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Chapter 7: Problem 9

A \(150 \mathrm{~g}\) baseball pitched at a speed of \(40 \mathrm{~m} / \mathrm{s}\) is hit straight back to the pitcher at a speed of \(60 \mathrm{~m} / \mathrm{s}\). What is the magnitude of the average force on the ball from the bat if the bat is in contact with the ball for \(5.0 \mathrm{~ms}\) ?

Short Answer

Expert verified
The magnitude of the average force is 3000 N.

Step by step solution

01

- Convert units

Convert the mass of the baseball from grams to kilograms. Given mass is 150 g. So, \[150 \text{ g} = 0.150 \text{ kg}\]. Convert the contact time from milliseconds to seconds: \[5.0 \text{ ms} = 0.005 \text{ s}\].
02

- Calculate initial and final velocities

Let the initial velocity (before being hit) be \( u = 40 \text{ m/s} \), and the final velocity (after being hit) be \( v = -60 \text{ m/s} \). Note that the final velocity is negative because the ball is hit back in the opposite direction.
03

- Calculate change in velocity

Since the final velocity \( v \) is \(-60 \text{ m/s} \) and the initial velocity \( u \) is \( 40 \text{ m/s} \), the change in velocity \( \Delta v \) is: \[ \Delta v = v - u = (-60) - 40 = -100 \text{ m/s} \].
04

- Calculate acceleration

Acceleration \( a \) is given by the formula: \[ a = \frac{\Delta v}{t} \]. Here, \( \Delta v = -100 \text{ m/s} \) and \( t = 0.005 \text{ s} \). Therefore, \[ a = \frac{-100}{0.005} = -20000 \text{ m/s}^2 \].
05

- Calculate the average force

According to Newton's second law, the average force \( F \) is given by \[ F = m \cdot a \]. Using \( m = 0.150 \text{ kg} \) and \( a = -20000 \text{ m/s}^2 \), we get: \[ F = 0.150 \times (-20000) = -3000 \text{ N} \]. The magnitude of the force is \( 3000 \text{ N} \).

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Key Concepts

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

Newton's second law
Newton's second law is a fundamental principle in physics describing the relationship between an object's mass, its acceleration, and the applied force. The law can be summed up by the formula: \[ F = m \cdot a \] where
  • \( F \) is the force applied, measured in newtons (N).
  • \( m \) is the mass of the object, measured in kilograms (kg).
  • \( a \) is the acceleration of the object, measured in meters per second squared (\( \text{m/s}^2 \)).
In our exercise, we used Newton's second law to find the average force exerted on a baseball hit by a bat. This involved determining the mass of the baseball, calculating its change in velocity due to the bat's impact, and then computing the acceleration from this change in velocity. Finally, we applied Newton's second law to solve for the force.
Unit conversion
Unit conversion is crucial in physics to ensure consistency in calculations and proper application of formulas. In our exercise, we performed two key unit conversions:
  • Mass: The baseball's mass was given as 150 grams. Since standard units for mass in physics are kilograms (kg), we converted 150 grams into kilograms: \[ 150 \text{g} = 0.150 \text{kg} \]
  • Time: The bat was in contact with the ball for 5 milliseconds. The standard unit of time is seconds (s), so we converted milliseconds to seconds: \[ 5 \text{ms} = 0.005 \text{s} \]
Correctly converting these units prepared the values for accurate use in further calculations.
Acceleration calculation
Acceleration is the rate of change of velocity of an object. It is calculated using the formula: \[ a = \frac{\Delta v}{t} \]where
  • \(a\) is the acceleration,
  • \(\Delta v\) is the change in velocity,
  • \(t\) is the time interval over which the change occurs.
In the problem, we calculated the change in velocity of the baseball when it was hit: \[ \Delta v = v - u = -60 \text{ m/s} - 40 \text{ m/s} = -100\text{ m/s} \] Using the given time interval (\( t = 0.005\text{s}\) ), we then determined the acceleration: \[ a = \frac{-100}{0.005} = -20000\text{ m/s}^2. \] This high value is typical of short time intervals involving changes in momentum, like a bat hitting a ball.
Force magnitude
Force magnitude refers to the strength or size of a force exerted on an object, without considering its direction. Even though force has both magnitude and direction, sometimes it's useful to discuss just its magnitude. Using Newton's second law, \[ F = m \cdot a \] we calculated the force magnitude exerted by the bat on the baseball. Given:
  • \( m = 0.150 \text{kg} \)
  • \(a = -20000 \text{ m/s}^2 \)
we found: \[ F = 0.150 \cdot (-20000) = -3000 \text{N}. \] The negative sign indicates direction, but when asked for the magnitude, we take the absolute value: \( 3000 \text{N}. \) This represents how strongly the bat acted on the ball during contact.

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