Question

A barge with mass \(1.50 \times 10^{5} \mathrm{~kg}\) is proceeding downriver at \(6.20 \mathrm{~m} / \mathrm{s}\) in heavy fog when it collides broadside with a barge heading directly across the river; see Fig. 6-25. The second barge has mass \(2.78 \times 10^{5} \mathrm{~kg}\) and was moving at \(4.30 \mathrm{~m} / \mathrm{s}\). Immediately after impact, the second barge finds its course deflected by \(18.0^{\circ}\) in the downriver direction and its speed increased to \(5.10 \mathrm{~m} / \mathrm{s}\). The river current was practially zero at the time of the accident. What is the speed and direction of motion of the first barge immediately after the collision?

Short answer

The speed and direction of motion of the first barge immediately after the collision can be calculated using the law of conservation of momentum and the given values for the barges' masses, velocities before and after collision, as well as the deflection angle of the second barge. The resulting equation for the speed of the first barge can be solved for a numerical value, and the equation for its direction will yield an angle in terms of degrees.

Step-by-step solution

Identifying Given Variables

The given variables in the exercise are: mass of the first barge \(m_1 = 1.50 \times 10^{5} \mathrm{~kg}\), velocity of the first barge \(v_{1i} = 6.20 \mathrm{~m/s}\), mass of the second barge \(m_2 = 2.78 \times 10^{5} \mathrm{~kg}\), velocity of the second barge \(v_{2i} = 4.30 \mathrm{~m/s}\), the final velocity of second barge \(v_{2f} = 5.10 \mathrm{~m/s}\) and the deflected angle \(theta = 18.0^\circ\). Your task is to find the final velocity and direction of the first barge.

Applying Conservation of Momentum

As per the law of conservation of momentum, the total initial momentum of the system (just before collision) is equal to the total final momentum (after collision). Mathematically, \(m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = m_1 \cdot v_{1f} + m_2 \cdot v_{2f} \cdot \cos(\theta)\) where \(v_{1f}\) is the final velocity of the first barge.

Deriving the Equation for Final Velocity

Rearranging the equation from step 2 for \(v_{1f}\), you obtain \(v_{1f} = \frac{m_1 \cdot v_{1i} + m_2 \cdot v_{2i} - m_2 \cdot v_{2f} \cdot \cos(\theta)}{m_1}\). You can now plug in the given values into this equation to find the final velocity of the first barge.

Solving for Final Velocity

After plugging in the given values in step 3, you obtain \(v_{1f} = \frac{1.50 \times 10^{5} \mathrm{~kg} \cdot 6.20 \mathrm{~m/s} + 2.78 \times 10^{5} \mathrm{~kg} \cdot 4.30 \mathrm{~m/s} - 2.78 \times 10^{5} \mathrm{~kg} \cdot 5.10 \mathrm{~m/s} \cdot \cos(18.0^\circ)}{1.50 \times 10^{5} \mathrm{~kg}}\). This equation provides the magnitude of the final velocity of the first barge.

Finding the Final Direction

To find the direction of the first barge after the collision, use the y-components of the momentum. Again, applying conservation of momentum along y-axis, we get \(0 = m_{1} \cdot v_{1fy} + m_{2} \cdot v_{2f} \cdot \sin(\theta)\), where \(v_{1fy}\) is the y-component of the final velocity of the first barge. Solving for \(v_{1fy}\), we get \(v_{1fy} = - m_{2} \cdot v_{2f} \cdot \sin(\theta) / m_{1}\). Since this quantity is negative, the final direction of the motion of the first barge is upwards (or against the original direction of the second barge before collision). The angle can be found using the inverse tangent function (taking the absolute value because we know the direction already), or \(\alpha = \tan^{-1}(|v_{1fy} / v_{1f} |)\).

Solving for Final Direction

Plugging in the given values into the equation from step 5, you can calculate \(\alpha = \tan^{-1}(|- m_{2} \cdot v_{2f} \cdot \sin(\theta) / m_{1} / v_{1f} |)\). This equation gives the angle of the final velocity or the direction of the first barge after the collision.

Key concepts

Collision Mechanics

Collision mechanics is a fascinating part of physics that deals with the interactions between two or more bodies when they collide. During such events, the law of conservation of momentum comes into play. This principle states that the total momentum of a closed system remains constant before and after the collision, assuming no external forces like friction are involved.
In our example, two barges collide—a heavy one moving downriver and another moving across. Even though they move in different directions, we calculate the total momentum before and after the impact. This allows us to predict the outcome of the collision.

• Momentum is a vector quantity, meaning it has both magnitude and direction.
• The principle of momentum conservation helps in understanding post-collision velocities.

Each barge's momentum is affected by both its mass and its velocity, with the combined system of both barges required to balance out to the same momentum state post-collision. This involves complex interactions of forces and can lead to changes in both speed and direction.

Barge Motion

Barge motion, particularly in a collision scenario, involves dynamic changes that can be analyzed through physics. In our case of a barge moving through a heavy fog, the lack of visibility plays a significant role in the eventual collision.
The motion of a barge is contingent upon the forces exerted on it and its initial momentum. Before the collision, one barge is moving downriver and the other perpendicular to this direction. This difference in direction and speed contributes to the complexity of their interaction during the collision.
Post-collision, the barge that was moving across the river ends up with its course deflected and speed increased. This change can be attributed to transferring energy and momentum, a characteristic feature of collisional impacts. Understanding these principles is crucial in predicting the barge's new pathway and how its speed and direction are altered upon impact.

Vector Components

Vector components are crucial in understanding the momentum in situations like the collision between barges. A vector is a mathematical representation that has both magnitude and direction. When analyzing problems involving two dimensions, such as the path of a moving barge, splitting vectors into their x and y components simplifies calculations.

• The x-component refers to the horizontal aspect, often related to east-west directions.
• The y-component refers to the vertical aspect, usually associated with north-south directions.

In this barge collision scenario, using vector components allows us to decompose the momentum of each barge into respective directions. This decomposition simplifies understanding both the initial and final states of momentum in each direction, crucial for utilizing the conservation of momentum principle effectively.
When solving for the final velocity and direction of the barges after the collision, employing these vector techniques allows for a more straightforward integration of angles and trigonometric functions, showing the deflection and resulting path post-collision.