A 1234 kg freight car moving at 6 m/s runs into a 2468 kg freight car at rest. They stick together upon collision. What was the final combined speed?

2 m/s

Explanation:

Applying,

The law of conservation of momentum

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m') Equation 1

Where m = mass of the first freight car, m' = mass of the second freight car, u = initial velocity of the first freight car, u' = initial velocity of the second freight car, V = final combined velocity/ speed.

make V the subject of the equation

V = (mu+m'u')/(m+m') Equation 2

From the question,

Given: m = 1234 kg, m' = 2468 kg, u = 6 m/s, u' = 0 m/s (at rest)

Substitute these values into equation 2

V = [(1234×6)+(2468×0)]/(1234+2468)

V = 7404/3702

V = 2 m/s

truck y has a greater velocity than truck x.