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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 railroad car of mass \(2.00 \times 10^{4} \mathrm{~kg}\) moving at \(3.00 \mathrm{~m} / \mathrm{s}\) collides and couples with two coupled railroad cars, each of the same mass as the single car and moving in the same direction at \(1.20 \mathrm{~m} / \mathrm{s}\). (a) What is the speed of the three coupled cars after the collision? (b) How much kinetic energy is lost in the collision?

A neutron in a reactor makes an elastic head-on collision with a carbon atom that is initially at rest. (The mass of the carbon nucleus is about 12 times that of the neutron.) (a) What fraction of the neutron's kinetic energy is transferred to the carbon nucleus? (b) If the neutron's initial kinetic energy is \(1.6 \times 10^{-13} \mathrm{~J}\), find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision.

A \(2.0\) -g particle moving at \(8.0 \mathrm{~m} / \mathrm{s}\) makes a perfectly elastic head-on collision with a resting \(1.0\) -g object. (a) Find the speed of each particle after the collision. (b) Find the speed of each particle after the collision if the stationary particle has a mass of \(10 \mathrm{~g}\). (c) Find the final kinetic energy of the incident \(2.0-g\) particle in the situations described in parts (a) and (b). In which case does the incident particle lose more kinetic energy?

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.

Identical twins, each with mass \(55.0 \mathrm{~kg}\), are on ice skates and at rest on a frozen lake, which may be taken as frictionless. Twin A is carrying a backpack of mass \(12.0 \mathrm{~kg}\). She throws it horizontally at \(3.00 \mathrm{~m} / \mathrm{s}\) to Twin \(\mathrm{B}\). Neglecting any gravity effects, what are the subsequent speeds of Twin A and Twin B?
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