Problem 99 - Elementary Algebra

Question

\(P=\left(5.0 \times 10^{-2} \mathrm{~kg}\right)\left(4.4 \times 10^{1} \frac{\mathrm{m}}{\mathrm{s}}\right)\)

Step-by-Step Solution

Answer

The momentum is 2.2 kg m/s.

Step 1: Understand the formula

The given exercise involves multiplying the mass and the velocity of an object to find its momentum (P). The formula for momentum is given by: \[ P = m \times v \] where \( m \) is the mass and \( v \) is the velocity.

Step 2: Identify the values

From the problem statement, identify the values for mass and velocity. Here, \( m = 5.0 \times 10^{-2} \text{ kg} \) and \( v = 4.4 \times 10^{1} \frac{\text{m}}{\text{s}} \).

Step 3: Multiply the values

Use the provided values to calculate momentum. Multiply the mass by the velocity: \[ P = (5.0 \times 10^{-2} \text{ kg}) \times (4.4 \times 10^{1} \frac{\text{m}}{\text{s}}) \].

Step 4: Simplify the multiplication

First, multiply the significant figures: \[ 5.0 \times 4.4 = 22.0 \]. Then, multiply the powers of 10: \[ 10^{-2} \times 10^{1} = 10^{-1} \].

Step 5: Combine the results

Combine the results from the previous step to find the momentum: \[ P = 22.0 \times 10^{-1} \].

Step 6: Express the final answer

Simplify the expression to get the momentum in standard form: \[ P = 2.2 \text{ kg} \frac{\text{m}}{\text{s}} \].

Key Concepts

elementary algebra

Understanding elementary algebra is important for solving physics problems like calculating momentum. In this exercise, you'll see how basic algebra is applied. Let's focus on the equation given: ewline

• The formula for momentum is: ewline ewline\[ P = m \times v \]
• This means momentum (P) equals the mass (m) of an object times its velocity (v).
• An equation like this involves multiplication, and in elementary algebra, we learn how to handle such multiplications correctly, especially when variables and constants are involved.

First, you identify the values given: mass and velocity. Then, you use algebraic rules to multiply these values together. In this case, both the mass and velocity are given in scientific notation. Don't worry; even with scientific notation, the multiplication principles from elementary algebra still apply. If you understand how to perform basic multiplications and follow the right steps, solving equations like this one becomes straightforward.

scientific notation

When you deal with very large or very small numbers, scientific notation simplifies the process. Let's break that down with our example: ewline

• Mass is given as ewline\(5.0 \times 10^{-2} \text{ kg} \)
• Velocity is given as ewline\(4.4 \times 10^{1} \frac{\text{m}}{\text{s}} \)

We multiply the numbers step by step:

• First, focus on the significant figures: ewline 5.0 and 4.4. Multiply them to get ewline 22.0.
• Next, handle the powers of 10: Add the exponents ewline\[10^{-2} \times 10^{1} = 10^{-1} \]

So, you combine the results to get \(22.0 \times 10^{-1} \). Finally, express this value in standard form. Scientific notation allows us to handle and compute large or small numbers easily, making these calculations less error-prone.

multiplication of significant figures

When multiplying quantities in scientific notation, it’s crucial to handle the significant figures properly. Here’s how:

• First, multiply the significant digits only:
  
  \( 5.0 \times 4.4 = 22.0 \)
• Then focus on the exponents in the powers of 10:
  ewline\[10^{-2} \times 10^{1} = 10^{-1} \]

Remember, significant figures reflect the precision of a measurement. When you multiply 5.0 and 4.4, each with two significant figures, your product (22) should also be considered with the appropriate number of significant figures. Combining the products of both steps correctly achieves an accurate final result. This ensures your answer remains precise, reflecting the correct level of accuracy given by the original measurements.