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Chapter 7: Problem 6

A \(0.30 \mathrm{~kg}\) softball has a velocity of \(15 \mathrm{~m} / \mathrm{s}\) at an angle of \(35^{\circ}\) below the horizontal just before making contact with the bat. What is the magnitude of the change in momentum of the ball while it is in contact with the bat if the ball leaves the bat with a velocity of (a) \(20 \mathrm{~m} / \mathrm{s}\), vertically downward and (b) \(20 \mathrm{~m} / \mathrm{s}\), horizontally away from the batter and back toward the pitcher?

Short Answer

Expert verified
The change in momentum (a) is 9.34 kg鈰卪/s and (b) is 3.47 kg鈰卪/s.

Step by step solution

01

Determine initial momentum components

Use the given initial velocity and angle to find the horizontal and vertical components of the initial momentum. The initial velocity \[ v_i = 15 \text{ m/s} \text{ and } \theta = 35^{\text{掳}}\text{ below the horizontal} \]\[ m = 0.30 \text{ kg} \]Horizontal component of initial velocity: \[ v_{ix} = v_i \times \text{cos}(\theta) = 15 \times \text{cos}(35^{\text{掳}}) \text{ m/s} \]Vertical component of initial velocity: \[ v_{iy} = v_i \times \text{sin}(\theta) = 15 \times \text{sin}(35^{\text{掳}}) \text{ m/s} \]Compute: \[ v_{ix} \text{ and } v_{iy} \]\[ p_{ix} = m \times v_{ix} \]\[ p_{iy} = m \times v_{iy} \]
02

Calculate initial momentum values

First compute the trigonometric values:\[ \text{cos}(35^{\text{掳}}) \text{ is approximately } 0.819 \]\[ \text{sin}(35^{\text{掳}}) \text{ is approximately } 0.574 \]Using these values:\[ v_{ix} = 15 \times 0.819 = 12.285 \text{ m/s} \]\[ v_{iy} = 15 \times 0.574 = 8.61 \text{ m/s} \]Therefore, initial momentum components are:\[ p_{ix} = 0.30 \text{ kg} \times 12.285 \text{ m/s} = 3.6855 \text{ kg鈰卪/s} \]\[ p_{iy} = 0.30 \text{ kg} \times 8.61 \text{ m/s} = 2.583 \text{ kg鈰卪/s} \]
03

Compute final momentum for Case (a)

For case (a), the final velocity is vertically downward.\[ v_f = 20 \text{ m/s} \text{ downward} \]Since downward is negative on the vertical axis:\[ v_{fx} = 0 \text{ m/s} \]\[ v_{fy} = -20 \text{ m/s} \]Final momentum components:\[ p_{fx} = 0.30 \text{ kg} \times 0 \text{ m/s} = 0 \text{ kg鈰卪/s} \]\[ p_{fy} = 0.30 \text{ kg} \times (-20 \text{ m/s}) = -6.0 \text{ kg鈰卪/s} \]
04

Calculate change in momentum for Case (a)

Change in momentum for horizontal component:\[ \triangle p_x = p_{fx} - p_{ix} = 0 - 3.6855 = -3.6855 \text{ kg鈰卪/s} \]Change in momentum for vertical component:\[ \triangle p_y = p_{fy} - p_{iy} = -6.0 - 2.583 = -8.583 \text{ kg鈰卪/s} \]Magnitude of the total change in momentum:\[ \triangle p = \text{鈭殅(\triangle p_x^2 + \triangle p_y^2) \]\[ \triangle p = \text{鈭殅((-3.6855)^2 + (-8.583)^2) \]\[ \triangle p \text{ calculated value: } \text{鈭殅(13.5856 + 73.6609) = \text{鈭殅(87.2465) = 9.34 \text{ kg鈰卪/s} \]
05

Compute final momentum for Case (b)

For case (b), the final velocity is horizontally away from the batter.\[ v_f = 20 \text{ m/s} \text{ horizontally} \]Final momentum components:\[ v_{fx} = 20 \text{ m/s} \]\[ v_{fy} = 0 \text{ m/s} \]Final momentum components:\[ p_{fx} = 0.30 \text{ kg} \times 20 \text{ m/s} = 6.0 \text{ kg鈰卪/s} \]\[ p_{fy} = 0.30 \text{ kg} \times 0 = 0 \text{ kg鈰卪/s} \]
06

Calculate change in momentum for Case (b)

Change in momentum for horizontal component:\[ \triangle p_x = p_{fx} - p_{ix} = 6.0 - 3.685 = 2.315 \text{ kg鈰卪/s} \]Change in momentum for vertical component:\[ \triangle p_y = p_{fy} - p_{iy} = 0 - 2.583 = -2.583 \text{ kg鈰卪/s} \]Magnitude of the total change in momentum:\[ \triangle p = \text{鈭殅(\triangle p_x^2 + \triangle p_y^2) \]\[ \triangle p = \text{鈭殅((2.315)^2 + (-2.583)^2) \]\[ \triangle p \text{ calculated value: } \text{鈭殅(5.3622 + 6.6749) = \text{鈭殅(12.0371) = 3.47 \text{ kg鈰卪/s} \]

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Key Concepts

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

Momentum
Momentum is a fundamental concept in physics. It describes the quantity of motion of a moving object and is calculated as the product of an object's mass and its velocity. The formula is:
\(p = m \times v\)
Where:
  • \(p\) is momentum
  • \(m\) is mass
  • \(v\) is velocity
Momentum has both magnitude and direction, making it a vector quantity. In any collision or interaction, the total momentum of a closed system remains constant, as dictated by the law of conservation of momentum.
Trigonometry
Trigonometry helps determine the relationships between the angles and sides of triangles. In physics, especially when dealing with vector components, trigonometric functions such as sine, cosine, and tangent are crucial.
  • \(\text{cos}(\theta)\): For finding the adjacent side of a right triangle
  • \(\text{sin}(\theta)\): For finding the opposite side of a right triangle
For example, given an angle \(\theta\) and the hypotenuse length \(v\), you can find the horizontal (adjacent) and vertical (opposite) components of velocity:
  • Horizontal component: \(v_x = v \times \text{cos}(\theta)\)
  • Vertical component: \(v_y = v \times \text{sin}(\theta)\)
These calculations are essential for breaking down the vectors into manageable components.
Vector Components
Vectors are quantities that have both magnitude and direction. To simplify calculations, vectors are often broken down into their components along the x and y axes.
  • Horizontal Component: This part of the vector aligns with the x-axis.
  • Vertical Component: This part of the vector aligns with the y-axis.
The magnitude of a vector can be found using the Pythagorean theorem when components are known:
\[\text{Magnitude} = \sqrt{v_x^2 + v_y^2}\]
Understanding and calculating vector components allow one to work with vectors more easily in physics problems.
Physics Education
Physics education aims to explain how the world works through fundamental principles and mathematical formulations. Grasping basic physics concepts like momentum, forces, and energy is foundational in understanding more complex phenomena.
  • Use real-world examples to relate concepts
  • Break problems down into simple steps
  • Support learning with visual aids and diagrams
A solid physics education develops critical thinking and problem-solving skills, which are invaluable in scientific and engineering careers.
Problem-Solving Steps
Solving physics problems involves a series of logical steps:
  • Identify the problem: Understand what is being asked.
  • Gather information: Note down known values and given data.
  • Break down the problem: Use diagrams and equations.
  • Apply relevant formulas: Like \(p = mv\) for momentum.
  • Compute the solution: Perform the math carefully.
  • Verify results: Check if the answer makes physical sense.
Following these steps methodically increases accuracy and understanding. Always review each step to ensure nothing is missed.

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Most popular questions from this chapter

A \(2.65 \mathrm{~kg}\) stationary package explodes into three parts that then slide across a frictionless floor. The package had been at the origin of a coordinate system. Part \(A\) has mass \(m_{A}=0.500 \mathrm{~kg}\) and velocity \((10.0 \mathrm{~m} / \mathrm{s} \hat{\mathrm{i}}+12.0 \mathrm{~m} / \mathrm{s} \hat{\mathrm{j}}) .\) Part \(B\) has mass \(m_{B}=0.750 \mathrm{~kg}\), a speed of \(14.0 \mathrm{~m} / \mathrm{s}\), and travels at an angle \(110^{\circ}\) counterclockwise from the positive direction of the \(x\) axis. (a) What is the speed of part \(C ?\) (b) In what direction does it travel?

A \(10 \mathrm{~g}\) bullet moving directly upward at \(1000 \mathrm{~m} / \mathrm{s}\) strikes and passes through the center of a \(5.0 \mathrm{~kg}\) block initially at rest (Fig. 7-29). The bullet emerges from the block moving directly upward at \(400 \mathrm{~m} / \mathrm{s}\). To what maximum height does the block then rise above its initial position? (Hint: Use free-fall equations from Chapter 3.)

Suppose that your mass is \(80 \mathrm{~kg}\). How fast would you have to run to have the same translational momentum as a \(1600 \mathrm{~kg}\) car moving at \(1.2 \mathrm{~km} / \mathrm{h}\) ?

A \(6090 \mathrm{~kg}\) space probe, moving nose-first toward Jupiter at \(105 \mathrm{~m} / \mathrm{s}\) relative to the Sun, fires its rocket engine, ejecting \(80.0 \mathrm{~kg}\) of exhaust at a speed of \(253 \mathrm{~m} / \mathrm{s}\) relative to the space probe. What is the final velocity of the probe?

A pellet gun fires ten \(2.0 \mathrm{~g}\) pellets per second with a speed of \(500 \mathrm{~m} / \mathrm{s}\). The pellets are stopped by a rigid wall. What are (a) the momentum of each pellet and (b) the magnitude of the average force on the wall from the stream of pellets? (c) If each pellet is in contact with the wall for \(0.6 \mathrm{~ms}\), what is the magnitude of the average force on the wall from each pellet during contact? (d) Why is this average force so different from the average force calculated in (b)?
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