91Ó°ÊÓ

We have two barges colliding in a foggy environment. The first barge has mass \(1.50 \times 10^{5} \text{ kg}\) and an initial speed of \(6.2 \text{ m/s}\). The second barge has mass \(2.78 \times 10^{5} \text{ kg}\) and an initial speed of \(4.3 \text{ m/s}\). After the collision, the second barge's speed is \(5.1 \text{ m/s}\) at an angle of \(18^{\circ}\). We need to find the speed and direction of motion of the first barge after the collision.

Determine the x- and y-components of the initial momentum for both barges. Let the downriver direction be the x-axis, and the direction directly across the river be the y-axis. For barge 1 (x-direction): \[ p_{1x\text{initial}} = m_{1} v_{1x} = 1.50 \times 10^{5} \text{ kg} \times 6.2 \text{ m/s} = 9.3 \times 10^{5} \text{ kg} \cdot \text{m/s} \] and for barge 2 (y-direction): \[ p_{2y\text{initial}} = m_{2} v_{2y} = 2.78 \times 10^{5} \text{ kg} \times 4.3 \text{ m/s} = 1.1954 \times 10^{6} \text{ kg} \cdot \text{m/s} \] Since barge 2 initially has no x-component and barge 1 no y-component, it simplifies our calculations.

Post-collision, find the x- and y-components of the final momentum for barge 2. Using its given speed and deflected angle: \[ p_{2x\text{final}} = m_{2} v_{2\text{final}} \cos(18^{\circ}) = 2.78 \times 10^{5} \text{ kg} \times 5.1 \text{ m/s} \cos(18^{\circ}) = 1.34622 \times 10^{6} \text{ kg} \cdot \text{m/s} \] \[ p_{2y\text{final}} = m_{2} v_{2\text{final}} \sin(18^{\circ}) = 2.78 \times 10^{5} \text{ kg} \times 5.1 \text{ m/s} \sin(18^{\circ}) = 4.36431 \times 10^{5} \text{ kg} \cdot \text{m/s} \]

Since momentum is conserved in both x and y directions: In the x-direction: \[ p_{1x\text{initial}} + 0 = p_{1x\text{final}} + p_{2x\text{final}} \] \[ 9.3 \times 10^{5} \text{ kg} \cdot \text{m/s} = p_{1x\text{final}} + 1.34622 \times 10^{6} \text{ kg} \cdot \text{m/s} \] Rearrange to find \(p_{1x\text{final}}\): \[ p_{1x\text{final}} = (9.3 \times 10^{5} - 1.34622 \times 10^{6}) \text{ kg} \cdot \text{m/s} = -4.1622 \times 10^5 \text{ kg} \cdot \text{m/s} \] And in the y-direction: \[ 0 + p_{2y\text{initial}} = p_{1y\text{final}} + p_{2y\text{final}} \] \[ 1.1954 \times 10^{6} \text{ kg} \cdot \text{m/s} = p_{1y\text{final}} + 4.36431 \times 10^{5} \text{ kg} \cdot \text{m/s} \] Rearrange to find \(p_{1y\text{final}}\): \[ p_{1y\text{final}} = (1.1954 \times 10^{6} - 4.36431 \times 10^{5}) \text{ kg} \cdot \text{m/s} = 7.58969 \times 10^5 \text{ kg} \cdot \text{m/s} \]

Use the components of final momentum to find the speed of the first barge. Recall: \[ p = m \cdot v \text{, hence} \; \vec{v}_1 = \frac{\vec{p}_1}{m_1} \] Therefore, \( v_{1\text{final}} = \sqrt{(\frac{p_{1x\text{final}}}{m_{1}})^2 + (\frac{p_{1y\text{final}}}{m_{1}})^2} \) \[ v_{1\text{final}} = \sqrt{(\frac{-4.1622 \times 10^5}{1.50 \times 10^5})^2 + (\frac{7.58969 \times 10^5}{1.50 \times 10^5})^2} \] \[ v_{1\text{final}} = \sqrt{(-2.7748)^2 + (5.0598)^2} \] \[ v_{1\text{final}} \approx \sqrt{7.7 + 25.6} = \sqrt{33.3} \approx 5.77 \text{ m/s} \]

Find the angle of the first barge's velocity after the collision. \[ \theta = \tan^{-1}(\frac{p_{1y\text{final}}}{p_{1x\text{final}}}) = \tan^{-1}(\frac{7.58969 \times 10^5}{-4.1622 \times 10^5}) \approx \tan^{-1}(-1.824) \approx -61.5^{\circ} \] The negative angle indicates the direction is 180° - 61.5° = 118.5° counterclockwise from the positive x-axis.