/*! 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 2 A high-speed photograph of a clu... [FREE SOLUTION] | 91Ó°ÊÓ

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A high-speed photograph of a club hitting a golf ball is shown in Figure \(6.3\). The club was in contact with a ball, initially at rest, for about \(0.0020 \mathrm{~s}\). If the ball has a mass of \(55 \mathrm{~g}\) and leaves the head of the club with a speed of \(2.0 \times 10^{2} \mathrm{ft} / \mathrm{s}\), find the average force exerted on the ball by the club.

Short Answer

Expert verified
The average force exerted on the ball by the club is approximately 1677.5 N.

Step by step solution

01

Convert units

Given mass \(m\) of the ball is 55 g, which we need to convert to kilograms. Since 1 kg = 1000 g, therefore mass \(m = 55 / 1000 = 0.055 kg\). The ball leaves the head of the club with a speed of \(2.0 \times 10^{2} ft/s\). We should convert this speed to meters per second. Given that 1 ft = 0.3048 m, our speed in SI units will be \( V = 2.0 \times 10^{2} \times 0.3048 = 61.0 m/s.\)
02

Determine acceleration

Next we need to determine the acceleration of the golf ball. This can be done by dividing the final velocity, \( V \), by the time of contact, \( t \). The given time is \( t = 0.0020 s \). Since the golf ball was initially at rest, its acceleration \( a \) is \( V/t = 61.0 m/s / 0.0020 s = 3.05 \times 10^{4} m/s^{2}\)
03

Find force

Finally, we will use Newton's second law, \( F = ma \), to calculate the force exerted on the ball by the club. Here, \( m = 0.055 kg \) and \( a = 3.05 \times 10^{4} m/s^{2} \). Substituting these values in our equation, we find the force \( F = 0.055 kg \times 30500 m/s^{2} = 1677.5 N.\)

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Key Concepts

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

Momentum in Collisions
Understanding momentum in collisions is critical to analyzing the forces involved when two objects interact. In physics, the momentum of an object is the product of its mass and velocity. It is a vector quantity, meaning it has both magnitude and direction. During a collision, such as a golf club striking a ball, momentum is transferred from one object to another.

According to the principle of conservation of momentum, the total momentum of a system of objects (the club and the ball, in this case) remains constant if no external forces act upon it. In our exercise, when the club hits the stationary ball, it exerts a force over a short time interval, imparting momentum to the ball, sending it into motion. This is what enables us to determine the average force exerted by the club on the ball - it is the force necessary to change the ball's momentum from zero to its value upon leaving the club.
Newton's Second Law
Central to our understanding of forces and motion is Newton's second law of motion. This law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The relationship is encapsulated in the formula: \( F = ma \), where \( F \) is the net force, \( m \) is the mass, and \( a \) is the acceleration.

In the context of the golf ball scenario, we apply Newton's second law to determine the average force exerted on the ball. Once we know the ball's acceleration and mass, we can calculate the force. This relationship tells us that a larger force would result in a greater acceleration if the mass remains constant. Likewise, if we were to have two golf balls of different masses struck with the same force, the lighter ball would have a greater acceleration.
Unit Conversion
In physics problems, especially in mechanics, it’s essential to perform unit conversions to maintain consistency across the calculations. The International System of Units (SI) is commonly used, where the standard units of length, mass, and time are meters (m), kilograms (kg), and seconds (s), respectively.

In the solution of our golf ball problem, the unit conversion from grams to kilograms and from feet per second to meters per second ensures that we use the correct SI units for mass and velocity. This step is crucial because if these conversions are not done properly, it can lead to incorrect calculations of other quantities such as acceleration and force. For this reason, when solving physics problems, always start by converting all your measurements to SI units to avoid any confusion and errors in your final answer.
Acceleration Calculation
The calculation of acceleration is fundamental when evaluating the motion of objects. In physics, acceleration is defined as the rate of change of velocity of an object. It can be calculated by using the formula: \( a = \frac{\Delta v}{\Delta t} \), where \( \Delta v \) is the change in velocity and \( \Delta t \) is the change in time.

For our golf ball, since it starts from rest, its initial velocity is zero, and the change in velocity is equal to its final velocity after being struck by the club. By dividing this velocity by the time interval during which the force was applied, we find the ball's acceleration. This step is fundamental in applying Newton's second law to find the force, as once we have the acceleration, we can easily calculate the force by multiplying it by the mass of the ball. Remember, the larger the acceleration, the more force is required to achieve it over the same time interval.

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Most popular questions from this chapter

(a) A car traveling due east strikes a car traveling due north at an intersection, and the two move together as a unit. A property owner on the southeast corner of the intersection claims that his fence was torn down in the collision. Should he be awarded damages by the insurance company? Defend your answer. (b) Let the eastward-moving car have a mass of 1 300 kg and a speed of 30.0 km/h and the northward-moving car a mass of 1 100 kg and a speed of 20.0 km/h. Find the velocity after the collision. Are the results consistent with your answer to part (a)?

Identical twins, each with mass 55.0 kg, are on ice skates and at rest on a frozen lake, which may be taken as frictionless. Twin A is carrying a backpack of mass ats fractionless. Twin A is carrying a backpack of inass \(12.0 \mathrm{~kg}\). She throws it horizomtally at \(9.00 \mathrm{~m} / \mathrm{s}\) to Twin B. Neglecting any gravity effects, what are the subsequent speeds of Twin \(A\) and Twin \(B\) ?

A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to \(5.20 \mathrm{~m} / \mathrm{s}\) in \(0.832 \mathrm{~s}\). What are (a) the magnitudes of the linear impulse and (b) the average total force experienced by a \(70.0-\mathrm{kg}\) passenger in the car during the time the car accelerates?

A man claims he can safely hold on to a \(12.0-\mathrm{kg}\) child in a head-on collision with a relative speed of \(120-\mathrm{mi} / \mathrm{h}\) lasting for \(0.10 \mathrm{~s}\) as long as he has his seat belt on. (a) Find the magnitude of the average force needed to hold onto the child. (b) Based on the result to part (a), is the man's claim valid? (c) What does the answer to this problem say about laws requiring the use of proper safety devices such as seat belts and special toddler seats?

A \(0.280-\mathrm{kg}\) volleyball approaches a player horizontally with a speed of \(15.0 \mathrm{~m} / \mathrm{s}\). The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of \(22.0 \mathrm{~m} / \mathrm{s}\). (a) What impulse is delivered to the ball by the player? (b) If the player's fist is in contact with the ball for \(0.0600 \mathrm{~s}\), find the magnitude of the average force exerted on the player's fist.

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