/*! 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 ball falls vertically on to a ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 18

A ball falls vertically on to a horizontal plane and bounces. Find the impulse the ball exerts on the plane if: (a) the ball is initially \(2 \mathrm{~m}\) above the plane, (b) it rises after bouncing to a height \(1.2 \mathrm{~m}\), (c) the coefficient of restitution is \(\sqrt{\frac{2}{3}}\) (d) the mass of the ball is \(0.5 \mathrm{~kg}\).

Short Answer

Expert verified
The impulse exerted on the plane is 5.685 kg 路 m/s.

Step by step solution

01

- Calculate the initial velocity just before impact

The velocity just before impact can be found using the equation for a freely falling object with initial height 2 m:\[v_1 = \sqrt{2gh_1} = \sqrt{2 \times 9.8 \times 2} = \sqrt{39.2} \approx 6.26 \mathrm{\frac{m}{s}}\]
02

- Calculate the velocity after the bounce

Use the equation for the coefficient of restitution to find the velocity after the bounce. The coefficient of restitution (e) is given by:\[e = \sqrt{\frac{2}{3}} = \frac{v_2}{v_1}\]Solve for \(v_2\):\[v_2 = e \times v_1 = \sqrt{\frac{2}{3}} \times 6.26 = \sqrt{\frac{2}{3}} \times 6.26 \approx 5.11 \mathrm{\frac{m}{s}}\]
03

- Calculate the impulse

Impulse (I) is given by the change in momentum, which can be calculated using the initial and final velocities:If we take downward as positive (initial velocity), the initial momentum before impact is:\[p_1 = m \times v_1 = 0.5 \times 6.26 = 3.13 \mathrm{kg \cdot \frac{m}{s}}\]The final momentum after the bounce is (upward direction is taken as negative):\[p_2 = m \times -v_2 = 0.5 \times -5.11 = -2.555 \mathrm{kg \cdot \frac{m}{s}}\]The impulse exerted on the plane is the change in momentum:\[I = p_2 - p_1 = -2.555 - 3.13 = -5.685 \mathrm{kg \cdot \frac{m}{s}}\]Since we only need the magnitude for impulse:\[I = 5.685 \mathrm{kg \cdot \frac{m}{s}}\]

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.

Free Fall
To understand the concept of free fall, let鈥檚 start with the basic idea. When an object falls solely under the influence of gravity, it is in free fall. This means no other forces, like air resistance, are acting on it. In this case, the ball is initially dropped from a height of 2 meters.
To find the velocity just before the ball hits the ground, we use the kinematic equation: \(v_1 = \sqrt{2gh_1} = \sqrt{2 \times 9.8 \times 2} = \sqrt{39.2} \approx 6.26 \mathrm{m/s}\).
This calculation shows that at this height, the ball鈥檚 velocity just before impact is approximately 6.26 meters per second.
Coefficient of Restitution
The coefficient of restitution (commonly denoted as e) measures how 'bouncy' a collision is. It鈥檚 the ratio of the relative speed after collision to the relative speed before collision.
Here's the mathematical expression: \( e = \sqrt{\frac{2}{3}} = \frac{v_2}{v_1}\).
In this exercise, \( e = \sqrt{\frac{2}{3}} \) is given. We use this to find the velocity after the bounce (\(v_2\)). By rearranging the formula,\( v_2 = e \times v_1 \), we get \( v_2 = \sqrt{\frac{2}{3}} \times 6.26 \approx 5.11 \mathrm{m/s}\). Thus, the velocity after rebounding is about 5.11 meters per second.
Momentum Change
Momentum is the product of an object's mass and velocity. It鈥檚 a vector quantity, meaning it has both magnitude and direction. Before and after the bounce, the momentum of the ball changes.
Initially: \( p_1 = m \times v_1 = 0.5 \times 6.26 = 3.13 \mathrm{kg \cdot m/s}\).
After the bounce (considering the upward direction as negative): \( p_2 = m \times (-v_2) = 0.5 \times (-5.11) = -2.555 \mathrm{kg \cdot m/s}\).
The change in momentum is therefore: \( \Delta_p = p_2 - p_1 = -2.555 - 3.13 = -5.685 \mathrm{kg \cdot m/s}\). The magnitude of this momentum change results in impulse.
Kinematics
Kinematics deals with the motion of objects without considering the forces causing them. In this exercise, we use kinematic principles to find the velocities before and after the ball bounces.
For free fall, the formula used is: \( v_1 = \sqrt{2gh_1} \) for velocity just before impact. And for bounce, we use the coefficient of restitution to find: \( v_2 = e \times v_1 \). This helps explain the velocities at different stages of the ball's motion.
Impact Forces
When the ball hits the plane, it exerts a force during the impact, which results in an impulse. Impulse is essentially the change in momentum. \( I = \Delta_p \).
This translates to: \( I = p_2 - p_1 = -2.555 - 3.13 = -5.685 \mathrm{kg \cdot m/s} \).
This value signifies the total momentum change during the collision. The magnitude of the impulse tells us how much force was exerted on the plane by the ball.

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 particle of mass \(2 \mathrm{~kg}\) moving with speed \(4 \mathrm{~ms}^{-1}\) is given a blow which changes the speed to \(1 \mathrm{~ms}^{-1}\) without deflecting the particle from a straight line. The impulse of the blow is: (a) \(10 \mathrm{Ns}\), (b) \(6 \mathrm{Ns}\), (c) we do not know whether it is \(10 \mathrm{Ns}\) or \(6 \mathrm{Ns}\).

Two masses of 20 and 10 units. moving in the same direction at speeds of 16 and 12 units respectively collide and stick together. Find the velocity of the combined mass immediately afterwards.

A smooth sphere \(\mathrm{A}\) of mass \(2 m\), moving on a horizontal plane with speed \(u\), collides directly with another smooth sphere B of equal radius and of mass \(m\), which is at rest. If the coefficient of restitution between the spheres is \(e\), find their speeds after impact. The sphere B later rebounds from a perfectly elastic vertical wall, and then collides directly with A. Prove that after this collision the speed of B is \(9(1+e)^{2} u\) and find the speed of \(\mathrm{A}\).

A sphere of mass \(m\) falls from rest at a height \(h\) above a horizontal plane and rebounds to a height \(\frac{h}{2}\), Find the coefficient of restitution, the impulse exerted by the plane and the loss in \(K E\) due to impact.

A particle of mass \(m\) is projected vertically upward with speed \(u\) and when it reaches its greatest height a second particle, of mass \(2 m\), is projected vertically upward with speed \(2 u\) from the same point as the first. Prove that the time that elapses between the projection of the second particle and its collision with the first is \(\frac{u}{4 g}\), and find the height above the point of projection at which the collision occurs. If, on collision, the particles coalesce, prove that the combined particle will reach a greatest height of \(\frac{19 u^{2}}{18 g}\) above the point of projection.
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

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.