or backwards, the speed of the rock is an important part of the motion. Different speeds affect how the wagon moves after the rock is thrown. That change in speed is what the concept of kinematics captures.

Some key points about kinematics are:
  • It deals with linear motion, such as the straight path of the wagon.
  • The motion is relative to a reference point—in this case, the ground.
This understanding aids in calculating how the momentum exchange affects the final velocities.
Mass
Mass is a fundamental property of physical objects that highlights the amount of matter present in them. It is usually measured in kilograms (kg). In the exercise, we deal with both the total mass of the system (wagon, rider, and rock) and the mass of the rock itself.

Knowing the mass is crucial because it directly influences the momentum of the system. Here are some important considerations regarding mass:
  • The total mass of the wagon, rider, and rock is 95.0 kg.
  • The mass of just the rock is 0.300 kg.
The mass measure ensures that we can accurately apply the principle of conservation of momentum and solve for how the system's velocity changes upon the rock's release. This knowledge helps us determine how the speed of the wagon changes dependencies based on whether the rock is thrown forward or backward.
Velocity
Velocity is a vector quantity that describes the speed of an object in a specific direction. It is one of the significant parameters in the original exercise and is central to solving the problem at hand.

Understanding velocity involves:
  • Initial velocity: The starting speed of the wagon, which is 0.500 m/s.
  • Rock's velocity: The throw of the rock results in a velocity of 16.0 m/s.
In physics, velocity helps us understand the dynamics of a system, showing how speed and direction vary as changes are applied. In this particular exercise, it's crucial to track how the velocity of the wagon is altered depending on the rock's direction of throw.

Calculating the final velocity involves using the conservation of momentum, as it accounts for changes in velocity due to external releases like throwing the rock.
Momentum
Momentum is a key concept in physics, defined as the product of an object's mass and its velocity. It is represented by the formula: \( p = m imes v \), where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity.

In the exercise, momentum is crucial as it governs the changes in the motion of the wagon. The principle of conservation of momentum dictates that the total momentum before and after an event—in this case, throwing the rock—remains constant if no external forces are acting.

Some critical insights about momentum include:
  • Initial momentum is calculated using the initial speed and total mass of the system.
  • Final momentum considers the separation of the rock's mass and its velocity from the rest of the system.
Through the conservation principle, we apply the momentum balance to calculate how the velocity of the wagon changes based on the given scenarios: either throwing the rock forward or backward. This comprehensive understanding allows us to arrive at the accurate final velocities of the moving wagon.

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 ball is attached to one end of a wire, the other end beingfastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. The masses of the ball and block are, respectively, \(1.60 \mathrm{kg}\) and \(2.40 \mathrm{kg},\) and the length of the wire is \(1.20 \mathrm{m} .\) Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision.

A student \((m=63 \mathrm{kg})\) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of \(0.040 \mathrm{s} .\) The average force exerted on him by the ground is \(+18000 \mathrm{N},\) where the upward direction is taken to be the positive direction. From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground.

A two-stage rocket moves in space at a constant velocity of \(4900 \mathrm{m} / \mathrm{s} .\) The two stages are then separated by a small explosive charge placed between them. Immediately after the explosion the velocity of the \(1200-\mathrm{kg}\) upper stage is \(5700 \mathrm{m} / \mathrm{s}\) in the same direction as before the explosion. What is the velocity (magnitude and direction) of the \(2400-\mathrm{kg}\) lower stage after the explosion?

You and your crew must dock your \(25000 \mathrm{kg}\) spaceship at Spaceport Alpha, which is orbiting Mars. In the process, Alpha's control tower has requested that you ram another vessel, a freight ship of mass \(16500 \mathrm{kg},\) latch onto it, and use your combined momentum to bring it into dock. The freight ship is not moving with respect to the colossal Spaceport Alpha, which has a mass of \(1.85 \times 10^{7} \mathrm{kg} .\) Alpha's automated system that guides incoming spacecraft into dock requires that the incoming speed is less than \(2.0 \mathrm{m} / \mathrm{s}\). (a) Assuming a perfectly linear alignment of your ship's velocity vector with the freight ship (which is stationary with respect to Alpha) and Alpha's docking port, what must be your ship's speed (before colliding with the freight ship) so that the combination of the freight ship and your ship arrives at Alpha's docking port with a speed of \(1.50 \mathrm{m} / \mathrm{s} ?\) (b) How does the velocity of Spaceport Alpha change when the combination of your vessel and the freight ship successfully docks with it? (c) Suppose you made a mistake while maneuvering your vessel in an attempt to ram the freight ship and, rather than latching on to it and making a perfectly inelastic collision, you strike it and knock it in the direction of the spaceport with a perfectly elastic collision. What is the speed of the freight ship in that case (assuming your ship had the same initial velocity as that calculated in part (a))?

A car (mass \(=1100 \mathrm{kg}\) ) is traveling at \(32 \mathrm{m} / \mathrm{s}\) when it collides head-on with a sport utility vehicle (mass \(=2500 \mathrm{kg}\) ) traveling in the opposite direction. In the collision, the two vehicles come to a halt. At what speed was the sport utility vehicle traveling?

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.