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Chapter 8: Problem 22

During a soccer game a ball (of mass \(0.425 \mathrm{~kg}\) ), which is initially at rest, is kicked by one of the players. The ball moves off at a speed of \(26 \mathrm{~m} / \mathrm{s}\). Given that the impact lasted for \(8.0 \mathrm{~ms}\), what was the average force exerted on the ball?

Short Answer

Expert verified
The average force exerted on the ball is approximately 1381 N.

Step by step solution

01

Understand the Problem

We are given the mass of the ball, its final speed, and the duration of the impact. We need to find the average force exerted on the ball during the impact.
02

Define Key Variables

Let the mass of the ball be \( m = 0.425 \text{ kg} \), the initial velocity \( u = 0 \text{ m/s} \) (since it was initially at rest), the final velocity \( v = 26 \text{ m/s} \), and the time duration \( t = 8.0 \times 10^{-3} \text{ s} \).
03

Calculate the Change in Momentum

The change in momentum \( \Delta p \) of the ball can be calculated using the formula: \( \Delta p = m \times (v - u) \). Given that \( u = 0 \) and \( v = 26 \text{ m/s} \), then \( \Delta p = 0.425 \text{ kg} \times 26 \text{ m/s} = 11.05 \text{ kg m/s} \).
04

Use the Impulse-Momentum Theorem

The impulse-momentum theorem states that the change in momentum is equal to the impulse applied to it, which is the product of force and time: \( F \times t = \Delta p \).
05

Solve for Average Force

Rearrange the impulse-momentum formula to solve for the force: \( F = \frac{\Delta p}{t} = \frac{11.05 \text{ kg m/s}}{8.0 \times 10^{-3} \text{ s}} = 1381.25 \text{ N} \).
06

Consider Accuracy and Units

Make sure the units are consistent and appropriate for the context. The average force experienced by the ball is \( 1381.25 \text{ N} \).

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Key Concepts

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

Impulse-Momentum Theorem
In physics, the impulse-momentum theorem is a fundamental concept that links force, time, and motion. It describes how the momentum of an object changes when a force is applied over a period of time. This theorem is expressed mathematically as:
  • Impulse = Change in Momentum
  • Impulse is the product of force (F) and time (t): \( F \times t \)
This equation tells us that the change in momentum of an object is caused by the impulse applied to it. Impulse itself is a vector quantity, which means it has both magnitude and direction.
In the context of the soccer game problem, the player's kick applies a force over a short time (8.0 ms) to change the ball's momentum. As a result, the impulse-momentum theorem helps us calculate the force exerted on the ball during this dynamic interaction.
Change in Momentum
Momentum is the product of an object's mass and its velocity. It is a measure of an object's motion and is conserved in an isolated system. The change in momentum, represented as \( \Delta p \), is the difference between the final momentum and the initial momentum of an object.
In the soccer game example, we determine the change in momentum (\( m \times (v - u) \)) by considering:
  • The mass of the ball: 0.425 kg
  • Initial velocity: 0 m/s (since the ball was at rest)
  • Final velocity: 26 m/s (after being kicked)
The calculated change in momentum is \( 11.05 \text{ kg m/s} \). This value shows how much the motion of the ball has changed due to the player's kick. Understanding change in momentum is key to solving many physics problems, especially when analyzing collisions and interactions.
Physics Problems
Physics problems often require a methodical approach to solve effectively. In problems involving forces, like the soccer ball scenario, it's essential to:
  • Identify and define key variables, such as mass, velocity, and time.
  • Use relevant physics principles, like the impulse-momentum theorem.
  • Perform calculations step-by-step, ensuring consistency in units.
In our exercise, the challenge was to find the average force acting on a ball using the information given: 1. Recognize the initial conditions, such as the ball's rest state. 2. Calculate the change in velocity and momentum using the supplied data. 3. Apply the impulse-momentum theorem to solve for force.
This structured approach not only results in determining the force (1381.25 N) but also helps in understanding the real-world mechanics of actions like kicking a ball. By practicing problems methodically, students build confidence and skill in physics.

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Most popular questions from this chapter

A system consists of the following masses in the \(x y\) -plane: \(4.0 \mathrm{~kg}\) at coordinates \((x=0, y=5.0 \mathrm{~m})\), \(7.0 \mathrm{~kg}\) at \((3.0 \mathrm{~m}, 8.0 \mathrm{~m})\), and \(5.0 \mathrm{~kg}\) at \((-3.0 \mathrm{~m},-6.0 \mathrm{~m})\). Find the position of its center of mass. $$ \begin{array}{l} x_{\mathrm{cm}}=\frac{\sum x_{i} m_{i}}{\sum m_{i}}=\frac{(0)(4.0 \mathrm{~kg})+(3.0 \mathrm{~m})(7.0 \mathrm{~kg})+(-3.0 \mathrm{~m})(5.0 \mathrm{~kg})}{(4.0+7.0+5.0) \mathrm{kg}}=0.38 \mathrm{~m} \\ y_{\mathrm{cm}}=\frac{\sum y_{i} m_{i}}{\sum m_{i}}=\frac{(5.0 \mathrm{~m})(4.0 \mathrm{~kg})+(8.0 \mathrm{~m})(7.0 \mathrm{~kg})+(-6.0 \mathrm{~m})(5.0 \mathrm{~kg})}{16 \mathrm{~kg}}=2.9 \mathrm{~m} \end{array} $$ and \(z_{\mathrm{cm}}=0\). These distances are, of course, measured from the origin \((0,0,0,\), .

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 \(7.00-\mathrm{g}\) bullet moving horizontally at \(200 \mathrm{~m} / \mathrm{s}\) strikes and passes through a 150 -g tin can sitting on a post. Just after impact, the can has a horizontal speed of \(180 \mathrm{~cm} / \mathrm{s}\). What was the bullet's speed after leaving the can?

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 $$ M_{\text {rocket }}=\frac{\text { weight }}{g}=\frac{75 \times 10^{6} \mathrm{~N}}{9.81 \mathrm{~m} / \mathrm{s}^{2}}=7.7 \times 10^{6} \mathrm{~kg} $$

Three point masses are placed on the \(x\) -axis: \(200 \mathrm{~g}\) at \(x=0,500 \mathrm{~g}\) at \(x=30 \mathrm{~cm}\), and \(400 \mathrm{~g}\) at \(x=70 \mathrm{~cm} .\) Find their center of mass. We can make the calculation with respect to any point, but since all the data is measured from the \(x=0\) origin, that point will do nicely. $$ x_{\mathrm{cm}}=\frac{\sum x_{i} m_{i}}{\sum m_{i}}=\frac{(0)(0.20 \mathrm{~kg})+(0.30 \mathrm{~m})(0.50 \mathrm{~kg})+(0.70 \mathrm{~m})(0.40 \mathrm{~kg})}{(0.20+0.50+0.40) \mathrm{kg}}=0.39 \mathrm{~m} $$ The center of mass is located at a distance of \(0.39 \mathrm{~m}\), in the positive \(x\) -direction, from the origin. The \(y\) - and \(z\) -coordinates of the center of mass are zero.
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