Answered

Three identical train cars, coupled together, are rolling east at speed v0. A fourth car traveling east at 2v0 catches up with the three and couples to make a four-car train. A moment later, the train cars hit a fifth car that was at rest on the tracks, and it couples to make a five-car train.

Answer :

Answer:

[tex]v_f = v_o[/tex] Towards East

Explanation:

As we know that whole system is moving without any external force on it

So here we can apply momentum conservation

Now we have

[tex]P_i = P_f[/tex]

[tex](3m) v_o + m(2v_o) = 4m(v)[/tex]

[tex]v = \frac{5}{4} v_o[/tex]

so above is the speed of four coupled train

Now when this is coupled with fifth stationary car then we have

[tex]4m v = (4m + m) v_f[/tex]

[tex]4m(\frac{5}{4}v_o) = 5m v_f[/tex]

[tex]v_f = v_o[/tex] Towards East                          

okpalawalter8

The speed of the five-car train is mathematically given as

V=Vo

What is the speed of the five-car train?

Question Parameter(s):

Three identical train cars, coupled together, are rolling east at speed v0

A fourth car traveling east at 2v0

the train cars hit a fifth car that was at rest = 0

Generally, the equation for the Initial linear momentum  is mathematically given as

Pi

Pi= 3m x Vo + m x 2Vo + 5m x 0

Pi=5 m Vo

final linear momentum Pf

Pf= 5m x V

In conclusion, linear momentum will be conserved therefore

Pi=Pf

5 m Vo=5 m V

V=Vo

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