A 13000-kg railroad car travels alone on a level frictionless track with a constant speed of 24.0 m/s...

A 13000-kg railroad car travels alone on a level frictionless track with a constant speed of 24.0 m/s. A 7450-kg load, initially at rest, is dropped onto the car.

What will be the car's new speed?

Any and all help is much appreciated! Thanks so much in advance!

this is a conservation of momentum problem

the car that is dropped has no initial horizontal momentum, so the horizontal momentum of the system is conserved, this means that

M v = (M+m) V

where M = 13000kg, v = 24m/s , m = 7450 kg and V is the speed of the combined system after the car is dropped onto the first car

so we have

V = 1300kg*24m/s / (13000kg + 7450 kg) = 15.3m/s

Use the principle of momentum conservation:

Momentum of initial state (before load is in car) = Momentum of final state (load in car)

Using the x: direction we have

m1v1+m2v2=(m1+m2)v3
Where:
m1=mass of car
v1= speed of car in x-direction
m2 = mass of load
v2 = speed of load in x-direction (hint: it's zero)
V3 = speed of the car+load object