/*! 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 18 A \(0.70 \mathrm{~kg}\) ball mov... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(0.70 \mathrm{~kg}\) ball moving horizontally at \(5.0 \mathrm{~m} / \mathrm{s}\) strikes a vertical wall and rebounds with speed \(2.0\) \(\mathrm{m} / \mathrm{s}\). What is the magnitude of the change in its linear momentum?

Short Answer

Expert verified
The magnitude of the change in linear momentum is \(4.9 \mathrm{~kg\cdot m/s}\).

Step by step solution

01

Understand Initial and Final Velocity

The ball initially moves horizontally towards the wall with a velocity of \(5.0 \mathrm{~m/s}\). After striking the wall, it rebounds with a velocity of \(2.0 \mathrm{~m/s}\) in the opposite direction. This means we consider the final velocity to be negative because it's in the opposite direction of the initial velocity.
02

Calculate Initial Momentum

Use the formula for linear momentum \(p = mv\), where \(m\) is mass and \(v\) is velocity. The initial momentum \(p_i\) is given by:\[p_i = 0.70 \mathrm{~kg} \times 5.0 \mathrm{~m/s} = 3.5 \mathrm{~kg\cdot m/s}.\]
03

Calculate Final Momentum

For the final momentum \(p_f\), use the formula \(p = mv\), with the final velocity being negative:\[p_f = 0.70 \mathrm{~kg} \times (-2.0 \mathrm{~m/s}) = -1.4 \mathrm{~kg\cdot m/s}.\]
04

Determine Change in Momentum

The change in momentum \(\Delta p\) is the difference between the final and initial momentum:\[\Delta p = p_f - p_i = -1.4 \mathrm{~kg\cdot m/s} - 3.5 \mathrm{~kg\cdot m/s} = -4.9 \mathrm{~kg\cdot m/s}.\]
05

Calculate Magnitude of Change in Momentum

To find the magnitude of the change in momentum, we take the absolute value of \(\Delta p\):\[\text{Magnitude of } \Delta p = |-4.9 \mathrm{~kg\cdot m/s}| = 4.9 \mathrm{~kg\cdot m/s}.\]

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.

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 Conservation
Momentum conservation is a fundamental principle in physics. It states that the total momentum of a closed system remains constant over time, as long as no external forces are involved. When analyzing the problem of a ball striking a wall:

- The ball initially moves towards the wall, and then it changes direction after impact.
- If you consider the wall and the ball as a system, the wall exerts an equal and opposite force on the ball, influencing it.
- Hence, in an isolated environment without the wall's influence, the momentum in the system would be preserved.

While in this scenario, the wall affects the ball's motion, understanding momentum conservation helps in predicting outcomes in systems without external interference.
Physics Problem Solving
Physics problem solving requires a structured approach to understand and analyze different scenarios. With the exercise in question, the goal is to calculate the change in momentum.

Here's an essential strategy:

- **Identify known values:** the mass of the ball and both initial and final velocities.
- **Apply relevant formulas:** use the linear momentum formula, which involves mass and velocity.
- **Track directions:** velocities going in opposite directions must be distinguished, often by assigning negative or positive signs.

Breaking down the problem into manageable steps ensures a clear path toward the solution.
Kinetic Energy
Kinetic energy represents the energy an object possesses due to its motion. Although the problem focuses on linear momentum, it's helpful to understand how kinetic energy fits into dynamics.

The kinetic energy formula is \[ KE = \frac{1}{2}mv^2 \].

- **Initial kinetic energy** can be calculated as:\[ KE_i = \frac{1}{2}(0.70 \, \mathrm{kg})(5.0 \, \mathrm{m/s})^2 = 8.75 \, \mathrm{J} \].
- **Final kinetic energy** using the rebound speed:\[ KE_f = \frac{1}{2}(0.70 \, \mathrm{kg})(2.0 \, \mathrm{m/s})^2 = 1.4 \, \mathrm{J} \].

By examining these energy values, one can understand how kinetic energy is transformed during the collision. Energy is not conserved in this inelastic collision, unlike momentum.
Collision Dynamics
Collision dynamics encompass how objects interact during impact, including how momentum and energy are transferred. In the scenario of the ball and the wall:

- **Initial momentum** before collision came from the entire mass moving with a certain velocity.
- On **hitting the wall**, the ball's direction changed, leading to a different **momentum** and velocity post-collision.
- The **change in direction** illustrates a transfer of momentum to the wall, though the wall remains unaffected in this simple model.

Analyzing such dynamics helps in understanding the behavior during various types of collisions, such as elastic where kinetic energy remains unchanged, or inelastic where some energy is lost to deformation or sound.

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 \(1.2 \mathrm{~kg}\) ball drops vertically onto a floor, hitting with a speed of \(25 \mathrm{~m} / \mathrm{s}\). It rebounds with an initial speed of \(10 \mathrm{~m} / \mathrm{s}\). (a) What impulse acts on the ball during the contact? (b) If the ball is in contact with the floor for \(0.020 \mathrm{~s}\), what is the magnitude of the average force on the floor from the ball?

At time \(t=0\), force \(\vec{F}_{1}=(-4.00 \hat{\mathrm{i}}+5.00 \hat{\mathrm{j}}) \mathrm{N}\) acts on an initially stationary particle of mass \(2.00 \times 10^{-3} \mathrm{~kg}\) and force \(\vec{F}_{2}=(2.00 \hat{\mathrm{i}}-4.00 \hat{\mathrm{j}}) \mathrm{N}\) acts on an initially stationary particle of mass \(4.00 \times 10^{-3} \mathrm{~kg}\). From time \(t=0\) to \(t=2.00 \mathrm{~ms}\), what are the (a) magnitude and (b) angle (relative to the positive direction of the \(x\) axis) of the displacement of the center of mass of the twoparticle system? (c) What is the kinetic energy of the center of mass at \(t=2.00 \mathrm{~ms} ?\)

A body of mass \(2.0 \mathrm{~kg}\) makes an elastic 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 \(4.0 \mathrm{~m} / \mathrm{s}\) ?

In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land hard on the floor. Suppose the victim falls by \(0.50 \mathrm{~m}\), the mass that moves downward is \(70 \mathrm{~kg}\), and the collision on the floor lasts \(0.082\) s. What are the magnitudes of the (a) impulse and (b) average force acting on the victim from the floor during the collision?

A \(20.0 \mathrm{~kg}\) body is moving through space in the positive direction of an \(x\) axis with a speed of \(200 \mathrm{~m} / \mathrm{s}\) when, due to an internal explosion, it breaks into three parts. One part, with a mass of \(10.0 \mathrm{~kg}\), moves away from the point of explosion with a speed of \(100 \mathrm{~m} / \mathrm{s}\) in the positive \(y\) direction. A second part, with a mass of \(4.00 \mathrm{~kg}\), moves in the negative \(x\) direction with a speed of \(500 \mathrm{~m} / \mathrm{s}\). (a) In unit-vector notation, what is the velocity of the third part? (b) How much energy is released in the explosion? Ignore effects due to the gravitational force.

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.