Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 9

An object of mass \(3.0 \mathrm{kg}\) is allowed to fall from rest under the force of gravity for 3.4 s. What is the change in its momentum? Ignore air resistance.

Short Answer

Expert verified
Answer: The change in momentum is approximately 99.96 kg·m/s.

Step by step solution

01

Calculate the final velocity of the object

Using the equation of motion \(v = u + at\), the final velocity can be calculated as follows: - \(u = 0 \mathrm{m/s}\) (initial velocity) - \(a = 9.8 \mathrm{m/s^2}\) (acceleration due to gravity) - \(t = 3.4 \mathrm{s}\) (time) Plugging in these values, we have \(v = 0 + 9.8 \times 3.4 = 33.32 \mathrm{m/s}\).
02

Calculate the change in momentum

Now, we can calculate the change in momentum using the formula \(\Delta p = m(v - u)\). Plugging in the values we have: - \(m = 3.0 \mathrm{kg}\) (mass) - \(v = 33.32 \mathrm{m/s}\) (final velocity) - \(u = 0 \mathrm{m/s}\) (initial velocity) So, \(\Delta p = 3.0 \times (33.32 - 0) = 99.96 \mathrm{kg \cdot m/s}\). The change in momentum of the object is approximately \(99.96 \mathrm{kg \cdot m/s}\).

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

A 3.0-kg body is initially moving northward at \(15 \mathrm{m} / \mathrm{s}\) Then a force of \(15 \mathrm{N},\) toward the east, acts on it for 4.0 s. (a) At the end of the 4.0 s, what is the body's final velocity? (b) What is the change in momentum during the \(4.0 \mathrm{s} ?\)
A \(0.020-\mathrm{kg}\) bullet traveling at \(200.0 \mathrm{m} / \mathrm{s}\) east hits a motionless \(2.0-\mathrm{kg}\) block and bounces off it, retracing its original path with a velocity of \(100.0 \mathrm{m} / \mathrm{s}\) west. What is the final velocity of the block? Assume the block rests on a perfectly frictionless horizontal surface.
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)?
In a nuclear reactor, a neutron moving at speed \(v_{\mathrm{i}}\) in the positive \(x\)-direction strikes a deuteron, which is at rest. The neutron is deflected by \(90.0^{\circ}\) and moves off with speed $v_{\mathrm{i}} / \sqrt{3}\( in the positive \)y\(-direction. Find the \)x\(- and \)y$-components of the deuteron's velocity after the collision. (The mass of the deuteron is twice the mass of the neutron.)
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.)
See all solutions

Recommended explanations on Physics Textbooks

Turning Points in Physics

Read Explanation

Classical Mechanics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Linear Momentum

Read Explanation

Radiation

Read Explanation

Thermodynamics

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