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Chapter 6: Problem 3

A pitcher claims he can throw a \(0.145\) -kg baseball with as much momentum as a \(3.00-\mathrm{g}\) bullet moving with a speed of \(1.50 \times 10^{3} \mathrm{~m} / \mathrm{s}\). (a) What must the baseball's speed be if the pitcher's claim is valid? (b) Which has greater kinetic energy, the ball or the bullet?

Short Answer

Expert verified
To validate the pitcher's claim, the baseball's speed must be approximately \(31.03\, m/s\). The bullet has the greater kinetic energy, at \(3375\, J\), compared to the baseball's \(69.79\, J\).

Step by step solution

01

Calculate the momentum of the bullet

Given that the bullet has a mass of \(3.00\, g\), we convert this mass to kilograms by dividing by \(1000\). Thus, the mass of the bullet is \(0.003\, kg\). The speed of the bullet is given as \(1.50 \times 10^{3}\, m/s\). We compute the momentum of the bullet using the formula \(P = m * v\), where \(m = 0.003\, kg\), and \(v = 1.50 \times 10^{3}\, m/s\). Therefore, the momentum of the bullet is \(P = 0.003\, kg * 1.50 \times 10^{3}\, m/s = 4.5\, kg\, m/s\).
02

Calculate the speed of the baseball

The mass of the baseball is given as \(0.145\, kg\), and its momentum is the same as the bullet's momentum (from the claim of the pitcher). Therefore, to get the speed (velocity) of the baseball, we rearrange the momentum formula as follows: \(v = P / m\). So, the speed of the baseball is \(v = 4.5\, kg\, m/s / 0.145\, kg = 31.03\, m/s\).
03

Calculate the kinetic energy of the baseball and bullet

Next, we find the kinetic energy of both the baseball and the bullet using the formula for kinetic energy \(K.E = 1/2 * m * v^2\). For the baseball, we have \(K.E = 1/2 * 0.145\, kg * (31.03\, m/s)^2 = 69.79\, J\). For the bullet, we have \(K.E = 1/2 * 0.003\, kg * (1.50 \times 10^{3}\, m/s)^2 = 3375\, J\). Both values are rounded to two decimal places.
04

Compare the kinetic energy of the baseball and the bullet

When we compare \(69.79\, J\) for the baseball to \(3375\, J\) for the bullet, it is clear that the bullet has the greater kinetic energy.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy of motion. Anything that moves has kinetic energy. It depends on two factors: mass and velocity. The formula to calculate kinetic energy is:
  • \( K.E = \frac{1}{2} * m * v^2 \)
Here, \( m \) is the mass in kilograms and \( v \) is the velocity in meters per second. This formula shows that kinetic energy is proportional to the mass and the square of velocity. In simpler terms, if the speed doubles, the kinetic energy increases by four times. This is why even a small, fast-moving object like a bullet can have significant kinetic energy compared to a heavier, slower-moving object like a baseball.
Mass and Velocity Relationship
The relationship between mass and velocity is crucial when discussing momentum. Momentum is the product of an object's mass and its velocity:
  • \( P = m * v \)
Both higher mass and higher velocity can increase momentum. But if two objects have the same momentum, a heavier object moves slower than a lighter one. In our exercise, the pitcher claims the baseball has the same momentum as the bullet. So, even though the baseball is heavier, it must move slower to have equal momentum:
  • The bullet's mass: \(0.003\, kg\) and speed: \(1.50 \times 10^{3}\, m/s\)
  • The baseball鈥檚 mass: \(0.145\, kg\) and calculated speed: \(31.03\, m/s\)
This showcases how an increase in mass can be balanced by a decrease in velocity to maintain the same momentum.
Unit Conversion
Unit conversion is often necessary in physics problems to ensure consistency of units throughout calculations. For example, converting grams to kilograms or meters to kilometers simplifies calculations by aligning with the International System of Units (SI). In this exercise:
  • Bullet mass: converted from grams to kilograms by dividing by \(1000\).
This conversion is essential since the standard unit for mass in momentum and kinetic energy calculations is kilograms. Without proper unit conversion, inaccurate results can arise. Always check the units before starting any calculations to set a solid foundation for reliable outcomes.

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Most popular questions from this chapter

A pitcher throws a \(0.15\) -kg baseball so that it crosses home plate horizontally with a speed of \(20 \mathrm{~m} / \mathrm{s}\). The ball is hit straight back at the pitcher with a final speed of \(22 \mathrm{~m} / \mathrm{s}\). (a) What is the impulse delivered to the ball? (b) Find the average force exerted by the bat on the ball if the two are in contact for \(2.0 \times 10^{-3} \mathrm{~s}\).

A \(1200-\mathrm{kg}\) car traveling initially with a speed of \(25.0 \mathrm{~m} / \mathrm{s}\) in an easterly direction crashes into the rear end of a 9000 -kg truck moving in the same direction at \(20.0 \mathrm{~m} / \mathrm{s}\) (Fig. \(\mathrm{P} 6.42\) ). The velocity of the car right after the collision is \(18.0 \mathrm{~m} / \mathrm{s}\) to the east. (a) What is the velocity of the truck right after the collision? (b) How much mechanical energy is lost in the collision? Account for this loss in energy.

A baseball player of mass \(84.0 \mathrm{~kg}\) running at \(6.70 \mathrm{~m} / \mathrm{s}\) slides into home plate. (a) What magnitude impulse is delivered to the player by friction? (b) If the slide lasts \(0.750 \mathrm{~s}\), what average friction force is exerted on the player?

An object has a kinetic energy of \(275 \mathrm{~J}\) and a momentum of magnitude \(25.0 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\). Find the speed and mass of the object.

ecp An astronaut in her space suit has a total mass of \(87.0 \mathrm{~kg}\), including suit and oxygen tank. Her tether line loses its attachment to her spacecraft while she's on a spacewalk. Initially at rest with respect to her spacecraft, she throws her \(12.0\) -kg oxygen tank away from her spacecraft with a speed of \(8.00 \mathrm{~m} / \mathrm{s}\) to propel herself back toward it (Fig. P6.25). (a) Determine the maximum distance she can be from the craft and still return within \(2.00 \mathrm{~min}\) (the amount of time the air in her helmet remains breathable). (b) Explain in terms of Newton's laws of motion why this strategy works.
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