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Chapter 8: Problem 32

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 was 1381.25 N.

Step by step solution

01

Understanding the Problem

We need to find the average force exerted on the ball when it is kicked. We know the mass of the ball, the final speed, and the duration of impact. The ball was initially at rest.
02

Applying Newton's Second Law

Newton's Second Law states that the force is equal to the rate of change of momentum. The formula can be expressed as \( F = \frac{\Delta p}{\Delta t} \), where \( \Delta p \) is the change in momentum and \( \Delta t \) is the change in time.
03

Calculate Change in Momentum

Since the ball was initially at rest, the initial momentum is 0. The final momentum, \( p_f \), is given by \( m \cdot v_f \). Here, \( m = 0.425 \, \text{kg} \) and \( v_f = 26 \, \text{m/s} \). Calculate \( \Delta p = p_f - p_i \), where \( p_i = 0 \).
04

Calculate Change in Time

We need \( \Delta t \) in seconds. Since the duration given is in milliseconds, convert it: \( 8.0 \, \text{ms} = 8.0 \, \text{ms} \times 10^{-3} = 0.008 \, \text{s} \).
05

Calculate Average Force

Now, use the formula from Step 2: \( F = \frac{\Delta p}{\Delta t} = \frac{0.425 \, \text{kg} \times 26 \, \text{m/s}}{0.008 \, \text{s}} \). Calculate the value to find the average force.

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Key Concepts

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

Understanding Momentum
Momentum is a measure of how difficult it is to stop a moving object. It is calculated as the product of an object's mass and its velocity.
When a soccer ball is kicked, it goes from being at rest to moving at a certain speed. This change in motion is represented by a change in momentum, given by the formula:
  • Total Momentum = Mass \( \times \) Velocity
  • Initial momentum is zero, as the ball is initially at rest.
  • Final momentum can be calculated using the given mass \( (0.425 \text{ kg}) \) and final speed \( (26 \text{ m/s}) \).
Thus, the momentum after the kick is \( 0.425 \times 26 \), which equals \( 11.05 \text{ kg m/s} \).
To determine how forceful the kick was, we need to explore how quickly this change in momentum happened.
Calculating Average Force
The concept of average force allows us to understand how much force was applied over a specific time period.
Newton's Second Law states, force is the rate of change of momentum. Mathematically, we express it as:
  • \( F = \frac{\Delta p}{\Delta t} \)
      • Where:
      • \Delta p = Change in momentum
      • \Delta t = Change in time
    To find the average force exerted on the ball during the kick:
    • First, compute change in momentum: \( p_f - 0 = 11.05 \text{ kg m/s} \).
    • Next, calculate the time duration of the impact in seconds: \( 8.0 \text{ ms} = 0.008 \text{ s} \)
    • Finally, use these values in the average force equation: \( F = \frac{11.05}{0.008} \)
    • This yields an average force of approximately \( 1381.25 \text{ N} \)
    The average force tells us how strong the kick was during the short contact period.
Examining Impact Duration
Impact duration refers to the brief moment of contact between the player's foot and the ball. The duration is often measured in very small units, like milliseconds (ms).
In this example, the ball and foot were in contact for only \(8.0\) ms, equivalent to \(0.008\) seconds.
This tiny fraction of time is crucial鈥攊t determines how quickly the momentum changed and therefore, how strong the average force was. The shorter the time, the greater the force needed to achieve the same change in momentum.
  • Shorter impact time means greater force for a given change in momentum.
  • Longer impact time reduces the required force for the same momentum change.
Understanding impact duration helps in analyzing quick interactions in sports and other fields where high forces occur over short periods. In this case, the brief impact resulted in a significant force that propelled the ball rapidly forward.

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Most popular questions from this chapter

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 slow 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 from which $$ \begin{array}{c} F \Delta t=m v_{f}-m v_{i}=m(50000 \mathrm{~m} / \mathrm{s})-0 \\ F=(50000 \mathrm{~m} / \mathrm{s}) \frac{\mathrm{m}}{\mathrm{L}} \end{array} $$ 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 $$ F=(50000 \mathrm{~m} / \mathrm{s})(1500 \mathrm{~kg} / \mathrm{s})=75 \mathrm{MN} $$ 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 $$ M_{\text {rocket }}=\frac{\text { weight }}{g}=\frac{75 \times 10^{6} \mathrm{~N}}{9.81 \mathrm{~m} / \mathrm{s}^{2}}=7.7 \times 10^{6} \mathrm{~kg} $$

A 6000 -kg truck traveling north at \(5.0 \mathrm{~m} / \mathrm{s}\) collides with a \(4000-\mathrm{kg}\) truck moving west at \(15 \mathrm{~m} / \mathrm{s}\). If the two trucks remain locked together after impact, with what speed and in what direction do they move immediately after the collision?

A force of \(1000 \mathrm{~N}\) is applied to a small space satellite for a time of \(10.0\) minutes. If the craft has a mass of \(200 \mathrm{~kg}\), what will be its final speed? [Hint: Be careful with those exponents when using a calculator.]

Imagine a space vehicle floating in the void. It fires a small thruster that delivers a forward force of \(2000 \mathrm{~N}\) for \(25.0 \mathrm{~s}\). Determine the resulting change in momentum of the craft. Do you need the mass of the ship?

Four masses are positioned in the \(x y\) -plane as follows: \(300 \mathrm{~g}\) at \((x=\) \(0, y=2.0 \mathrm{~m}), 500 \mathrm{~g}\) at \((-20 \mathrm{~m},-3.0 \mathrm{~m}), 700 \mathrm{~g}\) at \((50 \mathrm{~cm}, 30 \mathrm{~cm})\) and \(900 \mathrm{~g}\) at \((-80 \mathrm{~cm}, 150 \mathrm{~cm})\). Find their center of mass.
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