Problem 11 - Applied Physics

Question

(a) Find the momentum of an automobile of mass \(2630 \mathrm{~kg}\) traveling \(21.0 \mathrm{~m} / \mathrm{s}\). (b) Find the velocity (in \(\mathrm{km} / \mathrm{h}\) ) of a light auto of mass \(1170 \mathrm{~kg}\) so that it has the same momentum as the auto in part (a).

Step-by-Step Solution

Answer

(a) Momentum: 55230 kg·m/s. (b) Velocity: 170.0 km/h.

Step 1: Understanding Momentum

Momentum (\( p \) ) is defined as the product of an object's mass (\( m \) ) and its velocity (\( v \) ). The equation is \( p = m \times v \). In this exercise, you have been given the mass and velocity of the automobile to calculate its momentum.

Step 2: Calculate Momentum for Automobile

Given the mass of the automobile as \( 2630 \, \text{kg} \) and its velocity as \( 21.0 \, \text{m/s} \), the momentum is calculated by:\[p = 2630 \, \text{kg} \times 21.0 \, \text{m/s} = 55230 \, \text{kg} \cdot \text{m/s}.\]

Step 3: Equate Momentums for Velocity Calculation

For the light auto, we know the momentum should be the same as the automobile from part (a), which is \( 55230 \, \text{kg} \cdot \text{m/s} \). Use the formula \( p = m \times v \) and solve for velocity \( v \), which gives:\[ v = \frac{p}{m} = \frac{55230 \, \text{kg} \cdot \text{m/s}}{1170 \, \text{kg}} = 47.1795 \, \text{m/s}. \]

Step 4: Convert Velocity to km/h

To convert the velocity from meters per second to kilometers per hour, use the conversion factor 1 \( \text{m/s} = 3.6 \, \text{km/h} \). Thus:\[ 47.1795 \, \text{m/s} \times 3.6 \, \text{km/h per m/s} = 169.8462 \, \text{km/h}. \] We round this to 170.0 \, \text{km/h} for simplicity.

Key Concepts

Understanding Mass and Velocity

Mass and velocity are two key elements in physics essential for determining an object's momentum. Mass is a measure of how much matter is in an object, usually measured in kilograms (kg). On the other hand, velocity is defined as the speed of the object in a given direction, measured in meters per second (m/s). Both are critical to calculating momentum as they contribute to how much motion an object carries.
- **Mass**: The weight of the object contributing to its inertia, and heavier objects generally have more momentum if moving at the same velocity. - **Velocity**: It's directional speed, meaning not just how fast but also in which direction an object is moving.
Understanding the interplay between mass and velocity is vital, because when you change either one, the momentum of the object will also change. In any physics problem involving motion, these two are usually the primary factors you need to consider.

Calculating Momentum

Momentum quantifies the motion of an object and is symbolized by \( p \). The formula to calculate the momentum of an object is simple: \( p = m imes v \), where \( m \) is the mass and \( v \) is the velocity.
In our exercise, the momentum of an automobile can be determined using its specific mass and velocity values. For instance, if an automobile has a mass of \( 2630 \, \text{kg} \) and travels at a velocity of \( 21.0 \, \text{m/s} \), the momentum would be:
\[ p = 2630 \, \text{kg} \times 21.0 \, \text{m/s} = 55230 \, \text{kg} \cdot \text{m/s}. \]
Momentum is crucial as it shows not just how fast an object is moving, but how much "oomph" or impact it would have on anything it encounters. This information is not only theoretical but practical, used in everyday scenarios such as car safety designs and traffic accident assessments.

The Importance of Unit Conversion

Unit conversion is a critical skill in physics as it allows you to translate measurements into different units without changing the quantity. It ensures that results are understood globally as different regions use different units. For example, velocity in our exercise was originally given in meters per second (m/s) but needed to be converted into kilometers per hour (km/h) since this is a common unit for speed on roads.
To convert, multiply the velocity by a conversion factor. The factor for changing m/s to km/h is 3.6, because one meter/second equals 3.6 kilometers/hour. Our calculated velocity of \( 47.1795 \, \text{m/s} \) converts to:
\[ 47.1795 \, \text{m/s} \times 3.6 = 169.8462 \, \text{km/h}. \]
Such an understanding easy conversion not only helps solve physics problems but also equips you with an essential tool for practical everyday calculations.

Approaching Physics Problem Solving

Solving physics problems efficiently involves a series of logical steps and a clear understanding of the principles at play. Start by thoroughly reading the problem. Identify given values, required outcomes, and the principles involved. For example, in our exercise, recognizing momentum's dependence on mass and velocity was crucial.
The steps are often:

• **Identify known and unknown factors**: What values are given? What is the problem asking for?
• **Understand the relationship**: Knowing which formulas connect your variables is key. Here, the relationship \( p = m \times v \) was central.
• **Apply correct calculations**: Carefully substitute known values into your equations to solve for unknowns.
• **Use unit conversions when necessary**: Consistently ensure all values are in compatible units to maintain consistency.
• **Check your work**: Always review steps to catch errors and verify that results are plausible.

These systematic approaches ensure precision and clarity, leading to accurate, dependable results whether in homework exercises or real-world applications.