Problem 53 - University Physics with Modern Physics

Question

When you open the door to an air-conditioned room, you mix hot gas with cool gas. Saying that a gas is hot or cold actually refers to its average energy; that is, the hot gas molecules have a higher kinetic energy than the cold gas molecules. The difference in kinetic energy in the mixed gases decreases over time as a result of elastic collisions between the gas molecules, which redistribute the energy. Consider a two-dimensional collision between two nitrogen molecules \(\left(\mathrm{N}_{2},\right.\) molecular weight \(=28.0 \mathrm{~g} / \mathrm{mol}\) ). One molecule moves at \(30.0^{\circ}\) with respect to the horizontal with a velocity of \(672 \mathrm{~m} / \mathrm{s} .\) This molecule collides with a second molecule moving in the negative horizontal direction at \(246 \mathrm{~m} / \mathrm{s}\). What are the molecules' final velocities if the one that is initially more energetic moves in the vertical direction after the collision?

Step-by-Step Solution

Answer

Answer: The final velocities of the two nitrogen molecules after the collision are: - Molecule A: \(672 \cdot \sin{30^{\circ}} \mathrm{m/s} \mathrm{~in~vertical~direction}\) - Molecule B: \(672 \cdot \cos{30^{\circ}} \mathrm{m/s} \mathrm{~in~negative~horizontal~direction}\)

Step 1: Label the molecules and their initial velocities

Let's label the first molecule as 'A' and the second molecule as 'B'. The initial velocities can be noted as: - Molecule A: \(\vec{v}_{A1} = 672 \mathrm{~m} / \mathrm{s} \mathrm{~at~} 30.0^{\circ}\) - Molecule B: \(\vec{v}_{B1} = -246 \mathrm{~m} / \mathrm{s}\) (in the negative horizontal direction)

Step 2: Break the velocity of molecule A into components

To work with the velocities in 2D, it is useful to break them into their horizontal (x-axis) and vertical (y-axis) components. For molecule A, this can be done as follows: \(\vec{v}_{Ax1} = 672 \cdot \cos{30^{\circ}} \mathrm{m/s}\) \(\vec{v}_{Ay1} = 672 \cdot \sin{30^{\circ}} \mathrm{m/s}\)

Step 3: Conserve the momentum in both x and y directions

Since there is no external force acting on the molecules, we can write the conservation of momentum for both x and y directions: \(x\)-direction: \(m_A v_{Ax1} = m_A v_{Ax2} - m_B v_{Bx2}\) \(y\)-direction: \(m_A v_{Ay1} = m_A v_{Ay2} + m_B v_{By2}\) As both molecules are Nitrogen, they have the same mass \(m_A = m_B\).

Step 4: Use the given condition for the final velocities

We are given that molecule A will move in the vertical direction after the collision. This means: \(v_{Ax2} = 0\) Now, using the \(x\)-direction momentum equation: \(0 = 672 \cdot \cos{30^{\circ}} - v_{Bx2}\) Solving for \(v_{Bx2}\), we get: \(v_{Bx2} = 672 \cdot \cos{30^{\circ}}\) Since molecule B was moving in the negative horizontal direction initially, the sign of \(v_{Bx2}\) remains the same.

Step 5: Solve for the final vertical velocities

Using the \(y\)-direction momentum equation with \(v_{Ax2} = 0\): \(672 \cdot \sin{30^{\circ}} = v_{Ay2} + v_{By2}\) As molecule A was initially more energetic, it is moving upwards. So, \(v_{Ay2} = 672 \cdot \sin{30^{\circ}}\) and \(v_{By2} = 0\).

Step 6: Write the final velocities

Now that we have computed the final velocities, we can write them as: - Molecule A: \(\vec{v}_{A2} = 672 \cdot \sin{30^{\circ}} \mathrm{m/s} \mathrm{~in~vertical~direction}\) - Molecule B: \(\vec{v}_{B2} = 672 \cdot \cos{30^{\circ}} \mathrm{m/s} \mathrm{~in~negative~horizontal~direction}\)

Key Concepts

Kinetic Energy

Kinetic energy fundamentally refers to the energy possessed by an object due to its motion. All moving objects have kinetic energy, which is directly proportional to the mass of the object, and the square of its velocity. It is given by the formula:
\[ KE = \frac{1}{2}mv^2 \]
where \(m\) is the object's mass and \(v\) is its velocity. In the context of gases, such as in our exercise, molecules are constantly on the move, and their kinetic energy reflects their temperature. The 'hotter' a molecule is, the faster it moves, and thus, the higher its kinetic energy. In the given exercise, the collision between two nitrogen molecules redistributes their kinetic energy. It's worth noting how the collision leads to a dispersion of energy, eventually leading to a thermal equilibrium where both molecules would end up with similar kinetic energies, assuming a perfectly elastic collision in which no kinetic energy is lost.

Conservation of Momentum

The conservation of momentum is a fundamental concept in physics which states that if no external forces act on a closed system of objects, the total momentum of that system remains constant. Momentum is a vector quantity, meaning it has both magnitude and direction, and is a product of an object's mass and its velocity:
\[ \vec{p} = m\vec{v} \]
For collisions, momentum conservation holds true for each direction separately. This is crucial in analyzing the outcome of collisions, as seen in our exercise. With two molecules, A and B, moving in two dimensions, we set up separate equations for the horizontal (x) and vertical (y) momentum conservation. This principle allows us to predict the final velocities of each molecule after the collision, provided that the collision is elastic. By ensuring momentum is conserved, we can isolate the final velocities and find a solution that respects both the physical law and the conditions of the problem.

Two-Dimensional Collisions

Collisions in two-dimensional space, like the one outlined in our exercise involving nitrogen molecules, are slightly more complex than one-dimensional collisions because they involve both the x (horizontal) and y (vertical) components of motion. The principles governing the collision—specifically, the conservation of momentum and kinetic energy—are the same, but must be applied to each direction independently.

For an elastic collision in two dimensions, both the total kinetic energy and the momentum of the system are conserved. As we solved in the exercise, breaking down the initial velocities into their component forms allows us to apply these conservation laws separately on the x and y axes. The final velocities are then determined by the requirement that the sum of the momenta and kinetic energies before the collision must equal the sum after the collision. The complexity of two-dimensional collisions often requires a systematic approach to problem-solving, such as clearly labeling the initial and final velocities, and carefully applying vector addition and resolution.