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Chapter 8: Problem 21

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 of the tennis ball is \( 2.958 \mathrm{~kg \, m/s} \).

Step by step solution

01

Understand Momentum Formula

Momentum is defined as the product of mass and velocity. The formula is given by \( p = m imes v \). Here, \( p \) is momentum, \( m \) is mass, and \( v \) is velocity.
02

Identify Initial Conditions

Initially, the tennis ball is at rest, which means its initial velocity \( v_i = 0 \mathrm{~m/s} \). Its mass \( m = 0.058 \mathrm{~kg} \).
03

Identify Final Conditions

When the ball is struck, it reaches a velocity \( v_f = 51 \mathrm{~m/s} \). The mass remains constant \( m = 0.058 \mathrm{~kg} \).
04

Calculate Initial Momentum

Using the formula \( p = m \times v \), the initial momentum is \( p_i = 0.058 \times 0 = 0 \mathrm{~kg \, m/s} \).
05

Calculate Final Momentum

Using the same formula, the final momentum is \( p_f = 0.058 \times 51 = 2.958 \mathrm{~kg \, m/s} \).
06

Determine Change in Momentum

The change in momentum \( \Delta p \) is the difference between final momentum and initial momentum: \( \Delta p = p_f - p_i = 2.958 - 0 = 2.958 \mathrm{~kg \, m/s} \).

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Key Concepts

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

Momentum Formula
Understanding the concept of momentum in physics is essential as it allows you to comprehend how objects move and interact with each other. Momentum is a vector quantity, which means it has both magnitude and direction. The momentum formula is expressed as:
  • \( p = m \times v \)
  • Where \( p \) represents momentum, \( m \) is mass, and \( v \) is velocity.
This formula suggests that the momentum of an object is directly proportional to its mass and velocity. This means, the more massive an object or the faster it moves, the greater its momentum.
To grasp momentum better, think about a truck and a bicycle moving at the same speed. The truck will have significantly more momentum due to its larger mass.
Mass and Velocity
When considering momentum, both mass and velocity are crucial components. Mass refers to the amount of matter contained in an object, typically measured in kilograms (kg). In our tennis ball example, the ball鈥檚 mass is \( 0.058 \text{ kg} \).
Velocity, on the other hand, describes the speed and direction of an object's motion, measured in meters per second (m/s). Initially, the tennis ball鈥檚 velocity is \( 0 \text{ m/s} \) when at rest.
  • The mass remains constant once the ball is in motion.
  • The velocity, however, changes from \( 0 \text{ m/s} \) to \( 51 \text{ m/s} \) after being hit.
This change in velocity is what significantly affects the momentum of the ball.
Change in Momentum
The change in momentum is an important concept, indicating how much the momentum of an object has altered due to an external force. To find the change in momentum, first calculate the initial and final momentum, then determine their difference.
In our tennis ball scenario:
  • Initial momentum: When the ball is at rest (velocity = 0), the momentum is \( 0.058 \times 0 = 0 \text{ kg}\, \text{m/s} \).
  • Final momentum: After being hit, the ball鈥檚 momentum becomes \( 0.058 \times 51 = 2.958 \text{ kg}\, \text{m/s} \).
  • Change in momentum \( (\Delta p) \): It's calculated by subtracting the initial momentum from the final momentum:\[ \Delta p = 2.958 - 0 = 2.958 \text{ kg}\, \text{m/s} \]
So, the ball's momentum changed by \( 2.958 \text{ kg m/s} \) as a result of being struck.
Momentum Calculation
Momentum calculation involves using the formula \( p = m \times v \). First, identify the mass and the velocity of the object at the specific event points you鈥檙e interested in. In physics problems like our tennis ball example, calculations for initial and final moments are essential to analyzing how forces influence motion.
Consider the following:
  • Initial calculation: With the ball at rest, velocity is \( 0 \text{ m/s} \). The momentum is \( 0.058 \times 0 = 0 \text{ kg}\, \text{m/s} \).
  • Final calculation: After the ball moves with a velocity of \( 51 \text{ m/s} \), the momentum is \( 0.058 \times 51 = 2.958 \text{ kg}\, \text{m/s} \).
Performing these calculations can show the effect of striking the ball and changing its velocity鈥攁 change vital to understanding physical interactions in sports and other real-world scenarios.
Initial and Final Conditions
Understanding initial and final conditions is critical as they help determine how forces alter the state of motion of an object. Initial conditions refer to the state before any external forces act, while final conditions are the states after such forces.
In our tennis example:
  • Initial Conditions: The ball's state when at rest. Here, both velocity and momentum are zero.
  • Final Conditions: The state after the ball is hit by the racket. The velocity becomes \( 51 \text{ m/s} \) and the momentum increases to \( 2.958 \text{ kg}\, \text{m/s} \).
These conditions are crucial for calculating changes in momentum and understanding how applied forces influence the motion in physics.

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Most popular questions from this chapter

A \(1200-\mathrm{kg}\) gun mounted on wheels shoots an \(8.00-\mathrm{kg}\) projectile with a muzzle velocity of \(600 \mathrm{~m} / \mathrm{s}\) at an angle of \(300^{\circ}\) above the horizontal. Find the horizontal recoil speed of the gun.

A ball of mass \(m\) sits at the coordinate origin when it explodes into two pieces that shoot along the \(x\) -axis in opposite directions. When one of the pieces (which has mass \(0.270 \mathrm{~m}\) ) is at \(x=70 \mathrm{~cm}\), where is the other piece? [Hint: What happens to the mass center?]

A \(1.0\) -kg ball moving at \(12 \mathrm{~m} / \mathrm{s}\) collides head-on with a \(2.0-\mathrm{kg}\) ball moving in the opposite direction at \(24 \mathrm{~m} / \mathrm{s}\). Determine the motion of each after impact if \((a) e=2 / 3,(b)\) the balls stick together, and (c) the collision is perfectly elastic. In all three cases the collision occurs along a straight line, and momentum is conserved. Hence, $$ \begin{aligned} \text { Momentum before } &=\text { Momentum after } \\ (1.0 \mathrm{~kg})(12 \mathrm{~m} / \mathrm{s})+(2.0 \mathrm{~kg})(-24 \mathrm{~m} / \mathrm{s}) &=(1.0 \mathrm{~kg}) v_{1}+(2.0 \mathrm{~kg}) v_{2} \end{aligned} $$ which becomes $$ -36 \mathrm{~m} / \mathrm{s}=v_{1}+2 v_{2} $$ (a) When \(e=2 / 3\), $$ e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}} \quad \text { becomes } \quad \frac{2}{3}=\frac{v_{2}-v_{1}}{(12 \mathrm{~m} / \mathrm{s})-(-24 \mathrm{~m} / \mathrm{s})} $$ from which \(24 \mathrm{~m} / \mathrm{s}=v_{2}-v_{1}\). Combining this with the momentum equation found above gives \(v_{2}=-4.0 \mathrm{~m} / \mathrm{s}\) and \(v_{1}=-28 \mathrm{~m} / \mathrm{s}\) (b) In this case \(v_{1}=v_{2}=v\), and so the momentum equation becomes $$ 3 v=-36 \mathrm{~m} / \mathrm{s} \quad \text { or } \quad v=-12 \mathrm{~m} / \mathrm{s} $$ (c) Here \(e=1\), and $$ e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}} \quad \text { becomes } \quad 1=\frac{v_{2}-v_{1}}{(12 \mathrm{~m} / \mathrm{s})-(-24 \mathrm{~m} / \mathrm{s})} $$ from which \(v_{2}-v_{1}=36 \mathrm{~m} / \mathrm{s}\). Adding this to the momentum equation yields \(v_{2}=0\). Using this value for \(v_{2}\) then leads to \(v_{1}=-36 \mathrm{~m} / \mathrm{s}\).

An \(8.0\) -g bullet is fired horizontally into a \(9.00\) -kg cube of wood, which is at rest on a frictionless air table. The bullet lodges in the wood. The cube is free to move and has a speed of \(40 \mathrm{~cm} / \mathrm{s}\) after impact. Find the initial velocity of the bullet. This is an example of a completely inelastic collision for which momentum is conserved, although \(\mathrm{KE}\) is not. Consider the system (cube \(+\) bullet). The velocity, and hence the momentum, of the cube before impact is zero. Take the bullet's initial motion to be positive in the positive \(x\) -direction. The momentum conservation law tells us that Momentum of system before impact = Momentum of system after impact $$ \begin{aligned} \text { (Momentum of bullet) }+\text { (Momentum of cube) } &=(\text { Momentum of bullet }+\text { Cube }) \\ m_{B} v_{B x}+m_{C} v_{C x} &=\left(m_{B}+m_{C}\right) v_{x} \\ (0.0080 \mathrm{~kg}) v_{B x}+0 &=(9.008 \mathrm{~kg})(0.40 \mathrm{~m} / \mathrm{s}) \end{aligned} $$ Solving yields \(v_{B x}=0.45 \mathrm{~km} / \mathrm{s}\) and so \(\overrightarrow{\mathbf{v}}_{B}=0.45 \mathrm{~km} / \mathrm{s}\) - POSITIVE \(X\) -DIRECTION.

A rocket standing on its launch platform points straight upward. Its engines are activated and eject gas at a rate of \(1500 \mathrm{~kg} / \mathrm{s}\). The molecules are expelled with an average speed of \(50 \mathrm{~km} / \mathrm{s}\). How much mass can the rocket initially have if it is slowly to rise because of the thrust of the engines? The problem provides mass flow and speed, the product of which is equivalent to the time rate-of-change of momentum. That should bring to mind the impulse- momentum relationship, which, of course, is Newton's Second Law. Since the initial motion of the rocket itself is negligible in comparison to the speed of the expelled gas, we can assume the gas is accelerated from rest to a speed of \(50 \mathrm{~km} / \mathrm{s}\). The impulse required to provide this acceleration to a mass \(m\) of gas is $$ F \Delta t=m v_{f}-m v_{i}=m(50000 \mathrm{~m} / \mathrm{s})-0 $$ from which $$ F=(50000 \mathrm{~m} / \mathrm{s}) \frac{m}{\Delta t} $$ But we are told that the mass ejected per second \((m / \Delta t)\) is \(1500 \mathrm{~kg} / \mathrm{s}\), and so the force exerted on the expelled gas is $$ F=(50000 \mathrm{~m} / \mathrm{s})(1500 \mathrm{~kg} / \mathrm{s})=75 \mathrm{MN} $$ An equal but opposite reaction force acts on the rocket, and this is the upward thrust on the rocket. The engines can therefore support a weight of \(75 \mathrm{MN}\), so the maximum mass the rocket could have is $$ M_{\text {rocket }}=\frac{\text { weight }}{g}=\frac{75 \times 10^{6} \mathrm{~N}}{9.81 \mathrm{~m} / \mathrm{s}^{2}}=7.7 \times 10^{6} \mathrm{~kg} $$
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