/*! 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 81 I Immediately after the collisio... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

I Immediately after the collision, the momentum of the club \(+\) ball system will be A. Less than before the collision. B. The same as before the collision. C. Greater than before the collision.

Short Answer

Expert verified
The answer is B. The total momentum will be the same as before the collision.

Step by step solution

01

Identify the calculation formula

The law of conservation of momentum states that the total momentum before the collision equals total momentum after the collision, as no external forces are acting on the system. The momentum is calculated using the formula: Momentum = mass * velocity.
02

Apply the conservation law to before and after the collision

Before the collision, only the club has momentum (as it's in motion) and the ball is at rest. So, before momentum = Mass_club * velocity_club. After the collision, both the ball and the club have momentum: so, after momentum = Mass_club * Velocity_club(new) + Mass_ball * Velocity_ball. According to the conservation law, before momentum = after momentum.
03

Compare the total momentum before and after the collision

The comparison of momentums will show that the total momentum before the collision is equal to the total momentum after the collision. The momentums are the same before and after the collision as no external forces are acting on the system.

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.

Momentum Calculation
Momentum is a fundamental concept in physics, essential for understanding the motion of objects. It is calculated using the formula: \[ \text{Momentum} = \text{mass} \times \text{velocity} \]This means that the momentum of an object depends directly on two factors:
  • Mass: the amount of matter in the object.
  • Velocity: the speed and direction the object is moving.
When two objects collide, the changes in their velocity and possibly their mass (if parts break off or combine) will affect their resulting momentum.
Keep in mind that momentum is a vector quantity, meaning it has both magnitude and direction. Always remember to consider both, especially in multi-dimensional problems. The vector nature becomes particularly important in collision problems, as the direction of motion can change.
Collision Physics
The study of collisions in physics is fascinating because it allows us to apply core principles like the conservation of momentum. In a closed system, where no external forces like friction or air resistance act, momentum is conserved during collisions.
There are two main types of collisions:
  • Elastic Collisions: Both momentum and kinetic energy are conserved. The total kinetic energy before and after the collision remains the same.
  • Inelastic Collisions: While momentum is still conserved, kinetic energy is not. These collisions often result in energy transformation into sound, heat, or deformation.
Understanding the type of collision helps us set up the equations needed to find out the post-collision velocities or other variables. Analyzing collisions enhances our understanding of fundamental physics laws and prepares us for solving complex real-world dynamics scenarios.
Physics Problem Solving
Solving physics problems effectively involves a systematic approach. Here's a simple guide to tackle problems involving momentum and collisions:
1. **Understand the Problem:** Carefully read the problem statement. Identify what is given and what needs to be determined.
2. **Visualize:** Draw diagrams if necessary. A clear visual representation can help simplify complex situations.
3. **Identify Relevant Laws:** Figure out what physics principles apply. For collisions, the conservation of momentum is key.
4. **Set up Equations:** Using the applicable laws, write equations that relate the given quantities to the unknowns.
  • Use \[ m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \] for momentum, where primed variables denote after the collision.
5. **Solve Mathematically:** Simplify and solve these equations via algebra or calculus if required. Check your solutions to ensure they are physically reasonable.
By applying this structured approach, tackling physics problems becomes not only manageable but also enjoyable, as you reveal the underlying laws governing our universe.

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 manufacturer makes a golf ball that compresses more than a traditional golf ball when struck by a club. How will this affect the average force during the collision? A. The force will decrease. B. The force will not be affected. C. The force will increase.

\(\mathrm{A} 71 \mathrm{kg}\) baseball player jumps straight up to catch a hard- hit ball. If the 140 g ball is moving horizontally at \(28 \mathrm{m} / \mathrm{s}\), and the catch is made when the ballplayer is at the highest point of his leap, what is his speed immediately after stopping the ball?

\(\|\) A spaceship of mass \(2.0 \times 10^{6} \mathrm{kg}\) is cruising at a speed of \(5.0 \times 10^{6} \mathrm{m} / \mathrm{s}\) when the antimatter reactor fails, blow- ing the ship into three pieces. One section, having a mass of \(5.0 \times 10^{5} \mathrm{kg},\) is blown straight backward with a speed of \(2.0 \times 10^{6} \mathrm{m} / \mathrm{s} .\) A second piece, with mass \(8.0 \times 10^{5} \mathrm{kg},\) contin- ues forward at \(1.0 \times 10^{6} \mathrm{m} / \mathrm{s} .\) What are the direction and speed of the third piece?

The carbon isotope \({ }^{14} \mathrm{C}\) is used for carbon dating of archeological artifacts. \({ }^{14} \mathrm{C}\) (mass \(2.34 \times 10^{-26} \mathrm{kg}\) ) decays by the process known as beta decay in which the nucleus emits an electron (the beta particle) and a subatomic particle called a neutrino. In one such decay, the electron and the neutrino are emitted at right angles to each other. The electron (mass \(9.11 \times 10^{-31} \mathrm{kg}\) ) has a speed of \(5.00 \times 10^{7} \mathrm{m} / \mathrm{s}\) and the neutrino has a momentum of \(8.00 \times 10^{-24} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} .\) What is the recoil speed of the nucleus?

A student throws a 120 g snowball at \(7.5 \mathrm{m} / \mathrm{s}\) at the side of the schoolhouse, where it hits and sticks. What is the magnitude of the average force on the wall if the duration of the collision is \(0.15 \mathrm{s} ?\)

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.