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A soccer player kicks a soccer ball of mass \(0.45 \mathrm{~kg}\) that is initially at rest. The foot of the player is in contact with the ball for \(3.0 \times 10^{-3} \mathrm{~s}\), and the force of the kick is given by $$ F(t)=\left[\left(6.0 \times 10^{6}\right) t-\left(2.0 \times 10^{9}\right) t^{2}\right] \mathrm{N}$$ for \(0 \leq t \leq 3.0 \times 10^{-3} \mathrm{~s}\), where \(t\) is in seconds. Find the magnitudes of (a) the impulse on the ball due to the kick, (b) the average force on the ball from the player's foot during the period of contact, (c) the maximum force on the ball from the player's foot during the period of contact, and (d) the ball's velocity immediately after it loses contact with the player's foot.

Step by step solution

01

Define Impulse

Impulse is the integral of force over the time interval the force is applied. It can be expressed as \( I = \int_{0}^{3.0\times10^{-3}} F(t) \, dt \).

02

Setup the Integral for Impulse

Substitute the force equation \( F(t) = \left(6.0 \times 10^{6}\right) t - \left(2.0 \times 10^{9}\right) t^2 \) into the integral: \[ I = \int_{0}^{3.0\times10^{-3}} \left( (6.0 \times 10^6)t - (2.0 \times 10^9)t^2 \right) \, dt \].

03

Solve the Integral for Impulse

Calculate the integral: \[ I = \left[ \frac{6.0 \times 10^6}{2} t^2 - \frac{2.0 \times 10^9}{3} t^3 \right]_0^{3.0\times10^{-3}} \= \frac{6.0 \times 10^6}{2} (3.0\times10^{-3})^2 - \frac{2.0 \times 10^9}{3} (3.0\times10^{-3})^3. \\approx 27.0 \mathrm{~Ns} \].

04

Calculate Average Force

The average force is the total impulse divided by the time interval: \( F_{\text{avg}} = \frac{I}{\Delta t} = \frac{27.0}{3.0\times10^{-3}} = 9.0 \times 10^3 \mathrm{~N} \).

05

Find Maximum Force

Differentiate the force function \( F(t) \) with respect to \( t \) to find the maximum: \( \frac{dF}{dt} = 6.0 \times 10^6 - 4.0 \times 10^9 t \). Set \( \frac{dF}{dt} = 0 \), solve for \( t \): \( t = \frac{6.0 \times 10^6}{4.0 \times 10^9} = 1.5 \times 10^{-3} \mathrm{~s} \). Substituting \( t \) into the original function gives \( F_{\text{max}} = \left(6.0 \times 10^6\right)(1.5 \times 10^{-3}) - \left(2.0 \times 10^9\right)(1.5 \times 10^{-3})^2 = 4.5 \times 10^3 \mathrm{~N} \).

06

Calculate Ball's Velocity

Using the impulse-momentum theorem, the change in momentum equals the impulse: \( I = m \Delta v = 0.45 \times v \). Solve for \( v \): \( v = \frac{I}{m} = \frac{27.0}{0.45} \approx 60 \mathrm{~m/s} \).

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Key Concepts

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

Impulse calculation

Impulse is a crucial concept to understand when studying force and motion. It is essentially the product of force applied over a time interval, often expressed as the integral of a force function with respect to time. In this exercise, the impulse given to the soccer ball can be calculated using the force equation provided:

• Given: \( F(t) = \left(6.0 \times 10^6\right) t - \left(2.0 \times 10^9\right) t^2 \), where \( t \) is time in seconds.
• Time interval: \( 0 \leq t \leq 3.0 \times 10^{-3} \, \mathrm{s} \).
• Impulse, \( I = \int_{0}^{3.0\times10^{-3}} F(t) \, dt \).

By integrating this function, we can find the impulse, which in this case is approximately 27.0 Ns. This signifies the change in momentum experienced by the ball due to the kick, thereby quantifying the effect of the force applied over time.

Average force

Average force is a simplified way to represent the effect of a force applied over a time period. It is useful when dealing with varying force, where exact measurement at every point isn't feasible.
The average force \( F_{\text{avg}} \) during the contact interval is determined by dividing the total impulse by the duration of contact:

• Total impulse: 27.0 Ns (from the impulse calculation).
• Time interval: \( \Delta t = 3.0 \times 10^{-3} \, \mathrm{s} \).

Let's calculate it:
\( F_{\text{avg}} = \frac{I}{\Delta t} = \frac{27.0}{3.0 \times 10^{-3}} = 9.0 \times 10^3 \, \mathrm{N} \).
This means the average force exerted by the player’s foot on the ball is 9000 N, presenting a simplified view of the entire force application.

Maximum force

Maximum force is the highest value of force experienced during the period the player's foot is in contact with the ball. Finding this value can involve analyzing the force function for its peak by differentiating with respect to time.
The function given is \( F(t) = \left(6.0 \times 10^6\right) t - \left(2.0 \times 10^9\right) t^2 \). To find its maximum:

• Differentiate the force function: \( \frac{dF}{dt} = 6.0 \times 10^6 - 4.0 \times 10^9 t \).
• Set \( \frac{dF}{dt} = 0 \) to find critical points: \( t = \frac{6.0 \times 10^6}{4.0 \times 10^9} = 1.5 \times 10^{-3} \, \mathrm{s} \).
• Substitute \( t \) back into \( F(t) \) to calculate \( F_{\text{max}}: \).
• \( F_{\text{max}} = \left(6.0 \times 10^6\right)(1.5 \times 10^{-3}) - \left(2.0 \times 10^9\right)(1.5 \times 10^{-3})^2 = 4.5 \times 10^3 \, \mathrm{N} \).

Thus, the maximum force on the ball is determined as 4500 N, showing how force peaks during the shot.

Velocity calculation

The velocity of the soccer ball right after being kicked is directly related to the impulse experienced. The impulse-momentum theorem links these concepts, stating that impulse equals the change in momentum.
Based on the problem, the formula used is:

• Impulse \( I = m \Delta v \), where \( m \) is the mass of the ball and \( \Delta v \) is the change in velocity.
• Given mass of the ball: \( 0.45 \, \mathrm{kg} \).
• Impulse from earlier is 27.0 Ns.

We solve for \( v \) as follows:
\( v = \frac{I}{m} = \frac{27.0}{0.45} \approx 60 \, \mathrm{m/s} \).
This velocity represents the speed and direction the ball travels immediately after contact, highlighting the power of the kick.