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Chapter 8: Problem 6

The mass of a regulation tennis ball is 57 g (although it can vary slightly), and tests have shown that the ball is in contact with the tennis racket for 30 ms. (This number can also vary, depending on the racket and swing.) We shall assume a 30.0-ms contact time. The fastest-known served tennis ball was served by "Big Bill" Tilden in 1931, and its speed was measured to be 73 m/s. (a) What impulse and what force did Big Bill exert on the tennis ball in his record serve? (b) If Big Bill's opponent returned his serve with a speed of 55 m/s, what force and what impulse did he exert on the ball, assuming only horizontal motion?

Short Answer

Expert verified
Big Bill exerted an impulse of 4.161 Ns and a force of 138.7 N; the opponent exerted an impulse of -7.29 Ns and a force of -243 N.

Step by step solution

01

Convert Units

First, note the given mass of the tennis ball is 57 g. We must convert this into kilograms to use it in calculations involving SI units. So, the mass of the ball in kg is:\[ m = 57 \, \text{g} = \frac{57}{1000} \, \text{kg} = 0.057 \, \text{kg} \]
02

Calculate Impulse for Big Bill

Impulse is given by the change in momentum. The initial velocity of the ball was 0 m/s (since the ball is at rest before the serve), and the final velocity is 73 m/s. Hence, the impulse can be calculated as:\[ J = \Delta p = m \cdot (v_f - v_i) = 0.057 \, \text{kg} \cdot (73 \, \text{m/s} - 0 \, \text{m/s}) = 4.161 \, \text{Ns} \]
03

Calculate Force for Big Bill

Since impulse also equals force times the time of contact, we can rearrange the formula for impulse to solve for force:\[ F = \frac{J}{\Delta t} = \frac{4.161 \, \text{Ns}}{0.03 \, \text{s}} = 138.7 \, \text{N} \]
04

Calculate Impulse for Opponent

For the opponent returning the serve, the initial velocity is 73 m/s (towards the opponent) and the final velocity is -55 m/s (assuming the direction back toward Big Bill is negative). Hence, the impulse is:\[ J = m \cdot (v_f - v_i) = 0.057 \, \text{kg} \cdot (-55 \, \text{m/s} - 73 \, \text{m/s}) = -7.29 \, \text{Ns} \]
05

Calculate Force for Opponent

Using the impulse value for the opponent, calculate the force using the time of contact:\[ F = \frac{J}{\Delta t} = \frac{-7.29 \, \text{Ns}}{0.03 \, \text{s}} = -243 \, \text{N} \]

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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 that describes the motion of an object. It is defined as the product of an object's mass and its velocity. Mathematically, momentum \( p \) is expressed as:\[ p = m imes v \]where \( m \) is the mass in kilograms (kg) and \( v \) is the velocity in meters per second (m/s). Momentum is a vector quantity, which means it has both a magnitude and a direction. This makes it crucial to consider the direction of motion, especially when dealing with problems involving different paths or angles.

In the context of our tennis ball problem, you look at the change in momentum to calculate 'impulse'. Impulse is a crucial concept as it defines how much momentum is changed due to the application of a force over a contact time.
Force Calculation
Force calculation often involves understanding Newton鈥檚 second law, which states that force is the rate of change of momentum. For cases involving constant mass and velocity changes, force \( F \) can be derived via impulse where:\[ F = \frac{J}{\Delta t} \]Here, \( J \) represents impulse and \( \Delta t \) represents the contact time. In the example, we calculated the force exerted by Big Bill on the tennis ball during a serve. We first found the impulse as the product of the mass and the change in velocity and then divided by the contact time (30 ms) to find the actual force exerted.

The resulting units for force are in newtons (N), which align with the International System of Units (SI), making it easier for calculations and comparisons.
Unit Conversion
Unit conversion is a pivotal part of solving physics problems accurately. Often, you might be given measurements in units that need converting into the appropriate form for calculation, especially when working with the SI system.

In our tennis ball example, the mass of the ball is initially given in grams. Since standard physics calculations often require kilograms, you need to convert grams to kilograms:- Divide the mass in grams by 1000 (since 1 kg = 1000 g).Thus, for the tennis ball's mass:\[ m = \frac{57}{1000} = 0.057 \text{ kg} \]Such conversions are critical, as incorrect units can result in completely erroneous calculations and interpretations, impacting the outcome severely.
Contact Time
Contact time refers to the duration in which two interacting objects are in direct contact. Understanding contact time is essential in calculating the force during interactions as it inversely affects the resulting force exerted during the event according to the relation:\[ F = \frac{J}{\Delta t} \]where \( \Delta t \) is the time period for which the contact is made. In the context of sports or collisions, this brief period can make significant impacts on objects due to varying forces applied in different contact times.

Consider the tennis example, where a contact time of 30 milliseconds is assumed. This time factor, though brief, is fundamental in determining how much force Big Bill and his opponent exerted on the ball during serving and returning showcases. Investigating varying contact times can provide insights into optimizing performance in sports, ensuring maximum efficiency within the game's rules.

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Most popular questions from this chapter

In Section 8.5 we calculated the center of mass by considering objects composed of a \(finite\) number of point masses or objects that, by symmetry, could be represented by a finite number of point masses. For a solid object whose mass distribution does not allow for a simple determination of the center of mass by symmetry, the sums of Eqs. (8.28) must be generalized to integrals $$x_{cm} = {1\over M}\int x \space dm \space y_{cm} = {1 \over M}\int y \space dm$$ where \(x\) and \(y\) are the coordinates of the small piece of the object that has mass \(dm\). The integration is over the whole of the object. Consider a thin rod of length \(L\), mass \(M\), and cross-sectional area \(A\). Let the origin of the coordinates be at the left end of the rod and the positive \(x\)-axis lie along the rod. (a) If the density \(\rho = M/V\) of the object is uniform, perform the integration described above to show that the \(x\)-coordinate of the center of mass of the rod is at its geometrical center. (b) If the density of the object varies linearly with \(x-\)that is, \(\rho = ax\), where a is a positive constant\(-\)calculate the \(x\)-coordinate of the rod's center of mass.

A steel ball with mass 40.0 \(g\) is dropped from a height of 2.00 m onto a horizontal steel slab. The ball rebounds to a height of 1.60 m. (a) Calculate the impulse delivered to the ball during impact. (b) If the ball is in contact with the slab for 2.00 ms, find the average force on the ball during impact.

A radio-controlled model airplane has a momentum given by \([(-0.75 kg \cdot m/s^3)t^2\) + \((3.0 kg \cdot m/s)]\hat{\imath}\) + \((0.25 kg \cdot m/s^2)t\hat{\jmath}\). What are the \(x\)-, \(y\)-, and \(z\)-components of the net force on the airplane?

At time \(t\) = 0 a 2150-kg rocket in outer space fires an engine that exerts an increasing force on it in the \(+x\)-direction. This force obeys the equation \(F_x = At^2\), where \(t\) is time, and has a magnitude of 781.25 N when \(t\) = 1.25 s. (a) Find the SI value of the constant \(A\), including its units. (b) What impulse does the engine exert on the rocket during the 1.50-s interval starting 2.00 s after the engine is fired? (c) By how much does the rocket's velocity change during this interval? Assume constant mass.

Two skaters collide and grab on to each other on frictionless ice. One of them, of mass 70.0 kg, is moving to the right at 4.00 m/s, while the other, of mass 65.0 kg, is moving to the left at 2.50 m/s. What are the magnitude and direction of the velocity of these skaters just after they collide?
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