Using the conservation of momentum principle in Physics, the final velocity of the three coupled boxcars after the collision is 2.0 m/s. Option 1 is correct.
The scenario described involves the conservation of momentum, which is a key principle in Physics. When two boxcars moving at a certain velocity collide with and couple to a third stationary boxcar, the final velocity of the three coupled boxcars can be determined by ensuring the total momentum before and after the collision is the same, assuming no external forces are acting on the system. Since the third boxcar is stationary, its initial momentum is zero.
Final velocity (Vf) can be calculated by the equation:
M₁V₁,i + M₂V₂,i + M₃V₃,i = (M₁ + M₂ + M₃)Vf
Assuming the mass of the boxcars (M₁, M₂, and M₃) is the same for all and using the given velocities, we can solve for the Vf.
2M(3.0 m/s) + M(0 m/s) = 3MVf
6.0M = 3MVf
Vf = 2.0 m/s
Therefore, the resulting velocity of the three coupled boxcars is 2.0 m/s.
Hence, 1. is the correct option.
--The given question is incomplete, the complete question is
"Two coupled boxcars are rolling along at 3.0 m/s when they collide with and couple to a third, stationary boxcar. What is the resulting velocity of the three coupled boxcars? 1) 2.0 m/s 2) 6.0 m/s 3) 9.0 m/s 4) 12.0 m/s"--