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Answer:
read explanation hope it helps btw
Explanation:
To find the velocity of the cars after the impact, we can use the principle of conservation of momentum. The total momentum before the impact is equal to the total momentum after the impact.
Let’s consider the momentum of the first car (car A) before the impact:
P1 = m1 * v1
where m1 is the mass of car A and v1 is its velocity due north.
Given:
m1 = 500 kg
v1 = 20 m/s due north
The momentum of car A before the impact is:
P1 = (500 kg) * (20 m/s) = 10000 kg·m/s due north
Now, let’s consider the momentum of the second car (car B) before the impact:
P2 = m2 * v2
where m2 is the mass of car B and v2 is its velocity due west.
Given:
m2 = 500 kg
v2 = 30 m/s due west
The momentum of car B before the impact is:
P2 = (500 kg) * (30 m/s) = 15000 kg·m/s due west
Since the cars stick together after the impact, their combined mass is m1 + m2 = 1000 kg (mass of car A + mass of car B).
Now, let’s calculate the resultant momentum after the impact:
P’ = (m1 + m2) * v’
where v’ is the velocity of the cars after the impact.
Given:
P’ = 18,030 N·s
v’ = 18 m/s at 34° north of west
We can break down the velocity v’ into its north and west components. The north component will be v’ * sin(34°) and the west component will be v’ * cos(34°).
Let’s calculate the north and west components of the velocity:
v’_north = 18 m/s * sin(34°)
v’_west = 18 m/s * cos(34°)
Now, let’s calculate the resultant velocity after the impact:
P’ = (m1 + m2) * v’
v’ = P’ / (m1 + m2)
Substituting the given values:
v’_north = 18 m/s * sin(34°) ≈ 9.68 m/s north
v’_west = 18 m/s * cos(34°) ≈ 14.88 m/s west
Finally, the velocity of the cars, after the impact, is approximately 9.68 m/s north and 14.88 m/s west.