Bumper car A (67.3 kg) moves at 5.65 m/s left before colliding elastically with Car B (65.2 kg), which was initially moving 3.20 m/s right. After the collision, Car B moves to the left at 5.38 m/s. Find the velocity of car A after the collision.

2.66 m/s to the right

Explanation:

Applying,

The Law of conservation of momentum,

For elastic collision,

mu+m'u' = mv+m'v' Equation 1

Where m = mass of Car A, m' = mass of car B, u = initial velocity of car A, u' = Initial velocity of car B, v = Final velocity of car A, v' = Final velocity of car B

make v the subject of the equation

v = (mu+m'u'-m'v')/m Euqation 2

Note: Assuming Left direction is positive and right direction is negative.

Given: m = 67.3 kg, m' = 65.2 kg, u = 5.65 m/s, u' = -3.2 m/s, v' = 5.38 m/s

Substitute these values into equation 2

v = [(67.3×5.65)+(65.2×[-3.2])-(65.2×5.38)]/67.3

v = (380.245-208.64-350.776)/67.3

v = -179.17/67.3

v = -2.66

Hence v = 2.66 m/s to the right

the total distance traveled by the car is

30km + 49km = 79 km .