**Title: Momentum of a Thrown Ball****Problem Statement:**A ball of mass 189 g is thrown with an initial velocity of 25 m/s at an angle of 30° with the horizontal direction. Ignore air resistance. What is the momentum of the ball after 0.3 s? (Solve this problem by finding the components of the momentum first and then constructing the magnitude and direction of the momentum vector from the components. Find the magnitude in kg · m/s and the direction in degrees counterclockwise from the +x-axis. Assume the +x-axis points horizontally to the right.)**Diagram Explanation:**- A ball is depicted with an arrow indicating its initial velocity vector, labeled as \(\vec{v_i} = (25 \text{ m/s})\hat{v}\).- The velocity vector is inclined at an angle of 30° to the horizontal (x-axis).**Inputs Required:**- **Magnitude**: ________ kg · m/s- **Direction**: ________ ° counterclockwise from the +x-axis**Instructions:**1. Calculate the momentum components: - Break down the velocity into horizontal (\(v_{ix}\)) and vertical (\(v_{iy}\)) components: - \(v_{ix} = v_i \cdot \cos(30°)\) - \(v_{iy} = v_i \cdot \sin(30°)\) - Find the momentum components using: - Horizontal: \(p_{x} = m \cdot v_{ix}\) - Vertical: \(p_{y} = m \cdot v_{iy}\)2. Determine the magnitude of momentum: - \(p = \sqrt{(p_{x}^2 + p_{y}^2)}\)3. Determine the direction of momentum: - \(\text{Direction} = \tan^{-1}\left(\frac{p_{y}}{p_{x}}\right)\)**Note:** Enter the final calculated momentum magnitude in the first blank and the direction in degrees counterclockwise from the +x-axis in the second blank.

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A ball of mass 189 g is thrown with an initial velocity of 25 m/s at an angle of 30° with the horizontal direction. Ignore air resistance. What is the momentum of the ball after 0.3 s? (Do this problem by finding the components of the momentum first and then constructing the magnitude and direction of the momentum vector from the components. Find the magnitude in kg · m/s and the direction in degrees counterclockwise from the +x-axis. Assume the +x-axis points horizontally to the right.) v = (25 m/s)û 30° magnitude kg · m/s direction ° counterclockwise from the +x-axis

A ball of mass 189 g is thrown with an initial velocity of 25 m/s at an angle of 30° with the horizontal direction. Ignore air resistance. What is the momentum of the ball after 0.3 s? (Do this problem by finding the components of the momentum first and then constructing the magnitude and direction of the momentum vector from the components. Find the magnitude in kg · m/s and the direction in degrees counterclockwise from the +x-axis. Assume the +x-axis points horizontally to the right.) v = (25 m/s)û 30° magnitude kg · m/s direction ° counterclockwise from the +x-axis

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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