/*! 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 90 A hockey puck of mass \(4 m\) ha... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A hockey puck of mass \(4 m\) has been rigged to explode, as part of a practical joke. Initially the puck is at rest on a frictionless ice rink. Then it bursts into three pieces. One chunk, of mass \(m,\) slides across the ice at velocity \(v \hat{\mathbf{i}}\). Another chunk, of mass \(2 m\), slides across the ice at velocity \(2 v \hat{\mathbf{j}} .\) Determine the velocity of the third chunk.

Short Answer

Expert verified
The third chunk moves with velocity \(-v \hat{\mathbf{i}} - 4v \hat{\mathbf{j}}\).

Step by step solution

01

Set Up the Conservation of Momentum Equation

Initially, the puck is at rest, indicating that its total momentum is zero. After the explosion, momentum must still be conserved. Denote the velocity of the third chunk as \( \mathbf{v}_3 \). The total initial momentum is zero:\[ \mathbf{P}_{\text{initial}} = 0 \]The total final momentum involves the three chunks:\[ \mathbf{P}_{\text{final}} = m v \hat{\mathbf{i}} + 2m \cdot 2v \hat{\mathbf{j}} + (4m - m - 2m) \mathbf{v}_3 \]
02

Simplify the Mass of the Third Chunk

The mass of the third chunk is the total mass minus the sum of the masses of the other two chunks. Calculate this as:\[ 4m - m - 2m = m \]Thus, the third chunk also has a mass \(m\).
03

Write the Final Momentum of Each Component

Given the velocity vectors of the first two pieces, the total momentum vector in terms of \(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\) components can be expressed as:\[ m v \hat{\mathbf{i}} + 4m v \hat{\mathbf{j}} + m \mathbf{v}_3 \]
04

Solve for the Velocity of the Third Chunk

Set the zero initial momentum equal to the total final momentum and solve for \(\mathbf{v}_3\):\[ 0 = m v \hat{\mathbf{i}} + 4m v \hat{\mathbf{j}} + m \mathbf{v}_3 \]Isolate \( \mathbf{v}_3 \):\[ m \mathbf{v}_3 = -m v \hat{\mathbf{i}} - 4m v \hat{\mathbf{j}} \]Divide by \(m\) to solve for \( \mathbf{v}_3 \):\[ \mathbf{v}_3 = -v \hat{\mathbf{i}} - 4v \hat{\mathbf{j}} \]
05

Express the Velocity of the Third Chunk

Thus, the velocity of the third chunk is:\[ \mathbf{v}_3 = -v \hat{\mathbf{i}} - 4v \hat{\mathbf{j}} \]

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.

Hockey Puck Problem
In this problem, a hockey puck is subjected to a dramatic transformation due to an explosion that scatters it into three parts. Understanding this initial setup is crucial to solving the problem.
Initially, the puck is at rest on a frictionless surface. This means that the total momentum of the system before the explosion takes place is zero. Momentum, as a vector quantity, includes both magnitude and direction, which makes the concept of a frictionless surface an important one; it allows us to focus solely on the effects of the explosion.
  • The explosion divides the puck into three separate fragments.
  • Each fragment moves with a certain velocity, influenced by its mass and the force of the explosion.
  • Calculating the velocity of these fragments post-explosion requires applying the principle of the conservation of momentum.
In essence, this means that even though the pieces move in different directions with different velocities, their combined momentum will equal the original momentum of the puck, which was zero prior to the explosion.
Vector Components
Vector components are essential in this problem because they allow us to break down the velocities of the fragments into understandable and manageable parts.
The velocities given in the problem are vector quantities; they have both magnitude (speed) and direction. The direction in this scenario is expressed in terms of unit vector notation:
  • The first chunk moves along the x-axis (direction given by \(\hat{\mathbf{i}}\)),
  • while the second chunk moves along the y-axis (direction given by \(\hat{\mathbf{j}}\)).
Understanding vectors helps in setting up the equation for conservation of momentum. Each velocity vector can be split into its respective components, facilitating the solving process.
  • This breakdown leads to a system of equations that relates the components of momentum in different directions.
  • By balancing these equations, we find the velocity of the third fragment in terms of its vector components.
Overall, utilizing vector components simplifies complex multidirectional movement into a straightforward algebraic problem.
Explosion Mechanics
The explosion mechanics involved in this problem are a practical example of how force and mass interact in physical systems.
In the context of this problem, the explosion is what drives the subsequent movements of each fragment. Despite splitting into pieces, the system's total momentum remains constant, illustrating the key principle of the conservation of momentum.
  • Each fragment has its own mass and resulting velocity.
  • Conservation of momentum dictates that the entire system's momentum before and after the explosion is equal.
To solve for the unknowns, the explosion mechanics are translated into the language of physics calculations:
  • By assuming the initial total momentum of the puck was zero, balance the resulting momentums (in this case, all three).
  • This involves equating the sum of all resulting momentum vectors to the initial state.
Through these methods, we're able to determine the velocities of each resultant fragment, effectively converting the chaotic nature of an explosion into a quantifiable scenario governed by fundamental physics laws.

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

(III) \(\mathrm{A}\) neon atom \((m=20.0 \mathrm{u})\) makes a perfectly elastic collision with another atom at rest. After the impact, the neon atom travels away at a \(55.6^{\circ}\) angle from its original direction and the unknown atom travels away at a \(-50.0^{\circ}\) angle. What is the mass (in u) of the unknown atom? [Hint: You could use the law of sines.

A 4800 -kg open railroad car coasts along with a constant speed of 8.60 \(\mathrm{m} / \mathrm{s}\) on a level track. Snow begins to fall vertically and fills the car at a rate of 3.80 \(\mathrm{kg} / \mathrm{min}\) . Ignoring friction with the tracks, what is the speed of the car after 60.0 \(\mathrm{min}\) ? (See Section 2 of "Linear Momentum.")

(II) A neutron collides elastically with a helium nucleus (at rest initially) whose mass is four times that of the neutron. The helium nucleus is observed to move off at an angle \(\theta_{\mathrm{He}}^{\prime}=45^{\circ} .\) Determine the angle of the neutron, \(\theta_{\mathrm{n}}^{\prime},\) and the speeds of the two particles, \(v_{\mathrm{n}}^{\prime}\) and \(v_{\mathrm{He}}^{\prime},\) after the collision. The neutron's initial speed is \(6.2 \times 10^{5} \mathrm{~m} / \mathrm{s}\).

An extrasolar planet can be detected by observing the wobble it produces on the star around which it revolves. Suppose an extrasolar planet of mass \(m_{\mathrm{B}}\) revolves around its star of mass \(m_{\mathrm{A}}\) If no external force acts on this simple two-object system, then its \(\mathrm{CM}\) is stationary. Assume \(m_{\mathrm{A}}\) and \(m_{\mathrm{B}}\) are in circular orbits with radii \(r_{\mathrm{A}}\) and \(r_{\mathrm{B}}\) about the system's \(\mathrm{CM}\). (a) Show that \(r_{\mathrm{A}}=\frac{m_{\mathrm{B}}}{m_{\mathrm{A}}} r_{\mathrm{B}}\) (b) Now consider a Sun-like star and a single planet with the same characteristics as Jupiter. That is, \(m_{\mathrm{B}}=1.0 \times 10^{-3} \mathrm{~m}_{\mathrm{A}}\) and the planet has an orbital radius of \(8.0 \times 10^{11} \mathrm{~m} .\) Determine the radius \(r_{\mathrm{A}}\) of the star's orbit about the system's \(\mathrm{CM}\). (c) When viewed from Earth, the distant system appears to wobble over a distance of \(2 r_{\mathrm{A}}\). If astronomers are able to detect angular displacements \(\theta\) of about 1 milliarcsec \(\left(1 \operatorname{arcsec}=\frac{1}{3600}\right.\) of a degree), from what distance \(d\) (in light-years) can the star's wobble be detected \(\left(11 \mathrm{y}=9.46 \times 10^{15} \mathrm{~m}\right) ?(d)\) The star nearest to our Sun is about 4 ly away. Assuming stars are uniformly distributed throughout our region of the Milky Way Galaxy, about how many stars can this technique be applied to in the search for extrasolar planetary systems?

(II) The masses of the Earth and Moon are \(5.98 \times 10^{24} \mathrm{~kg}\) and \(7.35 \times 10^{22} \mathrm{~kg},\) respectively, and their centers are separated by \(3.84 \times 10^{8} \mathrm{~m} .\) (a) Where is the CM of this system located? (b) What can you say about the motion of the Earth-Moon system about the Sun, and of the Earth and Moon separately about the Sun?

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.