Arailroad freight car rolls on a track at 3.50 m/s toward two identical coupled freight cars, which are rolling in the same direction as the first, but at a speed of 1.20 m/s. the first reaches the second two and all couple together. the mass of each is 2.65 ✕ 104 kg. (a) what is the speed (in m/s) of the three coupled cars after the first couples with the other two?

Explanation:

Hello! To solve this problem we must be clear about the concept of energy conservation, and kinetic energy with the following sentence

The kinetic energy of the two cars (v = 1.2m / S) plus the kinetic energy of the third car (v = 3.5m / S) must be equal to the kinetic energy of the three cars together.

The kinetic energy is calculated by the following equation.

m= mass of the cars=26500kg

V=speed

E=kinetic energy

taking into account the above, the following equation is inferred

1=  the cars are separated

2= the cars are togheter

E1=E2

where

m= mass of each car

V1= 1.2m/s

Va=3.5,m/S

m= mass of each car

V=speed (in m/s) of the three coupled cars after the first couples with the other two

Solving

the speed  of the three coupled cars after the first couples with the other two is 2.245m/s

1.97 m/s

Explanation:

Recall that from the principle of conservation of energy, p = mv

----- eqn 1

Total momentum before collision equals total momentum after collision

sub mv for p in eqn 1

----- eqn 2

Parameters

Where is the speed of the three coupled cars

substitute for all m and v in eqn 2

=

= 1.9667 m/s

Therefore, the speed of the three cars after coupling is 1.97m/s

does anyone know or what?