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A pitcher claims he can throw a \(0.145-\mathrm{kg}\) baseball with as much momentum as a \(3.00-g\) 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
The baseball's speed must be \(6.21 m/s\) if the pitcher's claim is to be valid. The moving bullet has greater kinetic energy than the thrown baseball.

Step by step solution

01

Evaluate momentum of the bullet

Given: mass of bullet (m1) = \(3.00 g = 3.00 \times 10^{-3} kg\) (after converting grams to kilograms) and speed or velocity of bullet (v1) = \(1.50 \times 10^{3} m/s\). Now, find the momentum of the bullet using the formula of momentum: \(p1 = m1 \times v1\).
02

Evaluate speed of the baseball

The momentum of the bullet is equal to the momentum of the baseball since the pitcher claims he can throw the baseball with as much momentum as the bullet. Let's denote the mass of baseball as m2 = \(0.145 kg\) and the speed of baseball as v2 which we are to find. Therefore, \(p1 = p2 = m1 \times v1 = m2 \times v2\). Solve this equation for v2.
03

Evaluate kinetic energy of the bullet

We use the formula for kinetic energy (\(KE = \frac{1}{2} m v^2\)) to calculate the kinetic energy of the bullet: \(KE1 = \frac{1}{2} m1 v1^2\).
04

Evaluate kinetic energy of the baseball

Now, using the speed of the baseball that we found in step 2, we calculate the kinetic energy of the baseball: \(KE2 = \frac{1}{2} m2 v2^2\).
05

Compare kinetic energy

Compare KE1 and KE2 to find which is greater, thus determining which object has greater kinetic energy.

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Key Concepts

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

Momentum Formula
Momentum, often denoted as 'p', is a measure of an object's mass in motion. The momentum formula is given by the equation:
\[\begin{equation}p = mv\text{,}\text{where}\begin{aligned}p\text{ is momentum,}\m\text{ is the mass of the object, and}\v\text{ is the velocity of the object.}\end{aligned}\end{equation}\]
In physics, momentum is a vector quantity, possessing both magnitude and direction. It represents how difficult it is to stop a moving object. The SI unit for momentum is kilogram meters per second (kg·m/s).
In the given exercise, we compare the momentum of a baseball to that of a bullet. By using the momentum formula, we can determine how fast a baseball needs to travel to equal the momentum of a much smaller, yet significantly faster bullet. This concept lays the groundwork for the understanding of impulse and collisions in physics.
Kinetic Energy Calculation
Kinetic energy, represented as 'KE', is the energy an object possesses due to its motion. The kinetic energy calculation is performed using the formula:
\[\begin{equation}KE = \frac{1}{2}mv^2\text{,}\text{where}\begin{aligned}KE\text{ is kinetic energy,}\m\text{ is the mass of the object, and}\v\text{ is the velocity of the object.}\end{aligned}\end{equation}\]
The unit for kinetic energy in the International System of Units (SI) is the joule (J). When we compute the kinetic energy of any object, we observe the quadratic relationship between velocity and kinetic energy. This implies that a small increase in speed can cause a large increase in kinetic energy.
For example, as part of our provided exercise, we calculate both the kinetic energy of a bullet and a baseball, revealing which one has greater kinetic energy if they have the same momentum. This comparison highlights the distinct difference in how mass and velocity influence an object's kinetic energy.
Conservation of Momentum
The law of conservation of momentum states that in a closed system with no external forces, the total momentum before an event is equal to the total momentum after the event. This principle is fundamental in physics and applies to various scenarios such as collisions and explosions.
When two objects interact, their total momentum remains constant if no external forces interfere. The conservation of momentum is mathematically expressed as:
\[\begin{equation}\text{Total momentum before interaction} = \text{Total momentum after interaction},\end{equation}\]
This concept is used in our exercise when assuming the pitcher's claim is true. If the baseball thrown by the pitcher has the same momentum as the bullet, despite their vastly different masses and speeds, this demonstrates the conservation of momentum. Understanding this principle allows us to predict the results of collisions and can even be extended to complex systems in physics, including orbital mechanics and quantum particles.

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Most popular questions from this chapter

A high-speed photograph of a club hitting a golf ball is shown in Figure \(6.3\). The club was in contact with a ball, initially at rest, for about \(0.0020 \mathrm{~s}\). If the ball has a mass of \(55 \mathrm{~g}\) and leaves the head of the club with a speed of \(2.0 \times 10^{2} \mathrm{ft} / \mathrm{s}\), find the average force exerted on the ball by the club.

A billiand ball rolling across a table at \(1.50 \mathrm{~m} / \mathrm{s}\) makes a head-on elastic collision with an identical ball. Find the speed of each ball after the collision (a) when the second ball is initially at rest, (b) when the second ball is mowing toward the first at a speed of \(1.00 \mathrm{~m} / \mathrm{s}\), and (c) when the second ball is moving away from the first at a speed of \(1.00 \mathrm{~m} / \mathrm{s}\).

S Show that the kinetic energy of a particle of mass \(m\) is related to the magnitude of the momentum \(p\) of that particle by \(K E=p^{2} / 2 \mathrm{~m}\). (Note: This expression is invalid for particles traveling at speeds near that of light.)

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 \(9000-\mathrm{kg}\) truck mowing in the same direction at \(20.0 \mathrm{~m} / \mathrm{s}\) (Fig. P6.42). 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-

S This is a symbolic version of Problem 33. A railroad car of mass \(M\) moving at a speed \(v_{1}\) collides and couples with two coupled railroad cars, each of the same mass \(M\) and moving in the same direction at a speed \(t_{2}\). (a) What is the speed \(v_{y}\) of the three coupled cars after the collision in terms of \(v_{1}\) and \(v_{2}\) ? (b) How much kinetic energy is lost in the collision? Answer in terms of \(M\), \(v_{1}\), and \(v_{2}\).

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