Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 84

A 60.0 -kg woman stands at one end of a \(120-\) kg raft that is $6.0 \mathrm{m}\( long. The other end of the raft is \)0.50 \mathrm{m}$ from a pier. (a) The woman walks toward the pier until she gets to the other end of the raft and stops there. Now what is the distance between the raft and the pier? (b) In (a), how far did the woman walk (relative to the pier)?

Short Answer

Expert verified
Solution: (a) The distance between the raft and the pier after the woman walks to the other end of the raft is 3.50 m. (b) The woman has moved 3 m relative to the pier.

Step by step solution

01

1. List Given Information

- Mass of the woman, \(m_w = 60 \thinspace kg\) - Mass of the raft, \(m_r = 120 \thinspace kg\) - Length of the raft, \(L = 6.0 m\) - Initial distance between the raft and the pier, \(d_i = 0.50 m\)
02

2. Applying Conservation of Momentum

Since no external force acts on the woman-raft system, the momentum of the system is conserved. Initially, both the woman and the raft are at rest, so the initial momentum of the system is zero. As the woman walks towards the pier, both the woman and the raft will experience equal and opposite changes in their positions due to momentum conservation. Let the woman's final position be \(x_w\) m from her initial position, and the raft's final position be \(x_r\) m from its initial position. The total momentum before and after the woman walks should be equal: \(m_w * x_w = m_r * x_r\)
03

3. Finding the Woman's Final Position

The woman walks to the other end of the raft, so her final position is \(x_w = 6.0m\) from her initial position.
04

4. Finding the Raft's Final Position

Now, substitute the values of \(m_w\), \(m_r\), and \(x_w\) in the momentum conservation equation: \(60 * 6 = 120 * x_r\) Solve for \(x_r\): \(x_r = \frac{60 * 6}{120} = 3 m\) The raft has moved 3 m away from its initial position.
05

5. Calculate the Final Distance Between the Raft and the Pier

The initial distance between the raft and the pier is \(0.50m\). The raft moves away from the pier by \(3m\). So, the final distance between the raft and the pier is: \(d_f = d_i + x_r = 0.50 + 3 = 3.50m\) So the final distance between the raft and the pier is \(3.50m\). Answer to part (a): The distance between the raft and the pier after the woman.walks to the other end of the raft is 3.50 m.
06

6. Calculate the Woman's Distance Relative to the Pier

Since the woman moves \(6.0 m\) to the other end of the raft, and the raft moves away from the pier by \(3m\), the woman moves a total of: \(distance = 6 - 3 = 3m\) Answer to part (b): The woman has moved 3 m relative to the pier.

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

Dash is standing on his frictionless skateboard with three balls, each with a mass of \(100 \mathrm{g}\), in his hands. The combined mass of Dash and his skateboard is \(60 \mathrm{kg} .\) How fast should dash throw the balls forward if he wants to move backward with a speed of \(0.50 \mathrm{m} / \mathrm{s} ?\) Do you think Dash can succeed? Explain.
A uranium nucleus (mass 238 u), initially at rest, undergoes radioactive decay. After an alpha particle (mass \(4.0 \mathrm{u}\) ) is emitted, the remaining nucleus is thorium (mass \(234 \mathrm{u}) .\) If the alpha particle is moving at 0.050 times the speed of light, what is the recoil speed of the thorium nucleus? (Note: "u" is a unit of mass; it is not necessary to convert it to kg.)
A sports car traveling along a straight line increases its speed from $20.0 \mathrm{mi} / \mathrm{h}\( to \)60.0 \mathrm{mi} / \mathrm{h} .$ (a) What is the ratio of the final to the initial magnitude of its momentum? (b) What is the ratio of the final to the initial kinetic energy?
An object located at the origin and having mass \(M\) explodes into three pieces having masses \(M / 4, M / 3,\) and \(5 M / 12 .\) The pieces scatter on a horizontal frictionless xy-plane. The piece with mass \(M / 4\) flies away with velocity \(5.0 \mathrm{m} / \mathrm{s}\) at \(37^{\circ}\) above the \(x\)-axis. The piece with mass \(M / 3\) has velocity \(4.0 \mathrm{m} / \mathrm{s}\) directed at an angle of \(45^{\circ}\) above the \(-x\)-axis. (a) What are the velocity components of the third piece? (b) Describe the motion of the CM of the system after the explosion.
A \(58-\) kg astronaut is in space, far from any objects that would exert a significant gravitational force on him. He would like to move toward his spaceship, but his jet pack is not functioning. He throws a 720 -g socket wrench with a velocity of \(5.0 \mathrm{m} / \mathrm{s}\) in a direction away from the ship. After \(0.50 \mathrm{s}\), he throws a 800 -g spanner in the same direction with a speed of \(8.0 \mathrm{m} / \mathrm{s} .\) After another $9.90 \mathrm{s}\(, he throws a mallet with a speed of \)6.0 \mathrm{m} / \mathrm{s}$ in the same direction. The mallet has a mass of \(1200 \mathrm{g}\) How fast is the astronaut moving after he throws the mallet?
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Energy Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Force

Read Explanation

Conservation of Energy and Momentum

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