/*! 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 After a completely inelastic col... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

After a completely inelastic collision, two objects of the same mass and same initial speed move away together at half their initial speed. Find the angle between the initial velocities of the objects.

Short Answer

Expert verified
The angle between the initial velocities is 90 degrees.

Step by step solution

01

Understand the Collision Scenario

In a completely inelastic collision, two objects stick together after the collision. Here, both objects have the same mass, denoted as \( m \), and the same initial speed, denoted as \( v \). After the collision, they move together as one object with half their initial speed, \( \frac{v}{2} \).
02

Apply Conservation of Momentum

The conservation of momentum dictates that the total momentum before the collision equals the total momentum after the collision. Let the initial velocities make an angle \( \theta \) with each other. Thus, the momentum along each direction should be equal before and after collision, considering components due to the angle.
03

Set Up Equations for Momentum

The initial momentum of each object is \( mv \). Before collision, denote their momenta along a reference direction as \( mv \) and \( mv \). After the collision, the momentum of the combined mass \( 2m \) with velocity \( \frac{v}{2} \) is \( mv \) (along the resultant direction). We need to resolve the initial momenta into components and equate to the after-collision momentum.
04

Calculate Momentum Components

Resolve each initial momentum along horizontal and vertical components. For object 1: \( mv_x = mv \cos(\theta_1) \) and \( mv_y = mv \sin(\theta_1) \). Similarly for object 2, the components are \( mv_x = mv \cos(\theta_2) \) and \( mv_y = mv \sin(\theta_2) \). After collision: \( 2m \frac{v}{2} = mv \).
05

Find Angle \( \theta \) Using Momentum Components

Set up the equation \( 2m \cdot \frac{v}{2} = \sqrt{(mv \cos(\theta_1) + mv \cos(\theta_2))^2 + (mv \sin(\theta_1) + mv \sin(\theta_2))^2} \). This simplifies using angles, \( \theta_1 = \theta \) and \( \theta_2 = 0 \), or equivalently: \( \cos(\theta) + 1 = 1 \). This gives \( \cos(\theta) = 0 \).
06

Solve for \( \theta \)

Since \( \cos(\theta) = 0 \), \( \theta \) is \( 90^\circ \). The objects should be moving perpendicular to each other initially for their combined mass to have a speed \( \frac{v}{2} \) and satisfy conservation of momentum.

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 collisions, especially inelastic ones, the principle of conservation of momentum is crucial. Momentum is a property of a moving object and is calculated as the product of its mass and velocity, expressed as \( p = mv \). In an inelastic collision, the colliding objects stick together and move as one mass after the impact.
The total momentum before and after the collision must remain constant, meaning the initial momentum of each object combines to equal the final momentum of the unified mass. This principle allows us to predict the outcomes of collisions by understanding how momentum is transferred and conserved.
In our scenario, two objects with equal initial speeds collide and move together at half their original speed after collision. Applying momentum conservation helps us derive the necessary equations to solve for other unknowns in the problem, such as the angle of their initial velocities.
Angle Calculation in Collisions
When two objects collide, the direction in which they move after the collision is determined by their initial velocities. The angle between the initial velocities plays a critical role.
To find the angle, consider how the conservation of momentum applies across both horizontal and vertical components. Since both objects have the same speed initially and move together after the collision, analyzing their momentum directionally helps simplify the equation.
The angle \( \theta \) affects how much of each object's momentum contributes to the final shared movement. Through the use of trigonometric identities and resolving momentum into directional components, it was determined that the initial angle \( \theta \) between the two objects is \( 90^\circ \). This explains why their combined speed after collision is half of their initial speed.
Momentum Components
Breaking down momentum into horizontal and vertical components provides a clearer understanding of the collision dynamics. Before determining the post-collision movement, each object's momentum must be resolved into these components.
The initial momentum components are calculated using trigonometric functions:
  • For one object, the horizontal component is \( mv \cos(\theta_1) \) and vertical is \( mv \sin(\theta_1) \).
  • For the other object, these become \( mv \cos(\theta_2) \) and \( mv \sin(\theta_2) \).
After the collision, the combined mass moves in a new direction, with its momentum equal to the sum of the initial momentum components. Solving the resulting equation reveals how the initial angles impact the final trajectory. It also confirms that the angle between the objects was such that their momenta precisely counterbalanced each other out when combined, leading to their collective outcome.
Kinematic Equations
Kinematics, the study of motion, is useful for understanding how velocities and positions change over time during collisions. Although the problem focuses on momentum, kinematic equations offer additional insight into how these components interact.
The velocity of each object and their resulting velocity can be explored further using equations like \( v_f = v_i + at \) and \( s = ut + \frac{1}{2} at^2 \). These equations describe how objects accelerate, decelerate, or change direction during interactions.
In our exercise, the kinematics provide a backdrop for understanding how the momentary velocity halved post-collision and how the motion continued thereafter. They complement the momentum calculations, solidifying the understanding of various factors involved in inelastic collisions.

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

Two particles are launched from the origin of the coordinate system at time \(t=0 .\) Particle 1 of mass \(m_{1}=5.00 \mathrm{~g}\) is shot directly along the \(x\) axis on a frictionless floor, with constant speed \(10.0 \mathrm{~m} / \mathrm{s} .\) Particle 2 of mass \(m_{2}=3.00 \mathrm{~g}\) is shot with a velocity of magnitude \(20.0 \mathrm{~m} / \mathrm{s},\) at an upward angle such that it always stays directly above particle \(1 .\) (a) What is the maximum height \(H_{\max }\) reached by the com of the two-particle system? In unit- vector notation, what are the (b) velocity and (c) acceleration of the com when the com reaches \(H_{\max } ?\)

A \(2.00 \mathrm{~kg}\) particle has the \(x y\) coordinates \((-1.20 \mathrm{~m}, 0.500 \mathrm{~m})\) and a \(4.00 \mathrm{~kg}\) particle has the \(x y\) coordinates \((0.600 \mathrm{~m},-0.750 \mathrm{~m})\) Both lie on a horizontal plane. At what (a) \(x\) and (b) \(y\) coordinates must you place a \(3.00 \mathrm{~kg}\) particle such that the center of mass of the three-particle system has the coordinates \((-0.500 \mathrm{~m},-0.700 \mathrm{~m}) ?\)

A completely inclastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis. Just before the collision, one ball, of mass \(3.0 \mathrm{~kg},\) is moving upward at \(20 \mathrm{~m} / \mathrm{s}\) and the other ball, of mass \(2.0 \mathrm{~kg}\), is moving downward at \(12 \mathrm{~m} / \mathrm{s}\). How high do the combined two balls of putty risc above the collision point? (Ncglect air drag.)

Ricardo, of mass \(80 \mathrm{~kg}\), and Carmclita, 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 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 picr post, what is Carmelita's mass?

Block 1 of mass \(m_{1}\) slides along a frictionless floor and into a one- dimensional elastic collision with stationary block 2 of mass \(m_{2}=3 m_{1}\). Prior to the collision, the center of mass of the two-block system had a speed of \(3.00 \mathrm{~m} / \mathrm{s}\). Afterward, what are the speeds of (a) the center of mass and (b) block \(2 ?\)

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.