/*! 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 102 an \(80 \mathrm{~kg}\) man is on... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

an \(80 \mathrm{~kg}\) man is on a ladder hanging from a balloon that has a total mass of \(320 \mathrm{~kg}\) (including the basket passenger). The balloon is initially stationary relative to the ground. If the man on the ladder begins to climb at \(2.5 \mathrm{~m} / \mathrm{s}\) relative to the ladder, (a) in what direction and (b) at what speed does the balloon move? (c) If the man then stops climbing, what is the speed of the balloon?

Short Answer

Expert verified
(a) The balloon moves downward. (b) Speed is 0.625 m/s. (c) Speed is 0 m/s.

Step by step solution

01

Understand the System

We have a man and a balloon system initially at rest. The total mass of the system is the sum of the balloon with its basket and the mass of the man, equaling 400 kg. When the man climbs, he imparts a velocity to the balloon in the opposite direction according to the conservation of momentum.
02

Calculate Initial Momentum

The system is initially at rest, so the initial momentum is 0. The initial momentum can be expressed as: \( p_{initial} = (m_{man} + m_{balloon}) \times v_{initial} = 0 \).
03

Apply Conservation of Momentum

When the man climbs, the total momentum of the system must be conserved. Set the initial momentum equal to the final momentum: \( m_{man} \times v_{man} + m_{balloon} \times v_{balloon} = 0 \). We know \( v_{man} = 2.5 \mathrm{~m/s} \).
04

Solve for the Balloon's Velocity

Rearrange the equation from the previous step to solve for the balloon's velocity \( v_{balloon} \): \( v_{balloon} = - \frac{m_{man} \times v_{man}}{m_{balloon}} = - \frac{80 \mathrm{~kg} \times 2.5 \mathrm{~m/s}}{320 \mathrm{~kg}} = -0.625 \mathrm{~m/s} \).
05

Analyze Direction and Speed of Balloon

The negative sign indicates that the balloon moves in the opposite direction of the man's climbing, which is downward. The speed of the balloon is \( 0.625 \mathrm{~m/s} \).
06

If the Man Stops Climbing

If the man stops moving, the climbing motion ceases, and thus the induced velocity of the balloon also goes to zero. The system returns to a state of rest.

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.

The Balloon System
When we talk about a balloon system in physics, it's a good idea to think of it as a closed system that includes all its parts. In our particular exercise, this encompasses:
  • The balloon itself
  • The basket that might be attached
  • And importantly, the man climbing on the ladder
Initially, the entire system is stationary, meaning it's at rest with no movement in any direction relative to the ground. This stationary state is essential because it sets the stage for our analysis according to the conservation of momentum.

In problems involving a balloon system, you'll often find that one component of the system moves, like the man climbing the ladder in this case. As a result, movement is introduced through interaction among its parts. That is why understanding this closed balloon system and its total mass is crucial. Here, the combined mass is 400 kg, comprising the man's mass (80 kg) and the balloon's mass (320 kg). This setup helps us apply the laws of momentum effectively.
Understanding Relative Velocity
Relative velocity can seem a bit tricky at first, but it's all about the perspective from which you measure the speed. In this scenario, the man climbs the ladder with a velocity that is relative to the ladder itself.

To picture this, imagine you're on an escalator going up. If you walk up the escalator, your speed isn't just your walking speed; it's yours plus the speed of the escalator relative to stationary ground. In our exercise, the man's speed is given relative to the ladder, which is part of the balloon system.
  • His climbing speed = 2.5 m/s relative to the ladder
  • In terms of the overall system, it affects the balloon's motion
Once the man starts climbing, he doesn't just move himself; he causes motion in the entire system due to the conservation of momentum. The man imparts a velocity to the balloon in the opposite direction, which is why understanding relative velocity is vital for predicting what happens to the balloon.
Initial Rest Condition
The initial rest condition is a critical concept to grasp when dealing with conservation of momentum problems. It refers to the state of an object or system being completely still before any internal or external forces cause movement.

In our example, the balloon system is initially at rest. This means the sum of momentum for the whole system is zero. When the man starts climbing, he creates movement, but the overall momentum of the system should still add up to zero, according to the law of conservation of momentum:
  • Initial momentum = 0 (since the system is at rest)
  • Momentum must remain 0 as man climbs (change balanced by opposite velocity in the balloon)
Whenever you encounter such initial rest conditions in physics problems, it's a clue that you can use the conservation of momentum principle to predict or solve for subsequent movements. Essentially, this rule states that the total momentum before any action or movement is equal to the total momentum after, assuming no external forces interfere. In other words, the system’s initial resting state provides a baseline from which all actions can be measured.

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 of mass \(m=60 \mathrm{~g}\) is shot with speed \(v_{i}=\) \(22 \mathrm{~m} / \mathrm{s}\) into the barrel of a spring gun of mass \(M=240 \mathrm{~g}\) initially at rest on a frictionless surface. The ball sticks in the barrel at the point of maximum compression of the spring. Assume that the increase in thermal energy due to friction between the ball and the barrel is negligible. (a) What is the speed of the spring gun after the ball stops in the barrel? (b) What fraction of the initial kinetic energy of the ball is stored in the spring?

Shows a two-ended "rocket" that is initially stationary on a frictionless floor, with its center at the origin of an \(x\) axis. The rocket consists of a central block \(C\) (of mass \(M=6.00 \mathrm{~kg}\) ) and blocks \(L\) and \(R\) (cach of mass \(m=2.00 \mathrm{~kg}\) ) on the left and right sides. Small explosions can shoot either of the side blocks away from block \(C\) and along the \(x\) axis. Here is the sequence: (1) At time \(t=0,\) block \(L\) is shot to the left with a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) relative to the velocity that the explosion gives the rest of the rocket. (2) Next, at time \(t=0.80 \mathrm{~s},\) block \(R\) is shot to the right with a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) relative to the velocity that block \(C\) then has. At \(t=2.80 \mathrm{~s},\) what are (a) the velocity of block \(C\) and (b) the position of its center?

block 2 (mass \(1.0 \mathrm{~kg}\) ) is at rest on a frictionless surface and touching the end of an unstretched spring of spring constant \(200 \mathrm{~N} / \mathrm{m} .\) The other end of the spring is fixed to a wall. Block 1 (mass \(2.0 \mathrm{~kg}\) ), traveling at speed \(v_{1}=4.0 \mathrm{~m} / \mathrm{s}\), collides with block 2 , and the two blocks stick together. When the blocks momentarily stop, by what distance is the spring compressed?

Particle \(A\) and particle \(B\) are held together with a com- pressed spring between them. When they are released, the spring pushes them apart, and they then fly off in opposite directions, free of the spring. The mass of \(A\) is 2.00 times the mass of \(B,\) and the energy stored in the spring was 60 J. Assume that the spring has negligible mass and that all its stored energy is transferred to the particles. Once that transfer is complete, what are the kinetic energics of (a) particle \(A\) and (b) particle \(B ?\)

A body of mass \(2.0 \mathrm{~kg}\) makes an clastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed. (a) What is the mass of the other body? (b) What is the speed of the two-body center of mass if the initial speed of the \(2.0 \mathrm{~kg}\) body was \(4.0 \mathrm{~m} / \mathrm{s} ?\)

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.