Problem 105 - Chemistry Principles and Practice

Question

A The main weapon on a military tank is a cannon that fires a blunt projectile specially designed to cause a shock wave when it hits another tank. The projectile fits into a finned casing that improves its accuracy. Calculate the mass of the projectile, assuming it is a cylinder of uranium (density \(\left.=19.05 \mathrm{~g} / \mathrm{cm}^{3}\right)\) that is \(105 \mathrm{~mm}\) in diameter and \(30 \mathrm{~cm}\) in height. The volume of a cylinder is given by the equation \(V=\pi r^{2} b\).

Step-by-Step Solution

Answer

The mass of the projectile is approximately 49.49 kg.

Step 1: Recall the formula for the volume of a cylinder

The volume of a cylinder is calculated using the formula \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height of the cylinder.

Step 2: Convert diameter to radius and units

Given that the diameter of the cylinder is 105 mm, we need to find the radius. The radius \( r \) is half the diameter: \( r = \frac{105}{2} = 52.5 \) mm. Convert the dimensions to centimeters for consistent units: \( r = 5.25 \) cm and \( h = 30 \) cm.

Step 3: Calculate the volume of the cylinder

Substitute \( r = 5.25 \) cm and \( h = 30 \) cm into the volume formula: \( V = \pi (5.25)^2 (30) \). Calculate the square of the radius: \( (5.25)^2 = 27.5625 \). Calculate the volume: \( V = \pi (27.5625)(30) = 2596.96 \) cubic centimeters.

Step 4: Use the density to find the mass

The density \( \rho \) of uranium is given as \( 19.05 \) g/cm³. The mass \( m \) is calculated using the formula \( m = \rho \times V \). Substitute the values \( m = 19.05 \times 2596.96 \) to find the mass.

Step 5: Calculate the final mass

Calculate the mass: \( m = 19.05 \times 2596.96 = 49491.648 \). Therefore, the mass of the projectile is approximately 49491.65 grams, or 49.49 kilograms.

Key Concepts

Volume of a Cylinder

The first step in our problem is calculating the volume of the cylindrical projectile. Cylinders have a circular base and a specific height. The volume, which represents how much space the cylinder occupies, is determined by using the formula:\[V = \pi r^2 h\]Here, \( r \) stands for the radius of the circular base, and \( h \) denotes the height of the cylinder. These two measurements allow us to compute the three-dimensional space inside the cylindrical shape. It’s important to remember:

• Radius is half of the diameter.
• The constant \( \pi \) approximately equals 3.14159.

For our projectile, the diameter is 105 mm, making the radius 52.5 mm, which converts to 5.25 cm when using a consistent unit of measurement (centimeters in this case). With a height of 30 cm, you can plug these values into the formula to find the volume.

Density and Mass

Understanding density and mass is crucial in converting the volume of the projectile into its mass. Density is a measure of how much mass an object has in a given volume, expressed in units such as grams per cubic centimeter (g/cm³). In our case, the density of uranium used in the projectile is given as:\[\rho = 19.05 \text{ g/cm}^3\]Mass is linked to density and volume through the equation:\[m = \rho \times V\]In simpler terms, once you've determined the volume of the projectile, you simply multiply that by its density to find its mass. For students, remember:

• Mass indicates how much matter is in the object, typically in grams or kilograms.
• High density materials will have a significant mass without needing large volume.

After calculating the volume as 2596.96 cm³, we multiply by the given density to find the mass.

Unit Conversion

Unit conversion plays a pivotal role when dealing with measurements in calculations such as this. It's essential to have a standard unit to avoid errors. In this scenario, we converted the diameter from millimeters to centimeters. Here’s how you do it:

• Move from mm to cm by dividing by 10 (since 1 cm = 10 mm).
• Ensure all measurements are in the same unit for consistency.

For example, the diameter of 105 mm converts to a radius of 52.5 mm and further to 5.25 cm. Remember, consistent units enable easier calculations. When dealing with density, volume typically needs to be in cubic centimeters so that grams per cubic centimeter calculations are correct. If final mass is needed in kilograms, convert grams to kilograms by dividing by 1000.

Geometry in Physics

Geometry in physics often involves understanding shapes and how they influence physical properties like volume and mass. A cylinder is a common shape in physics problems, often dealing with volumes and areas. Recognizing how geometry provides a bridge between physical concepts and mathematical calculations is pivotal:

• Cylinders connect concepts of radius, height, and shape to real-world objects.
• Shapes and their properties determine how formulas are applied.

In this problem, the projectile’s cylindrical shape requires us to use the geometric formula for volume, which then ties directly into physics with density and mass calculation. Understanding both the mathematical and physical implications of geometry enhances problem-solving skills in physics.