/*! 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 64 A steel ball of mass \(0.500 \ma... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 64

A steel ball of mass \(0.500 \mathrm{~kg}\) is fastened to a cord that is \(70.0 \mathrm{~cm}\) long and fixed at the far end. The ball is then released when the cord is horizontal (Fig. 9-65). At the bottom of its path, the ball strikes a \(2.50 \mathrm{~kg}\) steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.

Short Answer

Expert verified
Ball: 1.11 m/s; Block: 2.22 m/s.

Step by step solution

01

Determine Initial Speed of the Ball Using Energy Conservation

When the ball is released from the horizontal, it falls freely under gravity. At the lowest point of its swing, all potential energy is converted to kinetic energy. The initial potential energy of the ball is given by \(PE = mgh\), where \(m = 0.500 \text{ kg}\), \(g = 9.81 \text{ m/s}^2\), and \(h = 0.70 \text{ m}\). Thus, \(PE = 0.500 \times 9.81 \times 0.70\). When the ball reaches the lowest point, its kinetic energy \(KE = \frac{1}{2}mv^2\) equals the initial potential energy. Set \(mgh = \frac{1}{2}mv^2\) to solve for \(v\), yielding \(v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.70} \approx 3.70 \text{ m/s}\).
02

Apply Conservation of Momentum for the Collision

In an elastic collision, both momentum and kinetic energy are conserved. The total initial momentum is only from the ball, given by \(p_i = mv = 0.500 \times 3.70\). The total final momentum is the sum of the momentum of the ball and the block, which gives us the equation \(0.500v' + 2.50v_{block} = 0.500 \times 3.70\).
03

Apply Conservation of Kinetic Energy for the Collision

For elastic collisions, kinetic energy is also conserved. Set the initial kinetic energy of the ball equal to the sum of the kinetic energies of both the ball and block after collision: \(\frac{1}{2} \times 0.500 \times (3.70)^2 = \frac{1}{2} \times 0.500 \times (v')^2 + \frac{1}{2} \times 2.50 \times (v_{block})^2\).
04

Solve the System of Equations

Now solve the two equations simultaneously: 1. From step 2: \(0.500v' + 2.50v_{block} = 0.500 \times 3.70\).2. From step 3: \(0.5 \times 0.500 \times 3.70^2 = 0.5 \times 0.500 \times v'^2 + 0.5 \times 2.50 \times v_{block}^2\). Solving these gives: - For the ball: \(v' \approx 1.11 \text{ m/s}\) - For the block: \(v_{block} \approx 2.22 \text{ m/s}\).

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 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Elastic collisions
In an elastic collision, both momentum and kinetic energy are conserved. This means that no kinetic energy is lost during the collision. When a steel ball strikes a steel block, as in our problem, the bounce-back effect is governed by these principles.
  • The kinetic energy before and after the collision remains unchanged.
  • The meeting point sees a perfect transfer of momentum between the colliding bodies.
These phenomena make elastic collisions fascinating, especially in physics problems. Understanding them is crucial in physics, allowing for predictable outcomes based on initial conditions.
Conservation of momentum
The law of conservation of momentum states that the total momentum of a closed system is constant unless acted upon by external forces. In our problem, the ball and block system is isolated at the moment of collision, meaning total momentum remains unchanged.
  • Initial momentum comes solely from the moving ball: \( p_i = mv \).
  • After collision, the momentum is shared between the ball and the block.
  • This results in the equation: \( 0.500v' + 2.50v_{block} = 0.500 \times 3.70 \).
Solving momentum equations helps find the post-collision speeds of both objects. This is fundamental to mastering collision problems in physics.
Conservation of energy
The principle of conservation of energy asserts energy is neither created nor destroyed but transformed. In our scenario, the ball's gravitational potential energy converts fully into kinetic energy as it swings down.
  • Initially, the ball has gravitational potential energy: \( PE = mgh \).
  • At the lowest point, this transfers entirely to kinetic energy: \( KE = \frac{1}{2}mv^2 \).
  • This interaction ensures the ball has maximum kinetic energy before impact.
Realizing this energy conversion is paramount to understanding the pre-collision speed of the ball.
Kinetic energy
Kinetic energy is the energy possessed by an object due to its motion, given by \( KE = \frac{1}{2}mv^2 \). When the ball reaches the bottom of its arc, it's at peak kinetic energy due to its conversion from potential energy.
  • The formula relates velocity directly to kinetic energy.
  • Before collision, the ball's kinetic energy is provided by this calculation.
Going from potential to kinetic energy is pivotal in solving how fast the ball moves initially and after impact.
Potential energy
Potential energy is the stored energy of position, readily observable in our ball's initial position. Here, the ball's potential energy is gravitational due to its height above the lowest swing point.
  • Calculated through: \( PE = mgh \).
  • It's the ball's energy while at rest horizontally.
  • This potential becomes kinetic as the ball descends.
Understanding this shift is critical for tracing the energy journey from the ball's start to its collision with the block.

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

Ricardo, of mass \(80 \mathrm{~kg}\), and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a \(30 \mathrm{~kg}\) canoe. When the canoe is at rest in the placid water, they exchange seats, which are \(3.0 \mathrm{~m}\) apart and symmetrically located with respect to the canoe's center. If the canoe moves \(40 \mathrm{~cm}\) horizontally relative to a pier post, what is Carmelita's mass?

A 75 kg man rides on a \(39 \mathrm{~kg}\) cart moving at a velocity of \(2.3 \mathrm{~m} / \mathrm{s}\). He jumps off with zero horizontal velocity relative to the ground. What is the resulting change in the cart's velocity, including sign?

An object, with mass \(m\) and speed \(v\) relative to an observer, explodes into two pieces, one three times as massive as the other; the explosion takes place in deep space. The less massive piece stops relative to the observer. How much kinetic energy is added to the system during the explosion, as measured in the observer's reference frame?

A stone is dropped at \(t=0\). A second stone, with twice the mass of the first, is dropped from the same point at \(t=100 \mathrm{~ms}\). (a) How far below the release point is the center of mass of the two stones at \(t=300 \mathrm{~ms} ?\) (Neither stone has yet reached the ground.) (b) How fast is the center of mass of the twostone system moving at that time?

A \(91 \mathrm{~kg}\) man lying on a surface of negligible friction shoves a \(68 \mathrm{~g}\) stone away from himself, giving it a speed of \(4.0 \mathrm{~m} / \mathrm{s}\). What speed does the man acquire as a result?
See all solutions

Recommended explanations on Physics Textbooks

Electricity and Magnetism

Read Explanation

Modern Physics

Read Explanation

Radiation

Read Explanation

Torque and Rotational Motion

Read Explanation

Circular Motion and Gravitation

Read Explanation

Energy Physics

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.