Problem 5 - College Physics

Question

The speed of the fastest-pitched baseball was \(45 \mathrm{~m} / \mathrm{s},\) and the ball's mass was \(145 \mathrm{~g}\). (a) What was the magnitude of the momentum of this ball, and how many joules of kinetic energy did it have? (b) How fast would a 57 gram ball have to travel to have the same amount of (i) kinetic energy, and (ii) momentum?

Step-by-Step Solution

Answer

Momentum: 6.525 kg·m/s; Kinetic Energy: 146.81 J. 57g ball needs 71.63 m/s for same KE and 114.47 m/s for same momentum.

Step 1: Calculate the Momentum of the Baseball

Momentum ( p ) is calculated by the formula \( p = m \times v \), where \( m \) is mass, and \( v \) is velocity. First, convert the mass of the baseball from grams to kilograms: \( 145 \text{ g} = 0.145 \text{ kg} \). Then, substitute the values into the formula: \( p = 0.145 \text{ kg} \times 45 \text{ m/s} = 6.525 \text{ kg} \cdot \text{m/s}\).

Step 2: Calculate the Kinetic Energy of the Baseball

Kinetic energy ( KE ) is given by the formula \( KE = \frac{1}{2} m v^2 \). Using the mass in kilograms, substitute into the formula: \( KE = \frac{1}{2} \times 0.145 \text{ kg} \times (45 \text{ m/s})^2 = 146.8125 \text{ J} \).

Step 3: Find Velocity for Equal Kinetic Energy with a 57g Ball

To find the velocity required for the 57g ball to have the same kinetic energy, use \( KE = \frac{1}{2} m v^2 \) and the KE found previously: \( 146.8125 = \frac{1}{2} \times 0.057 \times v^2 \). Solve for \( v \): \( v^2 = \frac{2 \times 146.8125}{0.057} \) leads to \( v \approx 71.63 \text{ m/s} \).

Step 4: Find Velocity for Equal Momentum with a 57g Ball

To find the velocity for the 57g ball to have the same momentum, use the formula \( p = m v \): \( 6.525 = 0.057 v \). Solve for \( v \): \( v = \frac{6.525}{0.057} \approx 114.47 \text{ m/s} \).

Key Concepts

Momentum Calculation

Momentum is often described as "mass in motion" and is a fundamental quantity in physics. It helps us understand how objects move and interact with each other. In this exercise, we're interested in calculating the momentum of baseballs. The formula for momentum is given as \( p = m \times v \), where \( m \) is the mass and \( v \) is the velocity of the object.
Let's look at the solution provided: the baseball has a mass of 145 grams, which is equivalent to 0.145 kilograms, and a velocity of 45 meters per second. Substituting these into the formula gives \( p = 0.145 \times 45 = 6.525 \) kg·m/s. This result tells us how much "oomph" the baseball carries while in motion.

• To calculate momentum, always ensure the unit of mass is in kilograms and the velocity in meters per second.
• Momentum has both direction and magnitude, just like velocity, making it a vector quantity.
• This principle of momentum is applicable across various domains, from sports to vehicular transport.

Kinetic Energy Calculation

Kinetic energy represents the energy of motion. It's a scalar quantity which means it only has magnitude and no direction. The formula to determine kinetic energy is \( KE = \frac{1}{2} m v^2 \). In our exercise, to find the kinetic energy of the baseball, we first ensure the mass is in kilograms.
The provided solution uses \( m = 0.145 \) kg and \( v = 45 \) m/s. Plugging these values into the kinetic energy formula we get \( KE = \frac{1}{2} \times 0.145 \times 45^2 \), which results in 146.8125 Joules. This energy is what allows the baseball to perform work when it hits something, such as a bat or a glove.

• While momentum is about mass and velocity, kinetic energy focuses on how movement translates to energy capacity.
• Kinetic energy depends on the square of the velocity, so small changes in speed significantly impact energy.
• Understanding both momentum and kinetic energy is crucial for solving physics problems accurately.

Physics Problem Solving

Solving physics problems involves a methodical approach of identifying what is known and what is required to find out. This specific problem explores both momentum and kinetic energy. The final parts of the exercise invite us to find the condition under which a lighter ball possesses the same kinetic energy and momentum as a heavier one.
For equal kinetic energy, the lighter 57g ball necessitates using the kinetic energy formula and solving for velocity. The steps are outlined as converting grams to kilograms, substituting \( KE = 146.8125 \) with \( m = 0.057 \) kg, resulting in a velocity approximately \( v = 71.63 \) m/s.
Similarly, for equal momentum, using \( p = m v \), we substitute \( p = 6.525 \) and solve for \( v \) with \( m = 0.057 \) kg, finding it requires approximately \( 114.47 \) m/s.

• Grasping the concept of converting units accurately is crucial for correct calculations.
• Each step of the formula should be followed carefully to avoid mistakes.
• Physics problem-solving enhances not only analytical skills but also assists in making connections between concepts like energy and momentum.