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Chapter 7: Problem 20

Rank the following objects from highest to lowest in terms of momentum and from highest to lowest in terms of energy. a) an asteroid with mass \(10^{6} \mathrm{~kg}\) and speed \(500 \mathrm{~m} / \mathrm{s}\) b) a high-speed train with a mass of \(180,000 \mathrm{~kg}\) and a speed of \(300 \mathrm{~km} / \mathrm{h}\) c) a 120 -kg linebacker with a speed of \(10 \mathrm{~m} / \mathrm{s}\) d) a 10 -kg cannonball with a speed of \(120 \mathrm{~m} / \mathrm{s}\) e) a proton with a mass of \(6 \cdot 10^{-27} \mathrm{~kg}\) and a speed of \(2 \cdot 10^{8} \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
Question: Rank the objects in terms of momentum and energy, from highest to lowest. Answer: Momentum Ranking (highest to lowest): 1. Asteroid 2. Train 3. Tie between Linebacker and Cannonball 4. Proton Energy Ranking (highest to lowest): 1. Asteroid 2. Train 3. Cannonball 4. Linebacker 5. Proton

Step by step solution

01

Convert units and calculate momentum and energy for each object.

a) Asteroid: - Mass: \(m_a = 10^6 \,\mathrm{kg}\) - Velocity: \(v_a = 500 \,\mathrm{m/s}\) b) Train: - Mass: \(m_b = 180,000 \,\mathrm{kg}\) - Velocity: \(v_b = 300 \,\mathrm{km/h}\) Convert to m/s: \(v_b = \frac{300\,\mathrm{km}}{3.6\,\mathrm{h}} = 83.33\,\mathrm{m/s}\) c) Linebacker: - Mass: \(m_c = 120 \,\mathrm{kg}\) - Velocity: \(v_c = 10\,\mathrm{m/s}\) d) Cannonball: - Mass: \(m_d = 10 \,\mathrm{kg}\) - Velocity: \(v_d = 120 \,\mathrm{m/s}\) e) Proton: - Mass: \(m_e = 6 \cdot 10^{-27} \,\mathrm{kg}\) - Velocity: \(v_e = 2 \cdot 10^8 \,\mathrm{m/s}\) Now, calculate the momentum and energy for each object: Object a: \(p_a = m_a \cdot v_a = 10^6 \,\mathrm{kg} \cdot 500 \,\mathrm{m/s} = 5 \cdot 10^8\,\mathrm{kg\cdot m/s}\) \(E_a = \frac{1}{2} \cdot m_a \cdot v_a^2 = \frac{1}{2} \cdot 10^6 \,\mathrm{kg} \cdot (500 \,\mathrm{m/s})^2 = 1.25 \cdot 10^{11}\,\mathrm{J}\) Object b: \(p_b = m_b \cdot v_b = 180,000 \,\mathrm{kg} \cdot 83.33 \,\mathrm{m/s} = 1.5 \cdot 10^7\,\mathrm{kg\cdot m/s}\) \(E_b = \frac{1}{2} \cdot m_b \cdot v_b^2 = \frac{1}{2} \cdot 180,000 \,\mathrm{kg} \cdot (83.33 \,\mathrm{m/s})^2 = 6.25 \cdot 10^8\,\mathrm{J}\) Object c: \(p_c = m_c \cdot v_c = 120 \,\mathrm{kg} \cdot 10 \,\mathrm{m/s} = 1200\,\mathrm{kg\cdot m/s}\) \(E_c = \frac{1}{2} \cdot m_c \cdot v_c^2 = \frac{1}{2} \cdot 120 \,\mathrm{kg} \cdot (10 \,\mathrm{m/s})^2 = 6,000\,\mathrm{J}\) Object d: \(p_d = m_d \cdot v_d = 10 \,\mathrm{kg} \cdot 120 \,\mathrm{m/s} = 1200\,\mathrm{kg\cdot m/s}\) \(E_d = \frac{1}{2} \cdot m_d \cdot v_d^2 = \frac{1}{2} \cdot 10 \,\mathrm{kg} \cdot (120 \,\mathrm{m/s})^2 = 72,000\,\mathrm{J}\) Object e: \(p_e = m_e \cdot v_e = 6 \cdot 10^{-27} \,\mathrm{kg} \cdot 2 \cdot 10^8 \,\mathrm{m/s} = 1.2 \cdot 10^{-18}\,\mathrm{kg\cdot m/s}\) \(E_e = \frac{1}{2} \cdot m_e \cdot v_e^2 = \frac{1}{2} \cdot 6 \cdot 10^{-27} \,\mathrm{kg} \cdot (2 \cdot 10^8 \,\mathrm{m/s})^2 = 1.2 \cdot 10^{-10}\,\mathrm{J}\)
02

Rank objects based on momentum and energy values.

Now that we have calculated the momentum and energy values for each object, let's rank the objects from highest to lowest based on these values. Momentum Ranking (highest to lowest): 1. Object a (Asteroid): \(5 \cdot 10^8\,\mathrm{kg\cdot m/s}\) 2. Object b (Train): \(1.5 \cdot 10^7\,\mathrm{kg\cdot m/s}\) 3. Object c (Linebacker): \(1200\,\mathrm{kg\cdot m/s}\) 4. Object d (Cannonball): \(1200\,\mathrm{kg\cdot m/s}\) 5. Object e (Proton): \(1.2 \cdot 10^{-18}\,\mathrm{kg\cdot m/s}\) Energy Ranking (highest to lowest): 1. Object a (Asteroid): \(1.25 \cdot 10^{11}\,\mathrm{J}\) 2. Object b (Train): \(6.25 \cdot 10^8\,\mathrm{J}\) 3. Object d (Cannonball): \(72,000\,\mathrm{J}\) 4. Object c (Linebacker): \(6,000\,\mathrm{J}\) 5. Object e (Proton): \(1.2 \cdot 10^{-10}\,\mathrm{J}\)

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Key Concepts

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

Conservation of Momentum
When we talk about the conservation of momentum, we're referring to a fundamental principle of physics indicating that if no external forces act on a system, the total momentum of that system remains constant over time. This concept is central to collision and explosion problems in mechanics, where two or more objects interact.

To apply conservation of momentum, we consider the momentum before and after an event and assert that the total remains unaltered, as long as the net external force is zero. Momentum, represented by the symbol \( p \), is the product of an object's mass \( m \) and its velocity \( v \), given by the equation \( p = m \times v \).

For instance, in the textbook exercise, the conservation of momentum would be crucial if these objects were to collide or interact in a closed system. The sum of their momenta before the interaction would equal the sum afterward, providing useful insights into their velocities post-interaction.
Kinetic Energy Calculations
Moving on to kinetic energy, this term describes the energy that an object possesses due to its motion. It's a scalar quantity, meaning it has magnitude but no direction, and is given by the formula \( E_k = \frac{1}{2} m v^2 \), where \( m \) is the mass and \( v \) is the velocity of the object.

As seen in the solution, calculating kinetic energy involves squaring the velocity, which makes it highly sensitive to changes in speed. This also explains why, even though a proton is moving at a tremendous speed, its tiny mass results in a relatively low kinetic energy compared to larger objects like asteroids or trains with much greater mass.

Comparing Momentum and Kinetic Energy

Importantly, momentum and kinetic energy are not equivalent. Momentum is a vector, which depends both on mass and velocity, while kinetic energy is directly proportional to the square of velocity. This is why in the exercise, objects with higher speeds may not necessarily have the most kinetic energy if their mass is significantly lower.
Unit Conversion
Unit conversion is a basic yet essential skill in physics, allowing us to compare and compute physical quantities in different units systematically. It's particularly important when dealing with equations where terms must be in consistent units to avoid errors.

In the step-by-step solutions provided, unit conversion is essential when dealing with the train's velocity, which was originally given in kilometers per hour (km/h). To use the momentum and kinetic energy formulas, velocity needs to be in meters per second (m/s). The conversion factor from km/h to m/s is \( \frac{1}{3.6} \) since there are 3.6 km in a mile and 3600 seconds in an hour. Without this crucial step of unit conversion, the subsequent calculations for momentum and kinetic energy would yield incorrect results.

Understanding and correctly implementing unit conversion helps assure that all calculations are made within a consistent framework, which is fundamental for accurate physics computations.

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Most popular questions from this chapter

A fireworks projectile is launched upward at an angle above a large flat plane. When the projectile reaches the top of its flight, at a height of \(h\) above a point that is a horizontal distance \(D\) from where it was launched, the projectile explodes into two equal pieces. One piece reverses its velocity and travels directly back to the launch point. How far from the launch point does the other piece land? a) \(D\) b) \(2 D\) c) \(3 D\) d) \(4 D\)

Although they don't have mass, photons-traveling at the speed of light-have momentum. Space travel experts have thought of capitalizing on this fact by constructing solar sails-large sheets of material that would work by reflecting photons. Since the momentum of the photon would be reversed, an impulse would be exerted on it by the solar sail, and-by Newton's Third Law-an impulse would also be exerted on the sail, providing a force. In space near the Earth, about \(3.84 \cdot 10^{21}\) photons are incident per square meter per second. On average, the momentum of each photon is \(1.30 \cdot 10^{-27} \mathrm{~kg} \mathrm{~m} / \mathrm{s}\). For a \(1000 .-\mathrm{kg}\) spaceship starting from rest and attached to a square sail \(20.0 \mathrm{~m}\) wide, how fast could the ship be moving after 1 hour? One week? One month? How long would it take the ship to attain a speed of \(8000 . \mathrm{m} / \mathrm{s}\), roughly the speed of the space shuttle in orbit?

A particle \(\left(M_{1}=1.00 \mathrm{~kg}\right)\) moving at \(30.0^{\circ}\) downward from the horizontal with \(v_{1}=2.50 \mathrm{~m} / \mathrm{s}\) hits a second particle \(\left(M_{2}=2.00 \mathrm{~kg}\right),\) which was at rest momentarily. After the collision, the speed of \(M_{1}\) was reduced to \(.500 \mathrm{~m} / \mathrm{s}\), and it was moving at an angle of \(32^{\circ}\) downward with respect to the horizontal. Assume the collision is elastic. a) What is the speed of \(M_{2}\) after the collision? b) What is the angle between the velocity vectors of \(M_{1}\) and \(M_{2}\) after the collision?

A small car of mass 1000 . kg traveling at a speed of \(33.0 \mathrm{~m} / \mathrm{s}\) collides head on with a large car of mass \(3000 \mathrm{~kg}\) traveling in the opposite direction at a speed of \(30.0 \mathrm{~m} / \mathrm{s}\). The two cars stick together. The duration of the collision is \(100 . \mathrm{ms}\). What acceleration (in \(g\) ) do the occupants of the small car experience? What acceleration (in \(g\) ) do the occupants of the large car experience?

In a severe storm, \(1.00 \mathrm{~cm}\) of rain falls on a flat horizontal roof in \(30.0 \mathrm{~min}\). If the area of the roof is \(100 \mathrm{~m}^{2}\) and the terminal velocity of the rain is \(5.00 \mathrm{~m} / \mathrm{s}\), what is the average force exerted on the roof by the rain during the storm?
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