Momentum: Freight cars problem

[Sandro Romualdez]

Homework Statement

Three freight cars of equal mass are coupled together. The freight cars are traveling at 6.7m/s down a straight track when they collide with two more stationary identical cars. If all five cars are coupled together after the collision, then what is their speed if they are on frictionless tracks?

Homework Equations

Law of conservation of momentum: p=p' or m₁v₁+m₂v₂=m₁v₁'+m₂v₂'

The Attempt at a Solution

From the question I can figure out that
m₁ = 3x
m₂ = 2x
v₁ = 6.7m/s [Down]
v₂ = 0m/s

and I need to find v_sys'
the m_sys after the collision will equal 5x

am I right if I use:
m₁v₁+m₂v₂=m_sysv_sys'
→ (3x)(6.7m/s [D])+(2x)(0m/s) = 5x(v_sys')
→ 20.1x kg·m/s = 5x(v_sys')
→ v_sys' = 4.02 m/s [D]

and the speed after the collision will be 4.02m/s [Down]?

 

[Merlin3189]

Homework Statement

Three freight cars of equal mass are coupled together. The freight cars are traveling at 6.7m/s down a straight track when they collide with two more stationary identical cars. If all five cars are coupled together after the collision, then what is their speed if they are on frictionless tracks?
...

The Attempt at a Solution

From the question I can figure out that
m₁ = 3x
m₂ = 2x
v₁ = 7.5m/s [Down]
v₂ = 0m/s

and I need to find v_sys'
the m_sys after the collision will equal 5x

am I right if I use:
m₁v₁+m₂v₂=m_sysv_sys'
→ (3x)(7.5m/s [D])+(2x)(0m/s) = 5x(v_sys')
→ 24.5 kg·m/s = 5x(v_sys')
→ v_sys' = 4.9 m/s [D]

and the speed after the collision will be 4.9m/s [Down]?

I don't see why you swap from 6.7 to 7.5 for the initial velocity.

 

[Sandro Romualdez]

I don't see why you swap from 6.7 to 7.5 for the initial velocity.

Whoops, my mistake. I changed the speed now and calculated accordingly.

 

[Merlin3189]

So you should be right when you correct that.

 

[Sandro Romualdez]

So you should be right when you correct that.

Alright, thank you.