Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A 2.0-kg block is moving to the right at \(1.0 \mathrm{m} / \mathrm{s}\) just before it strikes and sticks to a 1.0-kg block initially at rest. What is the total momentum of the two blocks after the collision?

Short Answer

Expert verified
Answer: The total momentum of the two blocks after the collision is 2.0 kg m/s.

Step by step solution

01

Express the given values and the conservation of momentum principle

First, express the given values and the momentum formula. Initially, let the mass of block A be \(m_A = 2.0 \,\text{kg}\) and its velocity be \(v_A = 1.0 \,\text{m/s}\). Let the mass of block B be \(m_B = 1.0 \,\text{kg}\) and its initial velocity be \(v_B = 0 \,\text{m/s}\) as it is at rest. The total momentum before collision is the sum of the momenta of individual blocks, i.e., \(p_{\text{total}} = m_A v_A + m_B v_B\).
02

Calculate the total momentum before collision

Use the formula and the given values to calculate the total momentum before the collision: \(p_{\text{total}} = m_A v_A + m_B v_B = (2.0 \,\text{kg})(1.0 \,\text{m/s}) + (1.0 \,\text{kg})(0 \,\text{m/s}) = 2.0 \,\text{kg m/s}\). The total momentum before the collision is \(2.0 \,\text{kg m/s}\).
03

Apply the conservation of momentum principle

The conservation of momentum principle states that the total momentum before the collision is equal to the total momentum after the collision. Since both blocks stick together after the collision, they move together as one single mass with the same velocity. The combined mass can be written as \(m_{\text{combined}} = m_A + m_B\). Let the velocity of the combined blocks after the collision be \(v_f\). Then the total momentum after the collision can be stated as \(p_{\text{total}} = m_{\text{combined}} v_f\).
04

Calculate the velocity of the combined blocks after the collision

To find the velocity of the combined blocks after the collision, set the total momentum before the collision equal to the total momentum after the collision and solve for the velocity \(v_f\): \(2.0 \,\text{kg m/s} = (2.0 \,\text{kg} + 1.0 \,\text{kg}) v_f\). Now, divide both sides by the combined mass: \(v_f = \frac{2.0 \,\text{kg m/s}}{3.0 \,\text{kg}} = 0.67\,\text{m/s}\). The velocity of the combined blocks after the collision is \(0.67\,\text{m/s}\).
05

Calculate the total momentum after the collision

Finally, use the velocity of the combined blocks after the collision and the combined mass to find the total momentum after the collision: \(p_{\text{total}} = m_{\text{combined}} v_f = (3.0 \,\text{kg})(0.67\,\text{m/s}) = 2.0\,\text{kg m/s}\). The total momentum of the two blocks after the collision is \(2.0\,\text{kg m/s}\).

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.

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

An automobile weighing \(13.6 \mathrm{kN}\) is moving at $17.0 \mathrm{m} / \mathrm{s}\( when it collides with a stopped car weighing \)9.0 \mathrm{kN} .$ If they lock bumpers and move off together, what is their speed just after the collision?
Two identical gliders, each with elastic bumpers and mass \(0.10 \mathrm{kg},\) are on a horizontal air track. Friction is negligible. Glider 2 is stationary. Glider 1 moves toward glider 2 from the left with a speed of $0.20 \mathrm{m} / \mathrm{s} .$ They collide. After the collision, what are the velocities of glider 1 and glider \(2 ?\)
Body A of mass \(M\) has an original velocity of \(6.0 \mathrm{m} / \mathrm{s}\) in the \(+x\)-direction toward a stationary body (body \(\mathrm{B}\) ) of the same mass. After the collision, body A has velocity components of $1.0 \mathrm{m} / \mathrm{s}\( in the \)+x\( -direction and \)2.0 \mathrm{m} / \mathrm{s}$ in the + y-direction. What is the magnitude of body B's velocity after the collision?
A 0.15-kg baseball is pitched with a speed of \(35 \mathrm{m} / \mathrm{s}\) \((78 \mathrm{mph}) .\) When the ball hits the catcher's glove, the glove moves back by \(5.0 \mathrm{cm}(2 \text { in. })\) as it stops the ball. (a) What was the change in momentum of the baseball? (b) What impulse was applied to the baseball? (c) Assuming a constant acceleration of the ball, what was the average force applied by the catcher's glove?
A system consists of three particles with these masses and velocities: mass \(3.0 \mathrm{kg},\) moving north at \(3.0 \mathrm{m} / \mathrm{s}\) mass $4.0 \mathrm{kg},\( moving south at \)5.0 \mathrm{m} / \mathrm{s} ;\( and mass \)7.0 \mathrm{kg}\( moving north at \)2.0 \mathrm{m} / \mathrm{s} .$ What is the total momentum of the system?
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.

Sign-up for free