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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 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 soccer ball with mass \(0.265 \mathrm{~kg}\) is initially at rest and is kicked at an angle of \(20.8^{\circ}\) with respect to the horizontal. The soccer ball travels a horizontal distance of \(52.8 \mathrm{~m}\) after it is kicked. What is the impulse received by the soccer ball during the kick? Assume there is no air resistance.

A sled initially at rest has a mass of \(52.0 \mathrm{~kg}\), including all of its contents. A block with a mass of \(13.5 \mathrm{~kg}\) is ejected to the left at a speed of \(13.6 \mathrm{~m} / \mathrm{s} .\) What is the speed of the sled and the remaining contents?

A golf ball of mass \(45.0 \mathrm{~g}\) moving at a speed of \(120 . \mathrm{km} / \mathrm{h}\) collides head on with a French TGV high-speed train of mass \(3.8 \cdot 10^{5} \mathrm{~kg}\) that is traveling at \(300 . \mathrm{km} / \mathrm{h}\). Assuming that the collision is elastic, what is the speed of the golf ball after the collision? (Do not try to conduct this experiment!)

Consider two carts, of masses \(m\) and \(2 m\), at rest on a frictionless air track. If you push the lower-mass cart for \(3 \mathrm{~s}\) and then the other cart for the same length of time and with the same force, which cart undergoes the larger change in momentum? a) The cart with mass \(m\) has the larger change b) The cart with mass \(2 m\) has the larger change. c) The change in momentum is the same for both carts. d) It is impossible to tell from the information given.
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