/*! 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 particle loses \(25 \%\) of it... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A particle loses \(25 \%\) of its energy during collision with another identical particle at rest. the coefficient of restitution will be (1) \(0.25\) (2) \(\sqrt{2}\) (3) \(\frac{1}{\sqrt{2}}\) (4) \(0.5\)

Short Answer

Expert verified
The coefficient of restitution is \( \frac{1}{\sqrt{2}} \), option (3).

Step by step solution

01

Understand the Energy Loss

The problem states the particle loses 25% of its energy during the collision. This means that 75% of the original kinetic energy is retained after the collision.
02

Initial and Final Kinetic Energy

Let the initial kinetic energy of the particle be denoted as \( E_i \). After the collision, the retained kinetic energy \( E_f \) is 75% of \( E_i \), so \( E_f = 0.75 E_i \).
03

Relate Kinetic Energy to Velocity

If the initial velocity of the particle is \( v_i \), then the initial kinetic energy \( E_i = \frac{1}{2}mv_i^2 \). Thus, the final kinetic energy \( E_f = \frac{1}{2}m v_f^2 = 0.75 \times \frac{1}{2}m v_i^2 \).
04

Solve for Final Velocity

By equating the expressions for the final kinetic energy, we have \( \frac{1}{2}mv_f^2 = 0.75 \times \frac{1}{2}mv_i^2 \). Simplifying, \( v_f^2 = 0.75 v_i^2 \). Hence, \( v_f = \sqrt{0.75} v_i = \frac{\sqrt{3}}{2} v_i \).
05

Definition of Coefficient of Restitution

The coefficient of restitution \( e \) is defined as the relative velocity of separation to the relative velocity of approach between two particles after and before collision, respectively. For this case, it is \( e = \frac{v_f}{v_i} \).
06

Calculate the Coefficient of Restitution

Using the relation between the final and initial velocities, \( e = \frac{v_f}{v_i} = \frac{\frac{\sqrt{3}}{2} v_i}{v_i} = \frac{\sqrt{3}}{2} \approx \frac{1}{\sqrt{2}} \).

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.

Energy Loss in Collisions
When two particles collide, the energy before and after the collision can change, primarily due to energy transformations and some irreversible processes, like sound and heat production. In this exercise, we examine a particle collision where a particle loses 25% of its energy. This means only 75% of the original energy remains available for kinetic activities.

Energy loss in such collisions is often quantified using the retained kinetic energy. When energy is lost, it does not mysteriously disappear; rather, it converts into other forms. Consider the following transformations:
  • Sound: Energy may produce sound waves during impact.
  • Heat: Friction in a collision can cause heat to be generated.
  • Deformation: Some energy might lead to bending or change of shape.
Understanding energy transformations is crucial when analyzing collisions as it helps in calculating key measures, such as velocities after the impact, and metrics like the coefficient of restitution.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. In the study of collisions, kinetic energy is the primary form we look at to understand how particles behave before and after they collide. For a particle with mass \( m \) moving at velocity \( v \), the kinetic energy \( E \) can be calculated using the formula:\[E = \frac{1}{2}mv^2\]
This equation tells us that kinetic energy is directly proportional to the mass and the square of the velocity.

In our exercise, the initial kinetic energy of the particle is lost by 25% due to the collision. Therefore, the final kinetic energy shrinks to 75% of the initial value. This change in kinetic energy can be rearranged as:
\[E_f = 0.75E_i = 0.75 \cdot \frac{1}{2}mv_i^2\]This equation shows how the retained kinetic energy is directly connected to the original velocity \( v_i \). It's critical to compare these energies to understand how energetic transformations influence the outcome of particle interactions.
Velocity
Velocity is a key factor affecting the dynamics of a collision. In this context, velocity not only determines the kinetic energy of the particles involved but also influences other parameters like the coefficient of restitution.

To find the final velocity \( v_f \), we first relate it to the kinetic energy retained, using:
  • Initial velocity \( v_i \)
  • The formula for kinetic energy
  • Conservation principles
Given that \( v_f^2 = 0.75v_i^2 \), the final velocity \( v_f \) is found to be:\[ v_f = \sqrt{0.75}v_i = \frac{\sqrt{3}}{2}v_i\]This derivation reveals that \( v_f \) is less than \( v_i \), representing the slowing down resulting from energy loss.

The coefficient of restitution, \( e \), defined as the ratio \( \frac{v_f}{v_i} \), further helps in understanding how velocities change after impact. In simpler terms, it describes how "bouncy" the collision is. Calculating for \( e \):\[e = \frac{\frac{\sqrt{3}}{2}v_i}{v_i} = \frac{\sqrt{3}}{2} \approx \frac{1}{\sqrt{2}}\]This number provides insights into both velocity and energy transformations during a collision.

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 ball of mass \(1 \mathrm{~kg}\) is dropped from a height of \(3.2 \mathrm{~m}\) on smooth inclined plane. The coefficient of restitution for the collision is \(e=1 / 2\). The ball's velocity become horizontal after the collision. (1) The angle \(\theta=\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)\). (2) The speed of the ball after the collision \(=4 \sqrt{2} \mathrm{~ms}^{-1}\). (3) The total loss in kinetic energy during the collision is \(8 \mathrm{~J}\) (4) The ball hits the inclined plane again while travelling vertically downward.

A body of mass \(2 \mathrm{~kg}\) moving with a velocity \(3 \mathrm{~m} / \mathrm{s}\) collides with a body of mass \(1 \mathrm{~kg}\) moving with a velocity of \(4 \mathrm{~m} / \mathrm{s}\) in opposite direction. If the collision is head on and completely inelastic, then (1) both particles move together with velocity \((2 / 3) \mathrm{m} / \mathrm{s}\) (2) the momentum of system is \(2 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) throughout (3) the momentum of system is \(10 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) (4) the loss of \(\mathrm{KE}\) of system is \((49 / 3) \mathrm{J}\)

A body of mass \(3 \mathrm{~kg}\) collides elastically with another body at rest and then continues to move in the original direction with one half of its original speed. What is the mass of the target body? (1) \(1 \mathrm{~kg}\) (2) \(1.5 \mathrm{~kg}\) (3) \(2 \mathrm{~kg}\) (4) \(5 \mathrm{~kg}\)

A stream of glass beads, each with a mass of 15 gram, comes out of a horizontal tube at a rate of 100 per second. The beads fall a distance of \(5 \mathrm{~m}\) to a balance pan and bounce back to their original height. How much mass (in \(\mathrm{kg}\) ) must be placed in the other pan of the balance to keep the pointer at zero?

A ball strikes a wall with a velocity \(\vec{u}\) at an angle \(\theta\) with the normal to the wall surface and rebounds from it at an angle \(\beta\) with the surface. Then (1) \((\theta+\beta)<90^{\circ}\), if the wall is smooth (2) if the wall is rough, coefficient of restitution \(=\tan \beta / \cos \theta\) (3) if the wall is rough, coefficient of restitution \(<\tan \beta / \cot \theta\) (4) none of the above

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.