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## Homework Statement

The problem says:

A mine provides a transport ystem constiting into carriages (of neglect mass) hanging on a riel free of friction.

In a specific moment two people of mass m=45kg are travelling when one of them says his breaks are not working. In that moment he was at V1=10m/s whilst the carriage ahead was at V2=8.0m/s.

Fortunately who travells in the second carriage can take a mass of 25Kg of sand and throw a fraction of it with a velocity V0=6.0m/s (respect to him).

Calculate the MINIMUM mass that the man will have to throw to avoid the colission.

** The "correct" answer is 23Kg

## Homework Equations

## The Attempt at a Solution

Well, obviously this is a conservation of linear momentum problem but I think it's not 23Kg. Here's what I did:

[tex] \displaystyle {{v}_{mass}}={{v}_{0}}+{{v}_{back}}[/tex]

That's the speed of the mass respect to the Earth. Vback is the additional speed the man should have when throwing the mass. If it asks the minimum then the man has to have a final speed of 10m/s, and therefore the "backward" speed should be 2m/s.(8+2=10):

[tex] \displaystyle {{v}_{mass}}=-6.0+2.0=-4.0[/tex]

[tex] \displaystyle {{p}_{i}}={{v}_{2i}}\left( m+{{m}_{x}} \right)[/tex]

[tex] \displaystyle {{p}_{f}}=\left( m+{{m}_{x}} \right){{v}_{2f}}+{{m}_{x}}\left( {{v}_{mass}} \right)[/tex]

m is 45Kg (man mass) and mx is the unknown thrown mass.

Notice that my initial mass is the unknown thrown mass plus the mass of the man, but solving for those values I get another answer BUT if I consider the initial mass as only the mass of the man I get 22.5 which is the correct answer. But should I consider both mass?

Thanks!

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