A railroad flatcar of mass 2000 kg Rolls to the right at 10 m/s and collides with a flatcar of mass 3000 kg that is rolling to the left at 5 m/s. The flatcars couple together. Their speed after the collision is

A 1 m/s

B 2.5 m/s

C 5 m/s

D 7 m/s

E 7.5 m/s

A (1 m/s)

Option A) 1m/s

Explanation:

For this case, we can consider flatcar going from left to right:

M₁ = 2000 Kg

V₁ = 10 m/s

And for flatcar going from right to left:

M₂ = 3000 Kg

V₂ = 5 m/s

We can first consider our reference system as: everything going from left to right will have a positive value, and everything going from right to left will have a negative value

When we have a plastic crash, and both bodies stick together, we can consider linear momentum conservation:

M₁V₁ + M₂V₂ = (M₁ + M₂) Vₓ where Vₓ = final speed of of both objects that couple together

Then: Vₓ = (M₁V₁ + M₂V₂) / (M₁ + M₂)

We now can replace our above numbers, considering that, for flatcar number 2, we should consider a speed with a negative value, according to our reference system (as it goes from right to left):

Vₓ = (2000Kgx10m/s - 3000Kgx5m/s) / (2000Kg + 3000Kg)

Vₓ = 1 m/s

So both flatcars will remain togheter, with a final speed (Vₓ) of 1 m/s, moving from left to right