/*! 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 73 The \(0.5-\mathrm{kg}\) ball is ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The \(0.5-\mathrm{kg}\) ball is fired from the tube at \(A\) with a velocity of \(v=6 \mathrm{~m} / \mathrm{s}\). If the coefficient of restitution between the ball and the surface is \(e=0.8\), determine the height \(h\) after it bounces off the surface.

Short Answer

Expert verified
The height \(h\) after the ball bounces off the surface is 0 meters.

Step by step solution

01

Find the initial vertical velocity

First determine the initial vertical velocity that the ball was fired with. Given that the velocity \(v\) is \(6 \mathrm{m/s}\), and the ball was fired horizontally (90° angle), the vertical component of the velocity is zero. So, \(v_{initial} = 0\).
02

Calculate the final vertical velocity at surface

Next, calculate the velocity at which the ball hits the ground. We'll use the equation of motion \(v_{final} = \sqrt{2 * g * h}\), where \(g = 9.81 \mathrm{m/ s^2}\) (acceleration due to gravity) and \(h\) is the height from which the ball was fired and for this case it is \(0\) (since the ball initially hits the ground). Thus, \(v_{final} = \sqrt{2 * 9.81 * 0} = 0 \mathrm{m/ s}\)
03

Use the coefficient of restitution

Now, we will use the coefficient of restitution which is defined as the ratio of final to initial relative velocity between two objects after they collide. The formula is \(e = \dfrac{v_{after collision} - v_{before collision}} {v_{before collision} - v_{after collision}}\)\nGiven, \(e = 0.8\ and\ v_{before collision} = 0 \mathrm{m/ s}\ and\ v_{after collision} = 0 \mathrm{m/ s}\), we can calculate the velocity after the ball bounces off the surface as \(v_{bounces} = e * v_{final} = 0.8 * 0 = 0 \mathrm{m/ s}\)
04

Find the height after the ball bounces

Finally, to find the height \(h_{bounces}\) after it bounces off the surface, we can use the formula \(h_{bounces} = \dfrac{v_{bounces}^2}{2g}\). After substituting the known values, we obtain \(h_{bounces} = \dfrac{0^2}{2*9.81} = 0 \mathrm{m}\)

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.

Conservation of Momentum
Understanding the conservation of momentum is crucial in analyzing scenarios like the bouncing ball problem. Momentum, a key concept in physics, represents the quantity of motion an object has and is the product of its mass and velocity.

The principle of conservation of momentum states that in an isolated system (where no external forces are acting), the total momentum of the objects before a collision is equal to the total momentum after the collision. For example, when our ball strikes the surface, the momentum it carries just before impact must be equal to the combined momentum of the ball and the surface after the collision, provided the collision is perfectly elastic and no other forces act on them.

This principle can help us predict the outcomes of interactions between objects, like in our exercise where the coefficient of restitution is crucial for determining the height after the ball's bounce. However, there might appear to be an inconsistency here, as the conservation of momentum can't be applied directly since we are only considering the ball's motion and not the Earth's, which is affected negligibly due to its massive size.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. The formula for kinetic energy is given by \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity.

In our exercise, when the ball is fired from the tube, it has a certain amount of kinetic energy based on its velocity. Upon hitting the surface, if the collision was perfectly elastic, the kinetic energy would be conserved; meaning the ball would bounce back with the same kinetic energy. However, in real-world scenarios, collisions are rarely perfectly elastic. The coefficient of restitution, which is less than 1 in the example, indicates that some kinetic energy is lost during the collision, usually converted to heat, sound or other forms of energy, and the ball does not bounce back to the original height.

Applying this concept to the problem helps us understand that the bounce height is dependent on the kinetic energy the ball retains after the impact.
Elastic Collision
An elastic collision is one in which both momentum and kinetic energy are conserved. In such collisions, objects bounce off each other with no loss in total kinetic energy. The coefficient of restitution, \( e \), is a dimensionless value that measures the elasticity of a collision between two objects. When \( e = 1 \) for a perfectly elastic collision, all kinetic energy is conserved.

In our exercise scenario, \( e = 0.8 \) suggests that the collision is not perfectly elastic and some kinetic energy is converted into other forms of energy. Here, \( e \) helps calculate the speed of the ball after the bounce, using the speed before collision. With that information, we can determine the bounce height, \( h \) after the collision, although the calculated bounce velocity in the provided solution is incorrect because the initial impact speed on the surface was not taken into account. Instead, it should be calculated using the first impact velocity derived from the horizontal launch speed and gravity influence.
Mechanics of Materials
Mechanics of materials is a branch of engineering and physics that deals with the behavior of solid objects subjected to stresses and strains. The area includes the study of forces and deformations in materials. In the context of our bouncing ball exercise, we would examine how the ball deforms upon impact, as well as the stress transferred to the surface.

Understanding the mechanics of materials helps in predicting how materials will behave in different situations, like a collision. A rigid and perfectly elastic material would allow the ball to bounce back without losing any kinetic energy. However, materials in the real world often absorb some energy and thus, the ball does not rebound to its original height. The actual coefficient of restitution between the ball and the surface is a result of material properties such as elasticity, rigidity, and energy absorption characteristics.

In conclusion, examining the mechanics of materials is essential for understanding and predicting material responses to collisions, which has direct implications for determining the outcome of impact events like the one described in the exercise.

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

Block \(A\) has a mass of \(2 \mathrm{~kg}\) and slides into an open ended box \(B\) with a velocity of \(2 \mathrm{~m} / \mathrm{s}\). If the box \(B\) has a mass of \(3 \mathrm{~kg}\) and rests on top of a plate \(P\) that has a mass of \(3 \mathrm{~kg}\), determine the distance the plate moves after it stops sliding on the floor. Also, how long is it after impact before all motion ceases? The coefficient of kinetic friction between the box and the plate is \(\mu_{k}=0.2,\) and between the plate and the floor \(\mu_{k}^{\prime}=0.4\). Also, the coefficient of static friction between the plate and the floor is \(\mu_{s}^{\prime}=0.5\)

A \(50-\mathrm{kg}\) crate rests against a stop block \(s\), which prevents the crate from moving down the plane. If the coefficients of static and kinetic friction between the plane and the crate are \(\mu_{s}=0.3\) and \(\mu_{k}=0.2,\) respectively, determine the time needed for the force \(\mathbf{F}\) to give the crate a speed of \(2 \mathrm{~m} / \mathrm{s}\) up the plane. The force always acts parallel to the plane and has a magnitude of \(F=(300 t) \mathrm{N},\) where \(t\) is in seconds. Hint: First determine the time needed to overcome static friction and start the crate moving.

An earth satellite of mass \(700 \mathrm{~kg}\) is launched into a free-flight trajectory about the earth with an initial speed of \(v_{A}=10 \mathrm{~km} / \mathrm{s}\) when the distance from the center of the earth is \(r_{A}=15 \mathrm{Mm}\). If the launch angle at this position is \(\phi_{A}=70^{\circ},\) determine the speed \(v_{B}\) of the satellite and its closest distance \(r_{B}\) from the center of the earth. The earth has a mass \(M_{e}=5.976\left(10^{24}\right) \mathrm{kg} .\) Hint: Under these conditions, the satellite is subjected only to the earth's gravitational force, \(F=G M_{e} m_{s} / r^{2},\) Eq. \(13-1\). For part of the solution, use the conservation of energy.

A hockey puck is traveling to the left with a velocity of \(v_{1}=10 \mathrm{~m} / \mathrm{s}\) when it is struck by a hockey stick and given a velocity of \(v_{2}=20 \mathrm{~m} / \mathrm{s}\) as shown. Determine the magnitude of the net impulse exerted by the hockey stick on the puck. The puck has a mass of \(0.2 \mathrm{~kg}\).

If the rod of negligible mass is subjected to a couple moment of \(M=\left(30 t^{2}\right) \mathrm{N} \cdot \mathrm{m},\) and the engine of the car supplies a traction force of \(F=(15 t) \mathrm{N}\) to the wheels, where \(t\) is in seconds, determine the speed of the car at the instant \(t=5 \mathrm{~s}\). The car starts from rest. The total mass of the car and rider is \(150 \mathrm{~kg}\). Neglect the size of the car.

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.