/*! 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 15 A \(0.50-\mathrm{kg}\) croquet b... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(0.50-\mathrm{kg}\) croquet ball is initially at rest on the grass. When the ball is struck by a mallet, the average force exerted on it is \(230 \mathrm{N}\). If the ball's speed after being struck is \(3.2 \mathrm{m} / \mathrm{s}\), how long was the mallet in contact with the ball?

Short Answer

Expert verified
The mallet was in contact with the ball for approximately 0.007 seconds.

Step by step solution

01

Understand the Given Problem

We are asked to find the time duration for which a mallet was in contact with a croquet ball. The force exerted by the mallet is given as 230 N, the mass of the ball is 0.50 kg, and its final speed is 3.2 m/s. The initial speed of the ball was 0 m/s since it was at rest.
02

Identify the Formula

Recall Newton's second law of motion, which states that Force equals mass times acceleration: \( F = ma \). Also, use the definition of acceleration as the change in velocity over time: \( a = \frac{\Delta v}{\Delta t} \). We can combine these two equations to find time: \( t = \frac{m \cdot \Delta v}{F} \).
03

Calculate Change in Velocity

The change in velocity (\( \Delta v \)) is the final velocity minus the initial velocity. Since the initial velocity is 0 m/s and the final velocity is 3.2 m/s, we have \( \Delta v = 3.2 \, \mathrm{m/s} \).
04

Substitute Values into the Formula

Substitute the given values into the formula for time: \( t = \frac{0.50 \, \mathrm{kg} \times 3.2 \, \mathrm{m/s}}{230 \, \mathrm{N}} \).
05

Calculate the Time of Contact

Perform the calculation: \( t = \frac{0.50 \times 3.2}{230} = \frac{1.6}{230} \approx 0.007 \mathrm{s} \). This is the time duration for which the mallet was in contact with the ball.

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Key Concepts

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

Impulse and Momentum
Impulse and momentum are two closely related concepts in physics, especially when analyzing moving objects. Impulse is defined as the change in momentum of an object when it is acted upon by a force for a particular length of time. This is expressed through the equation:\[ J = ext{Force} \times ext{Time} = ext{Change in Momentum} \]Momentum, on the other hand, is a measure of an object's motion. It's calculated as the product of an object's mass and its velocity, as shown below:\[ p = m \times v \]When an object like a croquet ball is struck by a mallet, the impulse exerted by the force from the mallet changes the ball's momentum. The impulse can be understood as the mechanism that leads to a change in the velocity of the ball, aligning perfectly with Newton's Second Law of Motion.

Think of impulse as the amount of push or pull applied over time to change how fast an object is moving.
Force Calculation
Newton’s Second Law of Motion is fundamental for calculating force, where the force exerted on an object equals its mass times its acceleration. This relationship is captured by the formula:\[ F = m \times a \]In the context of the croquet ball being hit by a mallet, it's important to understand how acceleration comes into play. Here, acceleration is determined by how the velocity changes over time. Since the initial velocity is zero, we can focus on the final velocity acquired due to the force applied.When solving such problems:
  • First, identify the force applied and the mass of the object.
  • Use the relationship between force and acceleration to find other missing values.
By understanding how forces result in changes of motion, you can predict the time duration the force was active. This is key to solving the problem of finding how long the mallet was in contact with the ball.
Velocity Change
The change in velocity is critical in understanding how quickly an object like a croquet ball speeds up or slows down after a force is applied. In our problem, we observe a change in velocity from 0 m/s (rest) to 3.2 m/s (after being struck), which indicates the acceleration caused by the force.To calculate the change in velocity, simply subtract the initial velocity from the final velocity:\[ \Delta v = v_{final} - v_{initial} \]This straightforward calculation helps in determining how effectively the force has impacted the object. In practical terms, it gives insight into the result of the force application over time or through different surfaces.Understanding velocity change tells us not only how an object moves, but also supports more advanced calculations involving kinetic energy, momentum, and further dynamics. Recognizing this change lets students grasp why certain formulas are structured the way they are and how to apply them in real-world scenarios.

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Most popular questions from this chapter

A piece of sheet metal of mass \(M\) is cut into the shape of a right triangle, as shown in Figure \(9-18\). A vertical dashed line is drawn on the sheet at the point where the mass to the left of the line \((M / 2)\) is equal to the mass to the right of the line (also \(M / 2\) ). The sheet is now placed on a fulcrum just under the dashed line and released from rest. (a) Does the metal sheet remain level, tip to the left, or tip to the right? (b) Choose the best explanation from among the following: I. Equal mass on either side will keep the metal sheet level. II. The metal sheet extends for a greater distance to the left, which shifts the center of mass to the left of the dashed line. III. The center of mass is to the right of the dashed line because the metal sheet is thicker there.

A cardboard box is in the shape of a cube with each side of length \(L\). If the top of the box is missing, where is the center of mass of the open box? Give your answer relative to the geometric center of the box.

A fireworks rocket is launched vertically into the night sky with an initial speed of \(44.2 \mathrm{m} / \mathrm{s}\). The rocket coasts after being launched, then explodes and breaks into two pieces of equal mass 2.50 s later. (a) If each piece follows a trajectory that is initially at \(45.0^{\circ}\) to the vertical, what was their speed immediately after the explosion? (b) What is the velocity of the rocket's center of mass before and after the explosion? (c) What is the acceleration of the rocket's center of mass before and after the explosion?

Suppose you throw a rubber ball at an elephant that is charging directly at you (not a good idea). When the ball bounces back toward you, is its speed greater than, less than, or equal to the speed with which you threw it? Explain.

Predict/Explain A stalactite in a cave has drops of water falling from it to the cave floor below. The drops are equally spaced in time and come in rapid succession, so that at any given moment there are many drops in midair. (a) Is the center of mass of the midair drops higher than, lower than, or equal to the halfway distance between the tip of the stalactite and the cave floor? (b) Choose the best explanation from among the following: I. The drops bunch up as they near the floor of the cave. II. The drops are equally spaced as they fall, since they are released at equal times. III. Though equally spaced in time, the drops are closer together higher up.

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