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Chapter 7: Problem 61

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.

Short Answer

Expert verified
(a) The angle is approximately \(40.58^\text{o}\).(b) The original speed of the cue ball is \(4.76 \, \text{m/s}\).

Step by step solution

01

Set Up the Momentum Equations

Since the collision is between two balls of the same mass, use the conservation of momentum in the x and y-directions. Let the original speed of the cue ball be denoted as \(v_{i}\), and its initial momentum will be \(mv_{i}\). After the collision, the x and y components of momentum for both balls need to be considered.
02

Write the Momentum Equations in the x-Direction

In the x-direction: \[ mv_i = m(3.50 \, \text{m/s} \, \text{cos} \, 22.0^\text{o}) + m(2.00 \, \text{m/s} \, \text{cos} \, \theta) \]Simplify to: \[ v_i = 3.50 \, \text{m/s} \, \text{cos} \, 22.0^\text{o} + 2.00 \, \text{m/s} \, \text{cos} \, \theta \]
03

Write the Momentum Equations in the y-Direction

In the y-direction: \[ 0 = m(3.50 \, \text{m/s} \, \text{sin} \, 22.0^\text{o}) - m(2.00 \, \text{m/s} \, \text{sin} \, \theta) \]Simplify to:\[ 3.50 \, \text{m/s} \, \text{sin} \, 22.0^\text{o} = 2.00 \, \text{m/s} \, \text{sin} \, \theta \]
04

Solve for the Angle \(\theta\)

Isolate \(\theta\) in the y-direction equation: \[ \text{sin} \, \theta = \frac{3.50 \, \text{m/s} \, \text{sin} \, 22.0^\text{o}}{2.00 \, \text{m/s}} \]Calculate: \[ \text{sin} \, \theta = 0.649 \to \theta = \text{arcsin}(0.649) \approx 40.58^\text{o} \]
05

Substitute \(\theta\) into the x-Direction Equation

Use \(\theta = 40.58^\text{o}\) in the x-direction equation: \[ v_i = 3.50 \, \text{m/s} \, \text{cos} \, 22.0^\text{o} + 2.00 \, \text{m/s} \, \text{cos} \, 40.58^\text{o} \]Calculate: \[ v_i = 3.50 \, \text{m/s} \, (0.927) + 2.00 \, \text{m/s} \, (0.758) \approx 3.24 \, \text{m/s} + 1.52 \, \text{m/s} = 4.76 \, \text{m/s} \]

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Key Concepts

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

elastic collision
In physics, an elastic collision is one in which both momentum and kinetic energy are conserved. This means that the total kinetic energy of the system before the collision is equal to the total kinetic energy after the collision. For example, imagine playing pool, where the cue ball strikes another ball of the same mass. If this is an elastic collision, the energy and momentum are preserved.

During this type of collision, two main principles are adhered to:
  • Conservation of momentum
  • Conservation of kinetic energy
Understanding these principles helps explain how and why the balls move the way they do after colliding.

Elastic collisions are common in atomic and subatomic particles interactions. They exhibit a perfect exchange of energy and momentum, like the stationary ball gaining speed while the cue ball slows down.
momentum in x-direction
Momentum in the x-direction refers to the horizontal components of the momentum. After the pool balls collide, their movements split into x and y components.

The initial momentum of the cue ball is given by: \ p_{initial} = m \, \ v_{i} . In the x-direction, conservation of momentum is applied, resulting in: \ m \, v_i = m \, (3.50 \, \text{m/s} \, \text{cos} \, 22.0^\text{o}) + m \, (2.00 \, \text{m/s} \, \text{cos} \, \theta). This equation can be simplified to:

\[ v_i = 3.50 \, \text{m/s} \, \text{cos} \, 22.0^\text{o} + 2.00 \, \text{m/s} \, \text{cos} \, \theta \]

By using trigonometry, you can substitute the values to keep momentum conserved in the x-direction, enabling you to find the original speed of the cue ball and the angle \ \theta .
momentum in y-direction
Momentum in the y-direction refers to the vertical components of the momentum after the collision. Like the x-direction, momentum must also be conserved in the y-direction.

Initially, the cue ball only moves horizontally. Hence, the total initial momentum in the y-direction is zero. After the collision, the y-components are given by:

0 = m \ (3.50 \, \text{m/s} \, \text{sin} \, 22.0^\text{o}) - m \ (2.00 \, \text{m/s} \, \text{sin} \, \theta)

This leads to: \[ 3.50 \, \text{m/s} \, \text{sin} \, 22.0^\text{o} = 2.00 \, \text{m/s} \, \text{sin} \, \theta \]

With this equation, you can find the angle \ \theta by isolating it and solving: \ \text{sin} \ \theta = \frac{3.50 \, \text{m/s} \, \text{sin} \, 22.0^\text{o}}{2.00 \, \text{m/s}} . The calculated angle helps confirm that the momentum in both directions has been correctly conserved.

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Most popular questions from this chapter

In Edmund Rostand's famous play, Cyrano de Bergerac, Cyrano, in an attempt to distract a suitor from visiting Roxanne, claims to have descended to Earth from the Moon and proclaims to have invented six novel and fantastical methods for traveling to the Moon. One is as follows. Sitting on an iron platform- thence To throw a magnet in the air. This is A method well conceived - the magnet flown, Infallibly the iron will pursue: Then quick! relaunch your magnet, and you thus Can mount and mount unmeasured distances! \(^{*}\) In an old cartoon, there is another version of this method. A character in the old West is on a hand-pumped, two-person rail car. After getting tired of pumping the handle up and down to make the car move along the rails, he takes out a magnet, hangs it from a fishing pole, and holds it in front of the cart. The magnet pulls the cart toward it, which pushes the magnet forward, and so on, so the cart moves forward continually. What do you think of these methods? Can some version of them work? Discuss in terms of the physics you have learned.

A projectile body of mass \(m_{A}\) and initial \(x\) component velocity \(v_{A x}\left(t_{1}\right)=10.0 \mathrm{~m} / \mathrm{s}\) collides with an initially stationary target body of mass \(m_{B}=2.00 m_{A}\) in a one-dimensional collision. What is the velocity of \(m_{B}\) following the collision if the two masses stick together?

A \(6100 \mathrm{~kg}\) rocket is set for vertical firing from the ground. If the exhaust speed is \(1200 \mathrm{~m} / \mathrm{s}\), how much gas must be ejected each second if the thrust (a) is to equal the magnitude of the gravitational force on the rocket and (b) is to give the rocket an initial upward acceleration of \(21 \mathrm{~m} / \mathrm{s}^{2}\) ?

A \(4.0 \mathrm{~kg}\) mess kit sliding on a frictionless surface explodes into two \(2.0 \mathrm{~kg}\) parts, one moving at \(3.0 \mathrm{~m} / \mathrm{s}\), due north, and the other at \(5.0 \mathrm{~m} / \mathrm{s}, 30^{\circ}\) north of east. What is the original speed of the mess kit?

A \(0.70 \mathrm{~kg}\) ball is moving horizontally with a speed of \(5.0 \mathrm{~m} / \mathrm{s}\) when it strikes a vertical wall. The ball rebounds with a speed of \(2.0 \mathrm{~m} / \mathrm{s}\). What is the magnitude of the change in translational momentum of the ball?
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