In procedure 3: assume cart 1 has mass 575 g and cart 2 has mass 815 g. Assume t

Question

In procedure 3: assume cart 1 has mass 575 g and cart 2 has mass 815 g. Assume the track is frictionless, and the initial speed of cart 1 is 2.07 m/s. Find:

- v1, the speed of cart 1 after the collision m/s

- v2, the speed of cart 2 after the collision m/s

Please explain thoroughly :)

Explanation / Answer

from momentum conservation; 575*2.07 + 0 = 575*v1 + 815*v2 => 575*v1 + 815*v2 = 1190.25 => v1 = (1190.25 - 815*v2)/575 from energy conservation; 0.5*575*2.07^2 = 0.5*575*v1^2 + 0.5*815*v2^2 => 2463.82 = 575*[(1190.25 - 815*v2)/575]^2 + 815* v2^2 => 815*v2^2 + (1416695.06 + 664225*v2^2 - 1940107.5*v2)/575 = 2463.82 => 815*v2^2 + 2463.82 + 1155.174*v2^2 - 3374.1*v2 = 2463.82 => 1970.174*v2^2 - 3374.1*v2 = 0 => v2 = 1.713 m/s v1 = (1190.25 - 815*1.713)/575 => v1 = - 0.358 m/s

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