Two barges collide in water. What happens to the second barge is given, what about the first one?
A barge with mass 1.50 X 10^5 kg is proceeding downriver at 6.20 m/s in heavy fog when it collides broadside with a barge heading directly across the river; see figure in link: http://physweb.bgu.ac.il/COURSES/PHYSICS_ExercisesPool/CONTRIBUTIONS/e_13_8_130_p.jpg
The second barge has mass 2.78 X 10^5 kg and was moving at 4.30 m/s. Immediately after impact, the second barge finds its course deflected by 18.0° in the downriver direction and its speed increased to 5.10 m/s. The river current was practically zero at the time of the accident.
What is the speed and direction of motion of the first barge immediately after the collision? Explain your answer. Thanks 
Resolving the final momentum of second barge into vertical and horizontal direction,
In vertical direction the momentum is 2.78 x10^5 x 5.1 cos 18.
In the horizontal direction the momentum is 2.78 x10^5 x 5.1sin 18.
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In the horizontal direction the initial momentum of the two barges is
1.5 x 10^5 x 6.2 + 0 .
Final momentum in this direction is 2.78 x10^5 x 5.1 sin 18 +1.5 x10^5 V.
1.5 x 10^5 x 6.2 + 0 = 2.78 x10^5 x 5.1 sin 18 +1.5 x10^5 V.
1.5 x 6.2 + 0 = 2.78 x 5.1sin 18 +1.5 x V.
9.3 =4.381 = 1.5 v
V = -3.2791 m/s
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In the vertical direction the initial momentum of the two barges is
2.78 x 10^5x 4.3 + 0 .
Final momentum in this direction is 2.78 x10^5 x 5.1 cos 18 +1.5 x10^5 U.
2.78 x 10^5x 4.3 = 2.78 x10^5 x 5.1 cos 18 + 1.5 x10^5 U.
2.78 x 4.3 = 2.78 x 5.1 cos 18 + U.
11.954 = 13.484 = 1.5 v
V = -1.02 m/s
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Resultant of these two velocities
v = √ { (-3.2791)^2 + (-1.02)^2} = 3.434 m/s
tanθ = -1.02 / = -3.2791
θ = 17.28° from the negative x-axis down
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