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Chapter 11: Problem 80

A \(2.50 \mathrm{~kg}\) particle that is moving horizontally over a floor with velocity \((-3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) undergoes a completely inelastic collision with a \(4.00 \mathrm{~kg}\) particle that is moving horizontally over the floor with velocity \((4.50 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}\). The collision occurs at \(x y\) coordinates \((-0.500 \mathrm{~m},-0.100 \mathrm{~m})\). After the collision and in unit- vector notation, what is the angular momentum of the stuck-together particles with respect to the origin?

Short Answer

Expert verified
\(\vec{L}_{\text{total}} = (0, 0, -1.950)\, \mathrm{kg}\, \mathrm{m}^2/\mathrm{s}\).

Step by step solution

01

Understanding Angular Momentum

Angular momentum \( \vec{L} \) of a particle moving with linear momentum \( \vec{p} \) at a position \( \vec{r} \) is given by the cross product \( \vec{L} = \vec{r} \times \vec{p} \). The total angular momentum of a system of particles is the sum of their individual angular momenta.
02

Calculate Linear Momentum Before Collision

Initially, the linear momentum \( \vec{p}_1 \) of the 2.50 kg particle is:\[ \vec{p}_1 = m_1 \vec{v}_1 = 2.50 \, \mathrm{kg} \times (-3.00 \, \mathrm{m/s}) \hat{\mathrm{j}} = (-7.50 \, \mathrm{kg} \, \mathrm{m/s}) \hat{\mathrm{j}} \]And the linear momentum \( \vec{p}_2 \) of the 4.00 kg particle is:\[ \vec{p}_2 = m_2 \vec{v}_2 = 4.00 \, \mathrm{kg} \times (4.50 \, \mathrm{m/s}) \hat{\mathrm{i}} = (18.00 \, \mathrm{kg} \, \mathrm{m/s}) \hat{\mathrm{i}} \].
03

Finding the Position Vectors

The position vector of the collision point is \( \vec{r} = (-0.500 \, \mathrm{m}) \hat{\mathrm{i}} + (-0.100 \, \mathrm{m}) \hat{\mathrm{j}} \). This vector is used to calculate the cross product with the linear momentum.
04

Compute Angular Momentum of Each Particle

For the 2.50 kg particle:\[ \vec{L}_1 = \vec{r} \times \vec{p}_1 = \begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ -0.500 & -0.100 & 0 \ 0 & -7.50 & 0 \end{vmatrix} = (0, 0, -3.750) \, \mathrm{kg} \, \mathrm{m}^2/\mathrm{s} \]For the 4.00 kg particle:\[ \vec{L}_2 = \vec{r} \times \vec{p}_2 = \begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ -0.500 & -0.100 & 0 \ 18.00 & 0 & 0 \end{vmatrix} = (0, 0, 1.800) \, \mathrm{kg} \, \mathrm{m}^2/\mathrm{s} \].
05

Adding Angular Momenta for Total Angular Momentum

The total angular momentum after the collision with respect to the origin is the vector sum of \( \vec{L}_1 \) and \( \vec{L}_2 \):\[ \vec{L}_{\text{total}} = (0, 0, -3.750) + (0, 0, 1.800) = (0, 0, -1.950) \, \mathrm{kg} \, \mathrm{m}^2/\mathrm{s} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inelastic Collision
An inelastic collision occurs when the colliding objects stick together after impact. This type of collision is characterized by the fact that kinetic energy is not conserved, although momentum is. Instead of rebounding off each other, the two objects move together as a single unit post-collision. This is crucial when determining subsequent motion and properties, such as angular momentum, as visualized in our current problem.

Key aspects to note about inelastic collisions include:
  • Momentum conservation: The total linear momentum before collision is equal to the total linear momentum after collision.
  • Loss of kinetic energy: Some kinetic energy is converted into other forms of energy, like heat or deformation.
  • Unified motion: Post-collision, the bodies move with a common velocity.
Understanding inelastic collisions helps in computing the motion and properties after impact, which are imperative in evaluating angular momentum sophisticatedly.
Cross Product
The cross product, also known as the vector product, is a mathematical operation used to determine a vector that is perpendicular to two given vectors. It is denoted by the symbol '脳'. In the context of our problem, the cross product is used to calculate the angular momentum of each particle due to their linear momentum and position vector.

The essential points to remember about the cross product include:
  • It results in a vector that is orthogonal to the original vectors involved.
  • It is crucial in determining quantities like angular momentum where directionality and rotational effects matter.
  • The magnitude of the cross product is calculated as \( |\vec{A} \times \vec{B}| = |\vec{A}||\vec{B}| \sin(\theta) \), where \( \theta\) is the angle between the vectors.
In physical scenarios like our exercise, the cross product of position vector and linear momentum fundamentally represents the angular momentum's tendency to cause rotational motion.
Linear Momentum
Linear momentum is a foundational concept in physics representing the quantity of motion an object possesses. It is a vector, having both magnitude and direction, defined mathematically as the product of an object's mass and velocity. In our problem, each particle possesses linear momentum, calculated before their collision.

Important features of linear momentum include:
  • Conservation in closed systems: The total linear momentum remains constant if no external forces act upon it.
  • It dictates the motion during and after collisions.
  • Expressed as \( \vec{p} = m \vec{v} \), where \( m\) is mass and \( \vec{v}\) is velocity.
In collisions, whether elastic or inelastic, examining linear momentum is critical in analyzing and predicting the movement of objects post-interaction.
Position Vector
A position vector defines the location of a point in space relative to a reference origin. In this context, the position vector of the collision point gives us the coordinates from which we measure other properties, like angular momentum.

Some key points regarding position vectors are:
  • It specifies the exact point in space via coordinates along the axes, typically expressed as \( \vec{r} = x \hat{\mathrm{i}} + y \hat{\mathrm{j}} + z \hat{\mathrm{k}} \).
  • In physics, knowing the position vector is essential for computing other vector quantities such as angular momentum.
  • In our problem, it is crucial for forming the cross product with linear momentum to find the rotational effects post-collision.
The position vector allows for the spatial quantification needed to analyze circumstances thoroughly, yielding crucial insights into motion dynamics.

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Most popular questions from this chapter

A small solid sphere with radius \(0.25 \mathrm{~cm}\) and mass \(0.56 \mathrm{~g}\) rolls without slipping on the inside of a large fixed hemisphere with radius \(15 \mathrm{~cm}\) and a vertical axis of symmetry. The sphere starts at the top from rest. (a) What is its kinetic energy at the bottom? (b) What fraction of its kinetic energy at the bottom is associated with rotation about an axis through its com? (c) What is the magnitude of the normal force on the hemisphere from the sphere when the sphere reaches the bottom?

A ball of mass \(M\) and radius \(R\) rolls smoothly from rest down a ramp and onto a circular loop of radius \(0.48 \mathrm{~m}\). The initial height of the ball is \(h=0.36 \mathrm{~m}\). At the loop bottom, the magnitude of the nor-mal force on the ball is \(2.00 \mathrm{Mg}\). The ball consists of an outer spherical shell (of a certain uniform density) that is glued to a central sphere (of a different uniform density). The rotational inertia of the ball can be expressed in the general form \(I=\beta M R^{2}\), but \(\beta\) is not \(0.4\) as it is for a ball of uniform density. Determine \(\beta\).

A top spins at \(30 \mathrm{rev} / \mathrm{s}\) about an axis that makes an angle of \(30^{\circ}\) with the vertical. The mass of the top is \(0.50 \mathrm{~kg}\), its rotational inertia about its central axis is \(5.0 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^{2}\), and its center of mass is \(4.0 \mathrm{~cm}\) from the pivot point. If the spin is clockwise from an overhead view, what are the (a) precession rate and (b) direction of the precession as viewed from overhead?

Is an overhead view of a thin uniform rod of length \(0.800 \mathrm{~m}\) and mass \(M\) rotating horizontally at angular speed \(20.0 \mathrm{rad} / \mathrm{s}\) about an axis through its center. A particle of mass \(M / 3.00\) initially attached to one end is ejected from the rod and travels along a path that is perpendicular to the rod at the instant of ejection. If the particle's speed \(v_{p}\) is \(6.00 \mathrm{~m} / \mathrm{s}\) greater than the speed of the rod end just after ejection, what is the value of \(v_{p}\) ?

An automobile traveling at \(80.0 \mathrm{~km} / \mathrm{h}\) has tires of \(75.0 \mathrm{~cm}\) diameter. (a) What is the angular speed of the tires about their axles? (b) If the car is brought to a stop uniformly in \(30.0\) complete turns of the tires (without skidding), what is the magnitude of the angular acceleration of the wheels? (c) How far does the car move during the braking?
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