Answer:
40 m/s
Step-by-step explanation:
We can use the principle of conservation of momentum to solve this problem. According to this principle, the total momentum of a closed system remains constant if there are no external forces acting on it. In this case, the two cars form a closed system, since they are the only objects involved in the collision and there are no external forces acting on them.
The total momentum of the system before the collision can be calculated as:
p1 = m1v1 + m2v2
where m1 and m2 are the masses of the two cars, v1 is the velocity of car A before the collision, and v2 is the velocity of car B before the collision. Plugging in the given values, we get:
p1 = (2000 kg)(10 m/s) + (500 kg)(v2)
p1 = 20,000 kg m/s + 500v2
The total momentum of the system after the collision is zero, since the cars come to a stop. Therefore, we have:
p2 = m1v1' + m2v2' = 0
where v1' and v2' are the velocities of the two cars after the collision. Since car B comes to a stop, we have v2' = 0. Substituting this into the above equation and solving for v1', we get:
v1' = -m2v2' / m1 = -500 kg × 0 / 2000 kg = 0 m/s
This means that car A also comes to a stop after the collision. Now we can use the conservation of momentum equation to solve for the initial velocity of car B:
p1 = p2
20,000 kg m/s + 500v2 = 0
500v2 = -20,000 kg m/s
v2 = -40 m/s
The negative sign for the velocity indicates that car B was traveling in the opposite direction to car A before the collision. To get the magnitude of the velocity, we can take the absolute value:
|v2| = 40 m/s
Therefore, the velocity of car B before the collision was 40 m/s in the opposite direction to car A.