/*! 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 79 Game of Pool In a game of pool, ... [FREE SOLUTION] | 91Ó°ÊÓ

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Game of Pool In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball moves at \(3.50 \mathrm{~m} / \mathrm{s}\) along a line making an angle of \(22.0^{\circ}\) with its original direction of motion, and the second ball has a speed of \(2.00 \mathrm{~m} / \mathrm{s}\). Find (a) the angle between the direction of motion of the second ball and the original direction of motion of the cue ball and (b) the original speed of the cue ball. (c) Is kinetic energy (of the centers of mass, don't consider the rotation) conserved?

Short Answer

Expert verified
(a) \(\theta_2 \approx 40.9^{\circ}\) (b) \(v_{1i} \approx 4.77 \text{ m/s}\) (c) Kinetic energy is not conserved.

Step by step solution

01

- Set up the problem using conservation of momentum

Since the collision involves only the two balls, and assuming no external forces act on the system, we can use the conservation of momentum. Let's denote the original speed of the cue ball as \(v_{1i}\), the final speed of the cue ball as \(v_{1f} = 3.50 \, m/s\), and the final speed of the second ball as \(v_{2f} = 2.00 \, m/s\). Use the angles given: angle of cue ball \(\theta_1 = 22.0^{\circ}\). Let's denote the unknown angle as \(\theta_2\), which is the angle of the second ball's motion with respect to the original direction.
02

- Apply conservation of momentum in the x-direction

The x-direction (original direction of cue ball): \[ m \cdot v_{1i} = m \cdot v_{1f} \cos(\theta_1) + m \cdot v_{2f} \cos(\theta_2) \] Simplifying (mass \(m\) cancels out): \[ v_{1i} = v_{1f} \cos(22.0^{\circ}) + v_{2f} \cos(\theta_2) \] Plug in the known values: \[ v_{1i} = 3.50 \cos(22.0^{\circ}) + 2.00 \cos(\theta_2) \]
03

- Apply conservation of momentum in the y-direction

The y-direction (perpendicular to original direction of cue ball): \[ 0 = m \cdot v_{1f} \sin(\theta_1) - m \cdot v_{2f} \sin(\theta_2) \] Simplifying (mass \(m\) cancels out): \[ 0 = v_{1f} \sin(22.0^{\circ}) - v_{2f} \sin(\theta_2) \] Solving for \(\theta_2\): \[ v_{2f} \sin(\theta_2) = 3.50 \sin(22.0^{\circ}) \] \[ \sin(\theta_2) = \frac{3.50 \sin(22.0^{\circ})}{2.00} \] Calculate \(\sin(\theta_2)\) first: \[ \sin(22.0^{\circ}) \approx 0.3746 \] \[ \sin(\theta_2) = \frac{3.50 \times 0.3746}{2.00} \approx 0.6555 \] Therefore, \(\theta_2 = \sin^{-1}(0.6555) \approx 40.9^{\circ}\)
04

- Solve for the original speed of the cue ball

Use the x-direction momentum equation with \(\theta_2\) value: \[ v_{1i} = 3.50 \cos(22.0^{\circ}) + 2.00 \cos(40.9^{\circ}) \] Calculate each term: \[ 3.50 \cos(22.0^{\circ}) \approx 3.50 \times 0.9272 \approx 3.2452 \] \[ 2.00 \cos(40.9^{\circ}) \approx 2.00 \times 0.7599 \approx 1.5198 \] Add these values: \[ v_{1i} = 3.2452 + 1.5198 \approx 4.765 \text{ m/s} \] Therefore, the original speed of the cue ball is approximately \( 4.77 \text{ m/s} \).
05

- Check if kinetic energy is conserved

Calculate the initial and final kinetic energies to see if they match. Initial kinetic energy: \[ KE_{initial} = \frac{1}{2} m v_{1i}^2 \approx \frac{1}{2} m (4.77)^2 \approx 11.37 m \] Final kinetic energy: \[ KE_{final} = \frac{1}{2} m (3.50)^2 + \frac{1}{2} m (2.00)^2 \approx \frac{1}{2} m (12.25 + 4.00) \approx 8.12 m \] Since \( KE_{initial} eq KE_{final} \), kinetic energy is not conserved.

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Key Concepts

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

Conservation of Momentum
When studying collisions, especially in physics problems like a pool shot, the conservation of momentum is key. Momentum is a vector quantity, meaning it has both magnitude and direction.

In a collision where no external forces are acting, the total momentum before the collision equals the total momentum after the collision. Mathematically, this is represented as:
\[ m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = m_1 \cdot v_{1f} + m_2 \cdot v_{2f} \]
For our pool problem, the second ball is initially at rest, so its initial velocity (\( v_{2i} \)) is zero. This simplification allows us to focus on the momentum of the moving cue ball.

By breaking down the motion into x and y components, we can solve for unknown variables using:
\[ v_{1i} = v_{1f} \cos(\theta_1) + v_{2f} \cos(\theta_2) \]
and
\[ 0 = v_{1f} \sin(22.0^{\circ}) - v_{2f} \sin(\theta_2) \]
These equations help us find the angle of the second ball and the original speed of the cue ball.
Kinetic Energy
Kinetic energy deals with the energy an object possesses due to its motion. It is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
In our situation, it's crucial to determine if kinetic energy is conserved during the collision.

For the pool balls, we calculated:
- Initial kinetic energy of the cue ball: \[ KE_{initial} = \frac{1}{2} m (4.77)^2 \]
- Final kinetic energy of both balls after collision: \[ KE_{final} = \frac{1}{2} m (3.50)^2 + \frac{1}{2} m (2.00)^2 \]
Our results showed that \( KE_{initial} e KE_{final} \), meaning kinetic energy is not conserved. This indicates an inelastic collision, where some kinetic energy is converted into other forms like heat or sound during the impact.
Collision Analysis
Analyzing a collision requires understanding several principles. The type of collision (elastic or inelastic) helps us understand energy transformations.

In an elastic collision, both momentum and kinetic energy are conserved, but in an inelastic collision, only momentum is conserved.

For our pool exercise, we verified momentum conservation through vector decomposition into x and y components. We concluded that our collision was inelastic because the kinetic energy was not conserved.

This analysis provides a clear picture of what happens during the collision. It shows energy transformation and validates our calculations through the conservation laws. This step is fundamental to accurately solve physics problems involving collisions.
Vector Decomposition
Vectors have both magnitude and direction. In collision problems like our pool example, it's essential to decompose these vectors into their components.

For a vector making an angle \( \theta \) with the x-axis, its components are:
- x-component: \[ v_x = v \cos(\theta) \]
- y-component: \[ v_y = v \sin(\theta) \]
We used these principles to break down the velocities of the pool balls after the collision into x and y components.

By setting up equations for each component separately, we solved for unknowns like the original speed of the cue ball and the angle of the second ball’s trajectory. This method simplifies complex 2D collision problems into manageable one-dimensional equations.

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Most popular questions from this chapter

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