Question

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)

Answer 1

Option A ) 1m/s

Step-by-step 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