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

with the invention of the automobile, the woman of the house could drive to the store and get groceries. women were also free to socialize more. during world war ii women were allowed to work in factories to the war effort and could drive to work or ride mass transit buses to their jobs.

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.