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Chapter 9: Problem 22

Gives an overhead view of the path taken by a \(0.165 \mathrm{~kg}\) cue ball as it bounces from a rail of a pool table. The ball's initial speed is \(2.00 \mathrm{~m} / \mathrm{s},\) and the angle \(\theta_{1}\) is \(30.0^{\circ}\). The bounce reverses the \(y\) component of the ball's velocity but docs not alter the \(x\) component. What are (a) angle \(\theta_{2}\) and (b) the change in the ball's linear momentum in unit-vector notation? (The fact that the ball rolls is irrelevant to the problem.)

Short Answer

Expert verified
a) \( \theta_2 = 30.0^\circ \); b) \( \Delta \mathbf{p} = 0 \mathbf{i} - 0.330 \mathbf{j} \text{ kg m/s} \).

Step by step solution

01

Determine Velocity Components

First, we determine the initial velocity components of the ball. The velocity components for an angle \( \theta_1 = 30.0^\circ \) with speed \( v = 2.00 \text{ m/s} \) are calculated as follows:\[v_{x} = v \cdot \cos(\theta_1) = 2.00 \cdot \cos(30.0^\circ) = 2.00 \cdot \frac{\sqrt{3}}{2} = 1.732 \text{ m/s}\]\[v_{y} = v \cdot \sin(\theta_1) = 2.00 \cdot \sin(30.0^\circ) = 2.00 \cdot 0.5 = 1.00 \text{ m/s}\]
02

Calculate Velocity Components Post-Bounce

After the bounce, the \( y \)-component of velocity is reversed but the \( x \)-component remains unchanged.\[v'_{x} = v_{x} = 1.732 \text{ m/s}\]\[v'_{y} = -v_{y} = -1.00 \text{ m/s}\]
03

Determine Angle \( \theta_2 \) After Bounce

The angle \( \theta_2 \) after the bounce is determined based on the new velocity components.\[\tan(\theta_2) = \frac{|v'_{y}|}{v'_{x}} = \frac{1.00}{1.732}\]\[\theta_2 = \tan^{-1}\left(\frac{1.00}{1.732}\right) \approx 30.0^\circ\]
04

Calculate Change in Linear Momentum

The change in linear momentum along the \( y \)-axis is given by:\[\Delta p_{y} = m(v'_{y} - v_{y}) = 0.165 \times (-1.00 - 1.00) = 0.165 \times (-2.00) = -0.330 \text{ kg m/s}\]There is no change in the \( x \)-axis momentum since the \( x \)-component of velocity remains the same:\[\Delta p_{x} = m(v'_{x} - v_{x}) = 0.165 \times (1.732 - 1.732) = 0\]Thus, the change in linear momentum in unit-vector notation is:\[\Delta \mathbf{p} = 0 \mathbf{i} - 0.330 \mathbf{j} \text{ kg m/s}\]
05

Compile Final Answers

a) The angle \( \theta_2 = 30.0^\circ \).b) The change in linear momentum is symbolically represented as \( \Delta \mathbf{p} = 0 \mathbf{i} - 0.330 \mathbf{j} \text{ kg m/s} \).

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Key Concepts

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

Linear Momentum
Linear momentum is a key principle in physics that describes the quantity of motion an object possesses. It's given by the product of an object's mass and its velocity. Linear momentum is a vector quantity, meaning it has both magnitude and direction.

This concept is vital in understanding collisions or interactions, such as the pool ball bouncing off a rail. Here, the process alters only the direction of the momentum along one axis, specifically the y-axis, while the x-axis remains unaffected.

When calculating the change in linear momentum, you should focus on the components of velocity that have altered due to an external force or interaction. In this exercise,
  • The mass of the cue ball is multiplied by the change in its y-component of velocity, since the x-component stays constant.
  • The final change in momentum can be expressed using unit vectors to denote its direction clearly.
Understanding these interactions through changes in linear momentum allows physicists to predict the outcomes of forces or interactions such as bounces or collisions.
Angle of Reflection
The angle of reflection is crucial when studying the bouncing of objects, as it relates to how these objects change direction after hitting a surface.

In physics, the angle of reflection is often equal to the angle of incidence, assuming no energy loss and ideal conditions. For the cue ball:
  • The angle at which it hits the pool table rail helps determine its path after bouncing.
  • Upon bouncing, although the direction of flight may change, some components, like the x-component, can remain consistent if the bounce is symmetrical.
When solving such problems, calculating these angles involves examining the velocity components before and after the impact. This exercise showed the angle of reflection being the same as the initial angle of incidence due to the symmetry of the problem, reflected in how the bounce solely reversed the y-component of velocity.

Understanding how angles of incidence and reflection work in practical scenarios enriches problem-solving skills in physics.
Velocity Components
Breaking down the velocity into components can simplify complex motion into more manageable parts, often along the x and y axes. This technique uses trigonometry to resolve a velocity vector into orthogonal components.

For example, in this pool table problem:
  • We determine the initial x and y components based on the ball's initial speed and angle.
  • The x-component, derived from the cosine of the angle, represents the horizontal motion, while the y-component, from the sine function, describes the vertical motion.
Post-bounce, only the y-component changes, specifically its direction, by becoming negative. The x-component remains unaltered, meaning the horizontal speed of the ball does not change due to the bounce.

This separation into components streamlines the calculation of phenomena like changes in momentum and the determination of angles post-bounce. By understanding velocity components, you gain an essential tool in physics for analyzing and predicting movement in a range of practical scenarios.

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Most popular questions from this chapter

A \(1400 \mathrm{~kg}\) car moving at \(5.3 \mathrm{~m} / \mathrm{s}\) is initially traveling north along the positive direction of a \(y\) axis. After completing a \(90^{\circ}\) right-hand turn in \(4.6 \mathrm{~s}\), the inattentive operator drives into a tree, which stops the car in \(350 \mathrm{~ms}\). In unit-vector notation, what is the impulse on the car (a) due to the turn and (b) due to the collision? What is the magnitude of the average force that acts on the car (c) during the turn and (d) during the collision? (e) What is the direction of the average force during the turn?

A \(5.0 \mathrm{~kg}\) block with a speed of \(3.0 \mathrm{~m} / \mathrm{s}\) collides with a \(10 \mathrm{~kg}\) block that has a speed of \(2.0 \mathrm{~m} / \mathrm{s}\) in the same direction. After the collision, the \(10 \mathrm{~kg}\) block travels in the original direction with a speed of \(2.5 \mathrm{~m} / \mathrm{s}\). (a) What is the velocity of the \(5.0 \mathrm{~kg}\) block immediately after the collision? (b) By how much does the total kinetic energy of the system of two blocks change because of the collision? (c) Suppose, instead, that the \(10 \mathrm{~kg}\) block ends up with a speed of \(4.0 \mathrm{~m} / \mathrm{s}\). What then is the change in the total kinetic energy? (d) Account for the result you obtained in (c).

A ball having a mass of \(150 \mathrm{~g}\) strikes a wall with a speed of \(5.2 \mathrm{~m} / \mathrm{s}\) and rebounds with only \(50 \%\) of its initial kinetic energy. (a) What is the speed of the ball immediately after rebounding? (b) What is the magnitude of the impulse on the wall from the ball? (c) If the ball is in contact with the wall for \(7.6 \mathrm{~ms}\), what is the magnitude of the average force on the ball from the wall during this time interval?

An electron undergoes a one-dimensional elastic collision with an initially stationary hydrogen atom. What percentage of the electron's initial kinetic energy is transferred to kinetic energy of the hydrogen atom? (The mass of the hydrogen atom is 1840 times the mass of the electron.)

two long barges are moving in the \(\operatorname{sam}\) dircetion in still water, one with a speed of \(10 \mathrm{~km} / \mathrm{h}\) and the othe with a speed of \(20 \mathrm{~km} / \mathrm{h}\). While they are passing each other, coal is shoveled from the slower to the faster one at a rate of \(1000 \mathrm{~kg} / \mathrm{min}\) How much additional force must be provided by the driving cngines of (a) the faster barge and (b) the slower barge if neithe is to change speed? Assume that the shoveling is always perfectly sideways and that the frictional forces between the barges and the water do not depend on the mass of the barges.
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