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Chapter 8: Problem 31

Typically, a tennis ball hit during a serve travels away at about 51 \(\mathrm{m} / \mathrm{s}\). If the ball is at rest mid-air when struck, and it has a mass of \(0.058 \mathrm{~kg}\), what is the change in its momentum on leaving the racket?

Short Answer

Expert verified
The change in momentum is 2.958 kg m/s.

Step by step solution

01

Identify Given Information

We are given the speed of the tennis ball after it is hit, which is \(51 \, \text{m/s}\), and its mass, which is \(0.058 \, \text{kg}\). The initial velocity of the ball is \(0 \, \text{m/s}\) as it is said to be at rest mid-air.
02

Understand the Concept of Momentum

Momentum (\(p\)) is defined as the product of an object's mass (\(m\)) and its velocity (\(v\)). The formula for momentum is \(p = mv\).
03

Calculate Initial Momentum

The initial momentum of the ball is calculated using the formula \(p = mv\), where \(v = 0 \, \text{m/s}\). So, the initial momentum is:\[ p_i = 0.058 \, \text{kg} \times 0 \, \text{m/s} = 0 \, \text{kg} \, \text{m/s} \]
04

Calculate Final Momentum

The final momentum of the ball is calculated using its velocity after the serve, \(v = 51 \, \text{m/s}\). So, the final momentum is:\[ p_f = 0.058 \, \text{kg} \times 51 \, \text{m/s} = 2.958 \, \text{kg} \, \text{m/s} \]
05

Calculate Change in Momentum

The change in momentum (\(\Delta p\)) is the final momentum minus the initial momentum:\[ \Delta p = p_f - p_i = 2.958 \, \text{kg} \, \text{m/s} - 0 \, \text{kg} \, \text{m/s} = 2.958 \, \text{kg} \, \text{m/s} \]

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Key Concepts

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

Conservation of Momentum
Conservation of momentum is a fundamental concept in physics that tells us how momentum behaves in isolated systems. According to this principle, the total momentum of a closed system remains constant if no external forces act upon it. This means that in any interaction, such as a collision or explosion, the total momentum before the interaction is equal to the total momentum after the interaction.
For instance, when a tennis ball is served, the interaction between the racket and the ball can be considered. If no external forces are acting on the racket-ball system, the momentum of the racket before and after hitting the ball as well as the ball's momentum must contribute to conserving the system's total momentum. This principle helps us analyze and predict the resulting speeds and directions of the objects involved.
Velocity
Velocity is a vector quantity that refers to the speed of an object in a particular direction. It is different from speed, which is a scalar and only tells you how fast an object is moving without any direction involved. In mathematical terms, velocity (\(v\)) is often expressed as distance divided by time: \(v = \frac{d}{t}\).
In the case of the tennis ball, its velocity after being hit by the racket is given as 51 m/s. This directional speed indicates how fast the ball travels away in a specific direction. Analyzing how velocity changes when a force acts on an object is critical to understanding how objects move in the world around us.
Changes in velocity, such as from 0 m/s to 51 m/s like in this case, indicate an increase, which directly influences the change in the ball's momentum.
Mass
Mass, in the context of physics, is a measure of the amount of matter in an object. It is a scalar quantity and remains unchanged regardless of an object's location or speed. Mass plays a crucial role in determining an object's momentum. The greater the object's mass, the larger its momentum for a given velocity.
For the problem at hand, the tennis ball's mass is 0.058 kg. This value is utilised in the momentum equation \(p = mv\), showing how integral mass is in calculating changes in an object's motion. Consistently, the mass remains constant whether the ball is stationary or in motion. However, it's the combination of this mass and the ball's velocity that offers insight into the dynamic properties of its movement.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. This energy is linked directly to an object's mass and the square of its velocity, formulated as \(KE = \frac{1}{2}mv^2\). An increase in either mass or velocity results in a higher kinetic energy.
In relation to the given problem, when the tennis ball goes from rest to a velocity of 51 m/s, its kinetic energy changes from zero to a larger amount proportional to its gained speed. Unlike momentum, which depends linearly on velocity, kinetic energy depends on the square of velocity, making it more sensitive to speed changes.
  • This means that a small increase in the ball's velocity results in a larger increase in its kinetic energy.
  • Understanding kinetic energy allows us to appreciate the work needed to accelerate the ball to its fast-moving state.
Comprehending both energy and momentum provides a comprehensive understanding of moving objects in physics.

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Most popular questions from this chapter

A 16 -g mass is moving in the \(+x\) -direction at \(30 \mathrm{~cm} / \mathrm{s}\), while a \(4.0\) -g mass is moving in the \(-x\) -direction at \(50 \mathrm{~cm} / \mathrm{s}\). They collide headon and stick together. Find their velocity after the collision. Assume negligible friction. This is a completely inelastic collision for which KE is not conserved, although momentum is. Let the 16 -g mass be \(m_{1}\) and the 4.0-g mass be \(m_{2}\). Take the \(+x\) -direction to be positive. That means that the velocity of the \(4.0\) -g mass has a scalar value of \(u_{2 x}\) \(=-50 \mathrm{~cm} / \mathrm{s}\). We apply the law of conservation of momentum to the system consisting of the two masses: $$ \begin{array}{l} \text { Momentum before impact = Momentum after impact }\\\ \begin{aligned} m_{1} v_{1 x}+m_{2} v_{2 x} &=\left(m_{1}+m_{2}\right) v_{x} \\ (0.016 \mathrm{~kg})(0.30 \mathrm{~m} / \mathrm{s})+(0.0040 \mathrm{~kg})(-0.50 \mathrm{~m} / \mathrm{s}) &=(0.020 \mathrm{~kg}) v_{x} \\ v_{x} &=+0.14 \mathrm{~m} / \mathrm{s} \end{aligned} \end{array} $$ (Notice that the \(4.0\) -g mass has negative momentum.) Hence, \(\overrightarrow{\mathbf{v}}=\) \(0.14 \mathrm{~m} / \mathrm{s}\) - POSITIVE \(X\) -DIRECTION

A \(0.25\) -kg ball moving in the \(+x\) -direction at \(13 \mathrm{~m} / \mathrm{s}\) is hit by a bat. Its final velocity leaving the bat is \(19 \mathrm{~m} / \mathrm{s}\) in the \(x\) -direction. The bat acts on the ball for \(0.010 \mathrm{~s}\). Find the average force \(F\) exerted on the ball by the bat. The problem provides the time over which a required force acts, as well as enough information to compute the change in momentum. That suggests the impulse equation (i.e., Newton's Second Law). We have \(v_{i}=13 \mathrm{~m} / \mathrm{s}\) and \(v_{f}=-19 \mathrm{~m} / \mathrm{s}\). Taking the initial direction of motion as positive, the impulse equation is $$ \begin{aligned} F \Delta t &=m v_{f}-m v_{i} \\ F(0.010 \mathrm{~s}) &=(0.25 \mathrm{~kg})(-19 \mathrm{~m} / \mathrm{s})-(0.25 \mathrm{~kg})(13 \mathrm{~m} / \mathrm{s}) \end{aligned} $$ from which \(F=-0.80 \mathrm{kN}\).

A billiard ball moving at a speed \(u_{1 i}\) strikes, head-on, another billiard ball that is at rest. Assuming the collision is completely elastic, show that $$ v_{1 i}^{2}=v_{\mathrm{lf}}^{2}+v_{2 f}^{2} $$

A ball is dropped from a height \(h\) above a tile floor and rebounds to a height of \(0.65 h\). Find the coefficient of restitution between ball and floor. Assign floor quantities the subscript 1 , and ball quantities the subscript \(2 .\) The initial and final velocities of the floor, \(u_{1}\) and \(v_{1}\), are zero. Therefore, $$ e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}}=-\frac{v_{2}}{u_{2}} $$ Since we know both the drop and rebound heights ( \(h\) and \(0.65 h\) ), we can write equations for the interchange of \(\mathrm{PE}_{\mathrm{G}}\) and \(\mathrm{KE}\) before and after the impact \(\sqrt{665}=0.8\) Notice that the coefficient of restitution equals the square root of the final rebound height over the initial drop height.

A 90 -g ball moving at \(100 \mathrm{~cm} / \mathrm{s}\) collides head-on with a stationary 10-g ball. Determine the speed of each after impact if \((a)\) they stick together, \((b)\) the collision is perfectly elastic, \((c)\) the coefficient of restitution is \(0.90\).
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