Final answer:
After a completely inelastic collision, the two cars move together at a speed of approximately 14.43 m/s at an angle of 40.2° north of east, determined by conservation of momentum and vector addition.
Step-by-step explanation:
The question pertains to the calculation of the final velocity and direction of two cars after a collision, using the principles of conservation of momentum. Since the two cars stick together after the collision, this is an example of a completely inelastic collision. To find the speed and direction of the cars after the collision, we can apply the conservation of linear momentum separately in the north-south and east-west directions.
First, we calculate the momentum of each car before the collision:
1400 kg car (northward): p1 = m1 * v1 = 1400 kg * 24.0 m/s = 33600 kg*m/s northward
2200 kg car (eastward): p2 = m2 * v2 = 2200 kg * 18.0 m/s = 39600 kg*m/s eastward
Since the cars stick together, their combined mass is m_total = 1400 kg + 2200 kg = 3600 kg.
Using vector addition, we calculate the combined momentum vector:
p_total = sqrt((33600^2) + (39600^2)) = sqrt(1131360000 + 1568160000) = sqrt(2699520000) = 51957 kg*m/s
For the direction, we calculate the angle θ with the east axis using arctan(northward momentum / eastward momentum):
θ = arctan(33600/39600) = arctan(0.8485) ≈ 40.2° north of east
Finally, we find the speed of the cars after the collision by dividing the total momentum by the total mass:
Speed after collision = p_total / m_total = 51957 kg*m/s / 3600 kg ≈ 14.43 m/s
Hence, the two cars move together at a speed of approximately 14.43 m/s at an angle of 40.2° north of east after the collision.