Problem 1 - College Physics

Question

If a \(60-\mathrm{kg}\) woman is riding in a car traveling at \(90 \mathrm{~km} / \mathrm{h}\), what is her linear momentum relative to (a) the ground and (b) the car?

Step-by-Step Solution

Answer

(a) 1500 kg m/s relative to the ground, (b) 0 kg m/s relative to the car.

Step 1: Understanding Linear Momentum

Linear momentum is defined as the product of an object's mass and its velocity. It is given by the formula: \[ p = m imes v \]where \( p \) is the momentum, \( m \) is the mass, and \( v \) is the velocity.

Step 2: Calculate the Momentum Relative to the Ground

First, use the woman's mass (60 kg) and the velocity of the car (90 km/h) to calculate her momentum relative to the ground. Convert the velocity from km/h to m/s by using the conversion factor (1 km/h = \( \frac{1}{3.6} \) m/s): \[ 90 \text{ km/h} = \frac{90}{3.6} \text{ m/s} = 25 \text{ m/s} \]Now apply the momentum formula: \[ p = 60 \times 25 = 1500 \text{ kg m/s} \]So, her linear momentum relative to the ground is 1500 kg m/s.

Step 3: Determine the Momentum Relative to the Car

When calculating linear momentum relative to the car, consider the fact that if both the woman and the car are moving together, her relative velocity to the car is 0 m/s. Thus: \[ p = m \times v = 60 \times 0 = 0 \text{ kg m/s} \]Thus, her linear momentum relative to the car is 0 kg m/s.

Key Concepts

Relative Motion

Relative motion is an important concept in physics, especially when analyzing the movement of objects from different perspectives. It describes how the velocity of one object appears to change when viewed relative to another moving object. In simpler terms, it represents how fast an object seems to be moving when observed from another moving object rather than from a stationary point.

• When dealing with relative motion, choose a reference point or object to compare movements.
• In the scenario of the woman in the car, we find the car itself and the ground as the two reference points.
• Compared to the ground, both the woman and the car are moving at 90 km/h. Therefore, her relative velocity to the ground is 90 km/h.
• With the car as a reference, however, her velocity becomes 0 km/h, because she is moving in unison with the car.

Understanding relative motion is essential to correctly apply physics equations and concepts, notably in exercises involving relative speeds or positions.

Momentum Calculation

Calculating momentum is straightforward once you grasp the fundamental formula of linear momentum, which is the product of mass and velocity: \[ p = m \times v \] where \( p \) represents momentum, \( m \) is mass, and \( v \) stands for velocity. In our example, the woman's mass is given as 60 kg and the velocity of the car is 90 km/h.

• Convert velocity to meters per second by dividing by 3.6: \[ \frac{90}{3.6} = 25 \text{ m/s} \]
• Insert these values into the momentum equation: \[ p = 60 \times 25 = 1500 \text{ kg m/s} \]
• Consequently, her momentum relative to the ground is 1500 kg m/s.

When shifting the frame of reference to the car, her velocity becomes 0 m/s as she is stationary relative to it, making her momentum \( 0 \text{ kg m/s} \).
Grasping momentum calculations is key to solving physics problems effectively, as they often form the basis for more complex computations.

Physics Problems

Physics problems, like the one about the woman's momentum, often aim to reinforce understanding of key principles like relative motion and momentum. They present real-world scenarios to practice the translation of physical concepts into mathematical expressions. Here's how to tackle similar problems:

• Identify Given Information: Clearly note the mass, velocity, and reference frames provided in the problem.
• Apply Relevant Equations: Utilize specific formulas, such as those for momentum, keeping units consistent.
• Think About Reference Frames: Consider different views (e.g., from the ground or from a moving object) to assess relative motion and momentum.
• Verify Results: Double-check calculations and examine if the results make physical sense.

By systematically applying these steps, students enhance their problem-solving skills, making physics more intuitive and less daunting.