Question

A 20,000 kg railroad car travels down the track at 30 m/s and hits a second stationary railroad car of 10,000 kg. The

Answer 1

Answer:

Explanation:

We have , A 20,000 kg railroad car travels down the track at 30 m/s and hits a second stationary railroad car of 10,000 kg. The  cars become coupled and continue traveling. The law of momentum conservation: For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. In this question we will use law of conservation of momentum as :

momentum before collision = momentum after collision

$m_{1} v_{1} + m_{2} v_{2} = m_{1} v_{12} + m_{2} v_{23}$

⇒ $20,000kg(30m/s) + 10,000kg(0m/s) = 20,000kg(v m/s)+ 10,000kg(v m/s)$

⇒ $600,000 = ( 20,000 +10,000 )(v)$

⇒ $30,000(v) = 600,000$

⇒ $(v) = 20m/s$

Therefore, final velocity of both cars are $(v) = 20m/s$.