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A volleyball is traveling toward you. (a) Which action will require a greater force on the volleyball, your catching the ball or your hitting the ball back? Why? (b) A 0.45 -kg volleyball travels with a horizontal velocity of \(4.0 \mathrm{~m} / \mathrm{s}\) over the net. You jump up and hit the ball back with a horizontal velocity of \(7.0 \mathrm{~m} / \mathrm{s}\). If the contact time is \(0.040 \mathrm{~s}\), what was the average force on the ball?

Short Answer

Expert verified
(a) Hitting requires more force. (b) The average force is -123.75 N.

Step by step solution

01

Understanding Impulse

Impulse is a concept that relates force and time. It's also equal to the change in momentum of an object. The formula is: \( \text{Impulse} = \text{Force} \times \text{Time} = \Delta \text{Momentum} \). This helps us understand why hitting the ball back requires greater force than just stopping it; you are changing both the speed and direction of the ball, which requires a greater impulse.
02

Comparison of Actions (Catching vs Hitting)

When you catch the ball, you bring it to rest, which means you only change its velocity from 4 m/s to 0 m/s. When hitting the ball back, you not only need to stop it (like catching) but also impart an additional speed in the opposite direction (from 0 m/s to 7 m/s). Thus, the change in velocity is greater when hitting.
03

Calculating Change in Velocity

For hitting the ball back, the change in velocity \( \Delta v \) is computed as: \( \Delta v = v_{\text{final}} - v_{\text{initial}} = -7.0 \text{ m/s} - 4.0 \text{ m/s} = -11.0 \text{ m/s} \). The negative sign indicates a reversal of direction.
04

Calculate Change in Momentum

Momentum change \( \Delta p \) for hitting is given by \( \Delta p = m \Delta v = 0.45 \text{ kg} \times (-11.0 \text{ m/s}) = -4.95 \text{ kg m/s} \). For catching, \( \Delta v = -4 \text{ m/s} \), so \( \Delta p = 0.45 \times (-4) = -1.8 \text{ kg m/s} \).
05

Calculate Average Force for Hitting

Using the impulse-momentum theorem, the average force \( F \) is: \( F = \frac{\Delta p}{\Delta t} = \frac{-4.95 \text{ kg m/s}}{0.040 \text{ s}} = -123.75 \text{ N} \). This indicates the force exerted on the ball when it is hit back.

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Key Concepts

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

Force Calculation
Calculating the force involved in interactions with objects, such as in volleyball, is crucial for understanding motion dynamics. Force is essentially what causes an object to accelerate, or in simpler terms, start, stop, or change direction. When playing volleyball, we can either catch the ball or hit it back, each requiring different forces.

  • Catching the ball: This action requires you to stop the ball, meaning you bring its velocity to zero. The force needed is associated only with decelerating the ball.
  • Hitting the ball back: Here, you must first stop the ball and then give it additional speed in the opposite direction. Therefore, this action requires a greater force because it involves a greater change in velocity.
Thus, hitting the ball back requires exerting a larger force because it involves more significant changes in the ball's momentum and direction. The force applied is not a fixed value but instead one that depends on the mass of the volleyball and the velocity change it undergoes.
Change in Velocity
Change in velocity is a key element when understanding motion and interactions. It is the difference between the final and initial velocities and is a vector quantity, which means it doesn't just have magnitude but also direction.

In the given volleyball scenario, the change in velocity can be expressed in two stages:
  • When catching, you reduce the ball's velocity from 4 m/s to 0 m/s. The change is simple, amounting to \[ \Delta v_{ ext{catching}} = 0 ext{ m/s} - 4 ext{ m/s} = -4 ext{ m/s}. \]
  • When hitting, you first bring the ball to a stop and then accelerate it in the opposite direction, increasing its speed to 7 m/s. This results in a much larger velocity change:\[ \Delta v_{ ext{hitting}} = -7 ext{ m/s} - 4 ext{ m/s} = -11 ext{ m/s}. \]
The negative sign highlights the reversal of direction, signifying a larger force requirement due to the more substantial change. Understanding this change helps explain why hitting the ball back has a more intense impact.
Impulse-Momentum Theorem
The impulse-momentum theorem is a fundamental concept in physics relating the change in an object's momentum to the impulse applied to it. The theorem is expressed by the equation:\[ \text{Impulse} = \text{Change in Momentum} \]or more formally:\[ F \times \Delta t = \Delta p \]where \(F\) is the force applied, \(\Delta t\) is the time duration over which the force acts, and \(\Delta p\) is the change in momentum. This relationship is crucial for calculating situations like that in the volleyball exercise.

  • Impulse: The duration and strength of your force determine how much you change the ball's momentum. In practice, a short, powerful hit alters the ball's motion more than a gentle stop.
  • Change in Momentum: Calculated as mass times change in velocity, it measures how much and in which direction an object's motion alters due to impulse.
  • In Volleyball: The formula allows us to determine the average force exerted when hitting the volleyball back. With a mass of 0.45 kg and a velocity change of -11 m/s in 0.040 s, the average force becomes \[ F = \frac{-4.95 \text{ kg m/s}}{0.040 \text{ s}} = -123.75 \text{ N}. \]
The impulse-momentum theorem clearly illustrates why a greater force is necessary to significantly change the ball's direction and speed, showcasing how physics governs real-world actions.

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