/*! 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 24 A billiard ball of mass \(m=0.25... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A billiard ball of mass \(m=0.250 \mathrm{~kg}\) hits the cushion of a billiard table at an angle of \(\theta_{1}=60.0^{\circ}\) at a speed of \(v_{1}=27.0 \mathrm{~m} / \mathrm{s}\) It bounces off at an angle of \(\theta_{2}=71.0^{\circ}\) and a speed of \(v_{2}=10.0 \mathrm{~m} / \mathrm{s}\). a) What is the magnitude of the change in momentum of the billiard ball? b) In which direction does the change of momentum vector point?

Short Answer

Expert verified
a) The magnitude of the change in momentum of the billiard ball is 4.315 kg.m/s. b) The direction of the change in momentum vector points at an angle of 54.1 degrees from the negative x-axis.

Step by step solution

01

Calculate the initial momentum components

Using the given information about the incoming angle and the speed of the billiard ball, we will find the components of the initial momentum (\(p_{1x}\) and \(p_{1y}\)) by using these equations: $$p_{1x} = m*v_{1}*cos(\theta_1)$$ $$p_{1y} = m*v_{1}*sin(\theta_1)$$ Plugging in the given values, $p_{1x} = 0.250 * 27.0 * cos(60^{\circ}) \\ = 0.250 * 27.0 * 0.5 \\ = 3.375 kg.m/s$ $p_{1y} = 0.250 * 27.0 * sin(60^{\circ}) \\ = 0.250 * 27.0 * \frac{\sqrt{3}}{2} \\ = 5.845 kg.m/s$
02

Calculate the final momentum components

Using the given information about the outgoing angle and speed of the billiard ball, we will find the components of the final momentum (\(p_{2x}\) and \(p_{2y}\)) by using these equations: $$p_{2x} = m*v_{2}*cos(\theta_2)$$ $$p_{2y} = m*v_{2}*sin(\theta_2)$$ Plugging in the given values, $p_{2x} = 0.250 * 10.0 * cos(71^{\circ}) \\ = 0.250 * 10.0 * 0.335 \\ = 0.837 kg.m/s$ $p_{2y} = 0.250 * 10.0 * sin(71^{\circ}) \\ = 0.250 * 10.0 * 0.942 \\ = 2.355 kg.m/s$
03

Calculate the change in momentum components

To find the change in momentum components, we will subtract the initial components from the final components: $$\Delta p_x = p_{2x} - p_{1x}$$ $$\Delta p_y = p_{2y} - p_{1y}$$ $\Delta p_x = 0.837 - 3.375 \\ = -2.538 kg.m/s$ $\Delta p_y = 2.355 - 5.845 \\ = -3.490 kg.m/s$
04

Calculate the magnitude of the change in momentum

We will use the Pythagorean theorem to find the magnitude of the change in momentum: $$|\Delta p| = \sqrt{(\Delta p_x)^2 + (\Delta p_y)^2}$$ $|\Delta p| = \sqrt{(-2.538)^2 + (-3.490)^2} \\ = \sqrt{6.433 + 12.180} \\ = \sqrt{18.613} \\ = 4.315 kg.m/s$ So, the magnitude of the change in momentum is 4.315 kg.m/s.
05

Calculate the direction of the change in momentum

We will find the angle of the change in momentum vector (\(\alpha\)) by using the arctangent function: $$\alpha = arctan\left(\frac{\Delta p_y}{\Delta p_x}\right)$$ $\alpha = arctan\left(\frac{-3.490}{-2.538}\right) \\ = arctan(1.375) \\ = 54.1^{\circ}$ So, the change in momentum vector points in a direction of \(54.1^{\circ}\) from the negative x-axis. To summarize: a) The magnitude of the change in momentum of the billiard ball is 4.315 kg.m/s. b) The change of momentum vector points in a direction of \(54.1^{\circ}\) from the negative x-axis.

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.

Newton's Laws of Motion
Newton's Laws of Motion are fundamental principles that describe how objects move. They are core to understanding momentum and collisions, like in our billiard ball problem. There are three laws:
- **First Law (Law of Inertia):** An object will remain at rest or move at a constant velocity unless acted upon by an external force.
- **Second Law:** It states that the force acting on an object is equal to the mass of that object times its acceleration (\( F = ma \) ). This law relates to the momentum concept because momentum (\( p \) ) is the product of mass and velocity, \( p = mv \).
- **Third Law:** For every action, there is an equal and opposite reaction. This is crucial in understanding how forces interact, such as when a billiard ball hits a cushion, exerting equal and opposite forces.
In the context of this exercise, when the billiard ball bounces off the cushion, its change in velocity reflects the application of Newton's laws, resulting in a change in momentum. This is particularly explained by the second law, where the change in velocity leads to a change in momentum due to external forces acting on the ball.
Pythagorean Theorem
The Pythagorean Theorem is a mathematical sacred tool used to find the lengths of sides in a right triangle. It fundamentally states that in a right triangle, the square of the length of the hypotenuse (\( c \) ) is equal to the sum of the squares of the other two sides (\( a \) and \( b \) ):\[c^2 = a^2 + b^2\]
In our problem, this theorem helps us calculate the magnitude of the momentum change. After finding the individual momentum components (\( \Delta p_x \) and \( \Delta p_y \) ), we apply the Pythagorean theorem to get the resultant change in momentum vector:\[|\Delta p| = \sqrt{(\Delta p_x)^2 + (\Delta p_y)^2}\]
This formula tells us the magnitude of the overall change in momentum, compactly solving for \( |\Delta p| \). By doing so, it allows us to measure how the ball's motion is altered after bouncing off the table cushion.
Vector Components
Breaking forces, velocities, or momentum into their vector components allows us to handle vector quantities as they interact in multiple dimensions. Any vector component describes a vector in terms of horizontal (x-axis) and vertical (y-axis) parts.
Vectors are essential in accurately modeling the motion in physics problems. For instance, the momentum of a billiard ball moving in a plane is not a straightforward value; it involves distinct movements along different axes.
In the example, the initial and final momenta are each broken down into x and y components:
  • Initial x-component: \( p_{1x} = m * v_1 * \cos(\theta_1) \)
  • Initial y-component: \( p_{1y} = m * v_1 * \sin(\theta_1) \)
  • Final x-component: \( p_{2x} = m * v_2 * \cos(\theta_2) \)
  • Final y-component: \( p_{2y} = m * v_2 * \sin(\theta_2) \)
This breakdown is vital because it allows us to calculate the changes in these individual components, which leads to understanding the overall vector change and magnitude through further calculations and applications like the Pythagorean Theorem.
Trigonometry in Physics
Trigonometry helps us use angles in physics to solve problems where components act in different directions. Through sine, cosine, and tangent functions, we can separate a vector like velocity or force into its horizontal and vertical parts.
In our billiard ball problem, trigonometry helps determine how much of the ball's velocity or momentum acts in the x and y directions. The angle \( \theta_1 \) is utilized to calculate initial component vectors:
  • \( p_{1x} = m*v_{1}*\cos(\theta_1) \)
  • \( p_{1y} = m*v_{1}*\sin(\theta_1) \)
The outgoing angle \( \theta_2 \) gives us the final vector components.
Further, trigonometry is used to determine the direction of the vector change, understood through the inverse tangent (arctan) function:\[\alpha = \arctan\left(\frac{\Delta p_y}{\Delta p_x}\right)\]The approach enables us to handle complex directional changes in physical systems, making sure we account for all forces and movements accurately.

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

To determine the muzzle velocity of a bullet fired from a rifle, you shoot the \(2.00-\mathrm{g}\) bullet into a \(2.00-\mathrm{kg}\) wooden block. The block is suspended by wires from the ceiling and is initially at rest. After the bullet is embedded in the block, the block swings up to a maximum height of \(0.500 \mathrm{~cm}\) above its initial position. What is the velocity of the bullet on leaving the gun's barrel?

A bungee jumper is concerned that his elastic cord might break if it is overstretched and is considering replacing the cord with a high-tensile- strength steel cable. Is this a good idea?

When hit in the face, a boxer will "ride the punch"; that is, if he anticipates the punch, he will allow his neck muscles to go slack. His head then moves back easily from the blow. From a momentum-impulse standpoint, explain why this is much better than stiffening his neck muscles and bracing himself against the punch.

A fireworks projectile is launched upward at an angle above a large flat plane. When the projectile reaches the top of its flight, at a height of \(h\) above a point that is a horizontal distance \(D\) from where it was launched, the projectile explodes into two equal pieces. One piece reverses its velocity and travels directly back to the launch point. How far from the launch point does the other piece land? a) \(D\) b) \(2 D\) c) \(3 D\) d) \(4 D\)

Two Sumo wrestlers are involved in an inelastic collision. The first wrestler, Hakurazan, has a mass of \(135 \mathrm{~kg}\) and moves forward along the positive \(x\) -direction at a speed of \(3.5 \mathrm{~m} / \mathrm{s}\). The second wrestler, Toyohibiki, has a mass of \(173 \mathrm{~kg}\) and moves straight toward Hakurazan at a speed of \(3.0 \mathrm{~m} / \mathrm{s} .\) Immediately after the collision, Hakurazan is deflected to his right by \(35^{\circ}\) (see the figure). In the collision, \(10 \%\) of the wrestlers' initial total kinetic energy is lost. What is the angle at which Toyohibiki is moving immediately after the collision?

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.