Problem 1 - College Physics

Question

Calculate the magnitude of the linear momentum for the following cases: (a) a proton with mass equal to \(1.67 \times 10^{-27} \mathrm{~kg}\), moving with a speed of \(5.00 \times 10^{6}\) \(\mathrm{m} / \mathrm{s} ;\) (b) a \(15.0-\mathrm{g}\) bullet moving with a speed of \(300 \mathrm{~m} / \mathrm{s} ;\) (c) a \(75.0-\mathrm{kg}\) sprinter running with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\); (d) the Earth (mass \(=5.98 \times 10^{24} \mathrm{~kg}\) ) moving with an orbital speed equal to \(2.98 \times 10^{4} \mathrm{~m} / \mathrm{s}\).

Step-by-Step Solution

Answer

The magnitude of the linear momentum for the proton is \(8.35 \times 10^{-21} \mathrm{kg \cdot m/s}\), for the bullet is \(4.5 \mathrm{kg \cdot m/s}\), for the sprinter is \(750.0 \mathrm{kg \cdot m/s}\), and for Earth is \(1.78 \times 10^{29} \mathrm{kg \cdot m/s}\).

Step 1: Calculate the Linear Momentum of a Proton

The linear momentum \(p\) of a particle can be calculated using the formula \(p=m \cdot v\), where \(m\) is the mass and \(v\) is the velocity. Let's find the momentum for a proton moving at \(5.00 \times 10^{6} \mathrm{m/s}\): \(p=1.67 \times 10^{-27} \mathrm{kg} \cdot 5.00 \times 10^{6} \mathrm{m/s} = 8.35 \times 10^{-21} \mathrm{kg \cdot m/s}\)

Step 2: Calculate the Linear Momentum of a Bullet

Next, calculate the momentum for a \(15.0-\mathrm{g}\) bullet moving at \(300 \mathrm{m/s}\). Remember that the mass of the bullet is given in grams, so we need to convert that into kilograms: \(15.0 \mathrm{g} = 0.015 \mathrm{kg}\). Now, we can compute the momentum: \(p=0.015 \mathrm{kg} \cdot 300 \mathrm{m/s} = 4.5 \mathrm{kg \cdot m/s}\)

Step 3: Calculate the Linear Momentum of a Sprinter

The third case involves a sprinter with a mass of \(75.0-\mathrm{kg}\), running at a speed of \(10.0 \mathrm{m/s}\). Using the formula, we find the momentum: \(p = 75.0 \mathrm{kg} \cdot 10.0 \mathrm{m/s} = 750.0 \mathrm{kg \cdot m/s}\)

Step 4: Calculate the Linear Momentum of the Earth

The last case involves the Earth moving in its orbit at a speed of \(2.98 \times 10^{4} \mathrm{m/s}\). The mass of the Earth is \(5.98 \times 10^{24} \mathrm{kg}\). Apply the formula to compute the momentum: \(p=5.98 \times 10^{24} \mathrm{kg} \cdot 2.98 \times 10^{4} \mathrm{m/s} = 1.78 \times 10^{29} \mathrm{kg \cdot m/s}\)

Key Concepts

Physics Problems

Physics problems often involve computations and conceptual understandings, making them a staple in learning the fundamentals. One of the common types of physics problems involves the calculation of physical quantities like velocity, force, or momentum. These problems may seem intricate, but they are all rooted in basic principles.

For the linear momentum problem provided, the core concept revolves around understanding how momentum serves as a key descriptor of moving bodies. In this exercise, different objects ranging from a proton to Earth were used to calculate linear momentum. This allows students to see the application of the formula across diverse scales, from microscopic particles to celestial bodies. Working through such problems enhances your problem-solving skills and increases your familiarity with scientific notation, units, and conversions.

• Physics problems teach you how to approach complex situations methodically.
• They help integrate theoretical concepts with practical calculations.

By mastering these exercises, students gain valuable insight into the workings of the physical world.

Momentum Calculation

Calculating momentum is an essential skill in physics, foundational to other topics like force interactions and motion analysis. The formula for linear momentum is simple: it is the product of the mass of an object and its velocity, written as:

1. Symbolically, this is expressed as: \( p = m \times v \). Here, \( p \) represents momentum, \( m \) stands for mass, and \( v \) denotes velocity.
2. It is crucial to maintain consistency with units when performing the calculation.

Understanding this concept was applied to different scenarios in our exercise, such as calculating the momentum of a proton compared to that of a sprinter or even the Earth. Each calculation varied significantly in magnitude due to differences in mass and velocity.

• For a proton: momentum was calculated using its miniscule mass and high velocity.
• For a bullet: a conversion from grams to kilograms was required to maintain unit consistency.
• For larger objects like a sprinter or Earth: calculations emphasize the influence of mass on momentum.

These calculations illustrate how momentum scales with the mass and speed of an object, emphasizing its significance in describing motion.

Conservation of Momentum

The conservation of momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant, provided no external forces act upon it. This principle is pivotal for understanding interactions in physics and has practical implications in everyday phenomena.

In real-world applications, the conservation of momentum helps us predict the outcomes of collisions and interactions in isolated systems. For instance, in collisions between two objects, the total momentum before the collision is equal to the total momentum after the collision. This is particularly useful in engineering and safety analysis, as well as in fields like astrophysics, where it aids in modeling the movements and interactions of celestial bodies.

• The conservation of momentum explains how rockets propel in space.
• It is essential for understanding car crashes and accident reconstruction.

By ensuring a grasp of momentum and its conservation, you unlock a deeper comprehension of many physical phenomena. This understanding lays the groundwork for further topics in mechanics and other branches of physics, where motion and energy transfer principles are applied.