Problem 16 - College Physics

Question

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.

Step-by-Step Solution

Answer

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

Step 1: 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.\[95.0 \text{ mi/h} = 95.0 \times \frac{1609.34 \text{ m}}{1 \text{ mi}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 42.47 \text{ m/s}\]

Step 2: 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.

Step 3: 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.

Key Concepts

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.