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A major league catcher catches a fastball moving at \(95.0 \mathrm{mi} / \mathrm{h}\) and his hand and glove recoil \(10.0 \mathrm{~cm}\) in bringing the ball to rest. If it took 0.00470 s to bring the ball (with a mass of \(250 \mathrm{~g}\) ) to rest in the glove, (a) what are the magnitude and direction of the change in momentum of the ball? (b) Find the average force the ball exerts on the hand and glove.

Short Answer

Expert verified
(a) Change in momentum: 10.6175 kg m/s backward; (b) Average force: 2259.04 N backward.

Step by step solution

01

Convert Units

First, convert the velocity of the fastball from miles per hour to meters per second, since we'll be working in SI units. Use the conversion factors: 1 mile = 1609.34 meters and 1 hour = 3600 seconds.\[95.0 \text{ mi/h} = 95.0 \times \frac{1609.34 \text{ m}}{1 \text{ mi}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 42.47 \text{ m/s}\]
02

Calculate the Change in Momentum

Momentum is given by the product of mass and velocity. The change in momentum (\(\Delta p\)) for the ball can be calculated using the initial (\(v_i\)) and final velocities (\(v_f = 0\)):\[\Delta p = m(v_f - v_i)\]The mass of the ball is 250 g = 0.25 kg, and the initial velocity is 42.47 m/s.\[\Delta p = 0.25 \times (0 - 42.47) = -10.6175 \text{ kg m/s}\]The negative sign indicates that the direction of the change in momentum is opposite to the initial velocity.
03

Calculate the Average Force

The average force exerted on the hand and glove can be determined using the impulse-momentum theorem, which relates the change in momentum to the product of average force (\(F_{avg}\)) and time (\(\Delta t\)):\[\Delta p = F_{avg} \times \Delta t\]Solving for \(F_{avg}\):\[F_{avg} = \frac{\Delta p}{\Delta t} = \frac{-10.6175}{0.00470} \approx -2259.04 \text{ N}\]The negative sign indicates the direction of the force is opposite to the ball's initial direction.

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Key Concepts

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

Momentum
Momentum is a measure of the motion of an object and is calculated as the product of its mass and velocity. It is a vector quantity, meaning it has both magnitude and direction. The formula for calculating momentum is:
  • \( p = mv \)
where:
  • \( p \) is momentum,
  • \( m \) is the mass of the object, and
  • \( v \) is its velocity.
When we look at changes in momentum, especially when an object comes to a stop or changes direction, it's important to note both the original direction and speed. In our example, a fastball initially traveling at 42.47 m/s was brought to rest, illustrating a complete reversal in its momentum. The negative sign during such calculations (as was used in \( \Delta p = -10.6175 \, \text{kg m/s} \)) reflects a change in direction opposite to the starting direction.
Impulse-Momentum Theorem
The impulse-momentum theorem is a fundamental concept connecting the impulse applied to an object to its change in momentum. Impulse is essentially the product of force and the time duration over which it acts:
  • \( J = F \times \Delta t \)
Here, \( J \) represents impulse, \( F \) is the force applied, and \( \Delta t \) is the time duration. The theorem is expressed as:
  • \( \Delta p = J \)
This relationship helps us understand how forces act over intervals to alter an object's motion. In practice, this means that the force needed to stop or change the motion of an object, like our example baseball, over a short time span is larger than if the force applied was drawn out over a longer period. Calculating \( \Delta p \) from the known factors, which in this case included converting units and understanding initial velocity, helps show exactly how the ball was brought to a stop within the 0.00470 seconds.
Force Calculation
Force calculation in motion problems often uses the impulse-momentum theorem as a tool to connect change in momentum to the average force applied over a time period. As derived from the impulse-momentum relationship, the formula for the average force is given by:
  • \( F_{avg} = \frac{\Delta p}{\Delta t} \)
Where:
  • \( F_{avg} \) is the average force,
  • \( \Delta p \) is the change in momentum, and
  • \( \Delta t \) is the time period during which the force acts.
In the provided problem, the average force exerted on the catcher's hand and glove as the ball comes to rest was calculated at -2259.04 N. The negative sign here signifies that this force acts in the opposite direction to the initial motion of the ball, indicative of slowing it down and stopping it in this case. Understanding these signs and what they suggest about the directions involved is crucial in physics calculations dealing with forces and motion.

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Most popular questions from this chapter

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?

A \(4.0-\mathrm{kg}\) ball with a velocity of \(4.0 \mathrm{~m} / \mathrm{s}\) in the \(+x\) -direction collides head-on elastically with a stationary \(2.0-\mathrm{kg}\) ball. What are the velocities of the balls after the collision?

At a basketball game, a 120 -lb cheerleader is tossed vertically upward with a speed of \(4.50 \mathrm{~m} / \mathrm{s}\) by a male cheerleader. (a) What is the cheerleader's change in momentum from the time she is released to just before being caught if she is caught at the height at which she was released? (b) Would there be any difference if she were caught \(0.30 \mathrm{~m}\) below the point of release? If so, what is the change then?

In football practice, two wide receivers run different pass receiving patterns. One with a mass of \(80.0 \mathrm{~kg}\) runs at \(45^{\circ}\) northeast at a speed of \(5.00 \mathrm{~m} / \mathrm{s}\). The second receiver (mass of \(90.0 \mathrm{~kg}\) ) runs straight down the field (due east) at \(6.00 \mathrm{~m} / \mathrm{s}\). (a) What is the direction of their total momentum: (1) exactly northeast, (2) to the north of northeast, (3) exactly east, or (4) to the east of northeast? (b) Justify your answer in part (a) by actually computing their total momentum.

A \(170-\mathrm{g}\) hockey puck sliding on ice perpendicularly impacts a flat piece of sideboard. Its incoming momentum is \(6.10 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\). It rebounds along its incoming path after having suffered a momentum change (magnitude) of \(8.80 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\). (a) If the impact with the board took \(35.0 \mathrm{~ms}\), determine the average force (including direction) exerted by the puck on the board. (b) Determine the final momentum of the puck. (c) Was this collision elastic or inelastic? Prove your answer mathematically.

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