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Force of a Baseball Swing. A baseball has mass 0.145 \(\mathrm{kg}\) . (a) If the velocity of a pitched ball has a magnitude of 45.0 \(\mathrm{m} / \mathrm{s}\) and the batted ball's velocity is 55.0 \(\mathrm{m} / \mathrm{s}\) in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat. (b) If the ball remains in contact with the bat for 2.00 \(\mathrm{ms}\) , find the magnitude of the average force applied by the bat.

Short Answer

Expert verified
Change in momentum and impulse: 14.5 kg·m/s, Average force: 7250 N.

Step by step solution

01

Identify the Initial and Final Velocities

The initial velocity of the baseball before it is hit by the bat is given as \( v_i = 45.0 \, \mathrm{m/s} \) and after being hit, its velocity in the opposite direction is \( v_f = -55.0 \, \mathrm{m/s} \). The negative sign indicates the change in direction.
02

Calculate the Change in Momentum

The change in momentum \( \Delta p \) is given by the formula \( \Delta p = m(v_f - v_i) \), where \( m \) is the mass of the ball. Substituting the known values: \( \Delta p = 0.145 \, \mathrm{kg} \times (-55.0 \, \mathrm{m/s} - 45.0 \, \mathrm{m/s}) = 0.145 \, \mathrm{kg} \times (-100.0 \, \mathrm{m/s}) = -14.5 \, \mathrm{kg \cdot m/s} \). The magnitude of the change in momentum is \( 14.5 \, \mathrm{kg \cdot m/s} \).
03

Establish the Relationship Between Impulse and Change in Momentum

The impulse \( I \) applied to the baseball is equal to the change in momentum. Thus, \( I = \Delta p \). Therefore, the magnitude of the impulse is also \( 14.5 \, \mathrm{kg \cdot m/s} \).
04

Calculate the Time of Contact in Seconds

The time of contact, given as 2.00 \( \mathrm{ms} \), needs to be converted to seconds: \( 2.00 \, \mathrm{ms} = 0.002 \, \mathrm{s} \).
05

Calculate the Average Force Applied by the Bat

The average force \( F \) can be calculated using the formula \( F = \frac{I}{\Delta t} \), where \( I \) is the impulse and \( \Delta t \) is the time of contact. Substituting the known values: \( F = \frac{14.5 \, \mathrm{kg \cdot m/s}}{0.002 \, \mathrm{s}} = 7250 \, \mathrm{N} \). The magnitude of the average force applied by the bat is \( 7250 \, \mathrm{N} \).

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Key Concepts

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

Impulse
When a bat hits a baseball, it exerts a certain amount of force over a specific period. This action is referred to as an impulse. Impulse is an important concept in physics because it directly relates to the change in an object's momentum. The impulse exerted on an object can be calculated using the formula:
  • Impule (\( I \)) = Force (\( F \)) \( \times \) Time (\( \Delta t \))
In our baseball scenario, the impulse applied by the bat is also equal to the change in momentum of the ball. Since the momentum has been roundly calculated as 14.5 \( \mathrm{kg \cdot m/s} \), the impulse is also 14.5 \( \mathrm{kg \cdot m/s} \). This tells us how much the velocity of the baseball has been altered by interacting with the bat. Understanding impulse can help us predict or calculate how objects will respond to forces over time. This is valuable in any situation involving impacts or sudden force changes.
Average Force
The concept of average force helps us understand the total impact of a force applied over time. It's particularly useful when the force varies or is applied quickly, like when a bat connects with a baseball. To calculate average force, we use the impulse-momentum theorem. This theorem states that the change in momentum of an object is equal to the impulse applied to it. Knowing this, we can calculate the average force using:
  • Average Force (\( F \)) = Impulse (\( I \)) / Time (\( \Delta t \))
In the exercise, the impulse was found to be 14.5 \( \mathrm{kg \cdot m/s} \) and the provided time of contact is 0.002 seconds. Plugging in these values gives us:
  • \( F = \frac{14.5 \, \mathrm{kg \cdot m/s}}{0.002 \, \mathrm{s}} = 7250 \, \mathrm{N} \)
This means that during the brief moment of contact, the bat exerted an average force of 7250 Newtons on the ball. Understanding average force gives insight into how much power was needed to change the ball's momentum.
Change in Momentum
Momentum, a fundamental concept in physics, represents the quantity of motion an object has. It depends on both the mass and velocity of the object. When a moving object encounters a force, its momentum can change. This change is crucial for understanding how the object's motion is affected.In our baseball scenario, the change in momentum is calculated by finding the difference between the final and initial momentum of the baseball. The formula for change in momentum (\( \Delta p \)) is:
  • \( \Delta p = m(v_f - v_i) \)
Where:
  • \( m \) is the mass of the baseball (0.145 kg)
  • \( v_f \) is the final velocity (-55.0 m/s, opposite direction)
  • \( v_i \) is the initial velocity (45.0 m/s)
Substituting the given numbers:
  • \( \Delta p = 0.145 \times (-55.0 - 45.0) = -14.5 \, \mathrm{kg \cdot m/s} \)
The negative sign indicates a direction change, but for magnitude, we ignore the sign and state the change as 14.5 \( \mathrm{kg \cdot m/s} \). Recognizing how momentum changes helps us decode the effects of forces in motion.

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Most popular questions from this chapter

A \(68.5-k g\) astronaut is doing a repair in space on the orbit ing space station. She throws a 2.25 -kg tool away from her at 3.20 \(\mathrm{m} / \mathrm{s}\) relative to the space station. With what speed and in what direction will she begin to move?

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Automobile Accident Analysis. You are called as an expert witness to analyze the following auto accident: Car \(B,\) of mass \(1900 \mathrm{kg},\) was stopped at a red light when it was hit from behind by car \(A,\) of mass 1500 \(\mathrm{kg}\) . The cars locked bumpers during the collision and slid to a stop with brakes locked on all wheels. Measurements of the skid marks left by the tires showed them to be 7.15 \(\mathrm{m}\) long. The coefficient of kinetic friction between the tires and the road was 0.65 . (a) What was the speed of car A just before the collision? (b) If the speed limit was 35 \(\mathrm{mph}\) , was car \(A\) speeding, and if so, by how many miles per hour was it exceeding the speed limit?

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