/*! 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 30 A particle of mass \(m_{1}\) ela... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A particle of mass \(m_{1}\) elastically collides with anote stationary particle of mass \(m_{2} .\) Then if (1) \(\frac{m_{1}}{m_{2}}=\frac{1}{2}\) and the collision is head on, then the particle fly apart in the opposite direction with unequal speed. (2) \(\frac{m_{1}}{m_{2}}=\frac{1}{3}\) and the collision is head on, then the particle fly apart in the opposite direction with equal speed. (3) \(\frac{m_{1}}{m_{2}}=\frac{2}{1}\) and the collision is head on, then the particle fly apart in the opposite direction with unequal speed (4) \(\frac{m_{1}}{m_{2}}=\frac{3}{1}\) and the collision is head on, then the particle fly apart in the opposite direction with equal speed.

Short Answer

Expert verified
Option (2) is correct: \(\frac{m_1}{m_2} = \frac{1}{3}\), the particles fly apart with equal speed.

Step by step solution

01

Understand the concept of elastic collision

In an elastic collision, both momentum and kinetic energy are conserved. This means we have two equations to describe the collision: one for the conservation of momentum and one for the conservation of kinetic energy.
02

Write the conservation of momentum equation

Before the collision, the particle with mass \(m_1\) is moving towards a stationary particle with mass \(m_2\). The total momentum before the collision is \(m_1 u_1\). After the collision, the particles have velocities \(v_1\) and \(v_2\) respectively. Therefore, the momentum conservation equation is \(m_1 u_1 = m_1 v_1 + m_2 v_2\).
03

Write the conservation of kinetic energy equation

The initial kinetic energy is \(\frac{1}{2}m_1 u_1^2\). After the collision, the kinetic energy is \(\frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2\). Therefore, the energy conservation equation is \(\frac{1}{2}m_1 u_1^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2\).
04

Solve for conditions with equal speed

For the particles to fly apart with equal speed, their masses have to be equal. Therefore, when \(m_1 = m_2\), the velocities after collision are \(v_1 = -v_2\). Check the given options for equal speeds. In option (2) and (4), the mass ratios allow for equality, but only in (2) \(m_1/m_2 = 1/3\) when proportioned correctly with velocities shows equality.
05

Determine the correct scenario

Evaluate the given scenarios: Only in option (2) with the scenario described, do the masses with a mathematical simplification allow the result of particles having equal velocity in opposite directions, verifying the condition for elastic collision with equal mass-derived velocity.

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
In physics, the concept of momentum conservation is pivotal, especially in understanding elastic collisions. Momentum is the product of mass and velocity, described by the equation:
  • Before collision: total momentum = \( m_1 u_1 + m_2 u_2 \)
  • After collision: total momentum = \( m_1 v_1 + m_2 v_2 \)
For a collision to be classified as elastic, the total momentum before and after the event remains constant. This is known as the conservation of momentum. It allows us to predict the final velocities of the colliding objects.
In the given problem, since the second particle is initially at rest, we simplify the momentum conservation equation to: \[ m_1 u_1 = m_1 v_1 + m_2 v_2 \]. This equation tells us how the momentum is divided between the two particles after they collide. By understanding that momentum is shared between the masses dependant on their proportions, we can begin to predict outcomes like the direction and speed of the particles post-collision.
Kinetic Energy Conservation
Kinetic energy conservation is another defining feature of elastic collisions. Kinetic energy is represented by: \[ KE = \frac{1}{2}mv^2 \]. In elastic collisions, the total kinetic energy before and after the collision remains unchanged. This principle means that no energy is lost to sound, heat, or deformation.
For the exercise under consideration, we use the kinetic energy equation:
  • Before collision: total energy = \( \frac{1}{2} m_1 u_1^2 \)
  • After collision: total energy = \( \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \)
The expression \( \frac{1}{2} m_1 u_1^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \) ensures that the sum of the kinetic energies post-collision is the same as it was pre-collision.
This preserved energy principle is what allows us to make predictions about the velocities of the particles. Adjusting mass ratios impacts how easily energy can be "transferred" from one particle to the other, determining whether or not equal speeds ensue.
Mass Ratio in Collisions
Mass ratio plays a crucial role in determining the outcome of a collision between two particles. Specifically, it influences whether the velocities after the collision will be equal or not.
If the masses are equal, it often leads to equal and opposite velocities upon colliding; however, this isn’t always the case when mass ratios deviate. In the exercise provided:
  • For \( \frac{m_1}{m_2} = \frac{1}{2} \): The particles move apart with unequal speeds but opposite directions.
  • For \( \frac{m_1}{m_2} = \frac{1}{3} \): Under specific simplifications, they can achieve equal speed, validating one of the provided options.
  • For \( \frac{m_1}{m_2} = \frac{2}{1} \) and \( \frac{m_1}{m_2} = \frac{3}{1} \): Unequal masses result in different speeds post-collision.
Through these scenarios, mass ratio dynamics become apparent, impacting how complementary and separate energy and momentum are distributed. This insight aids in understanding how to engineer real-world applications, like car safety features, which depend on controlled impact responses.

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 steel ball of mass \(2 \mathrm{~m}\) suffers one-dimensional elastic collision with a row of three steel balls, each of mass \(m\). If mass \(2 m\) has collided with velocity \(v\) and the three balls numbered \(1,2,3\) were initially at rest, then after the collision (1) balls 1,2 and 3 would start moving to the right, each \(_{w_{2}}\) velocity \(v / 3\) (2) balls 2 and 3 would start moving to the right, each \(w_{10}\) velocity \(v / 2\) (3) balls 2 and 3 would start moving to the right, each \(w_{2}\) velocity \(v\) (4) ball 1 and ball of mass \(2 m\) would remain at rest

Column I contains physical quantity/process while column contains formula/principle. Match columns I and II such that the formula/principle is correct corresponding to the quantity in column I. $$ \begin{array}{|l|l|l|} \hline {\text { Column I }} & {\text { Column II }} \\ \hline \text { i. } \quad \text { Momentum } & \text { a. } \quad m\left(v_{2}-v_{1}\right) \\ \hline \text { ii. } \text { Impulse } & \text { b. } \begin{array}{l} \text { only momentum is } \\ \text { conserved } \end{array} \\ \hline \text { iii. Elastic collision } & \text { c. } \begin{array}{l} \text { momentum and kinefie } \\ \text { energy both are conserved } \end{array} \\ \hline \text { iv. Inelastic collision } & \text { d. } m v \\ \hline \end{array} $$

0\. A smooth sphere \(A\) of mass \(m\) collides elastically with an identical sphere \(B\) at rest. The velocity of \(A\) before collision is \(10 \mathrm{~m} / \mathrm{s}\) in a direction making \(60^{\circ}\) with the line of centres at the time of impact. (1) The sphere \(A\) comes to rest after collision. (2) The sphere \(B\) will move with a speed of \(10 \mathrm{~m} / \mathrm{s}\) after collision. (3) The directions of motion \(A\) and \(B\) after collision are at right angles. (4) The speed of \(B\) after collision is \(5 \mathrm{~m} / \mathrm{s}\).

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) \(5 \mathrm{~kg}\)

Two small spheres ' \(A\) 'and ' \(B^{\prime}\) mass \(1 \mathrm{~kg}\) and \(2 \mathrm{~kg}\) are moving with speeds \(21 \mathrm{~m} / \mathrm{s}\) and \(4 \mathrm{~m} / \mathrm{s}\) respectively in opposite direction collide head on. After collision \(A\) moves with speed \(1 \mathrm{~m} / \mathrm{s}\) in its initial direction. Which is/are correct? (1) Velocity of \(B\) after collision is \(6 \mathrm{~m} / \mathrm{s}\) opposite to its direction before collision (2) The coefficient of restitution \(e=0.2\) (3) Loss of kinetic energy due to collision is \(200 \mathrm{~J}\) (4) lmpulse of force between two balls is \(400 \mathrm{Ns}\)

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.