- #1

henry3369

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## Homework Statement

A 59.8kg person holding two 0.902kg bricks stands on a 2.14kg skateboard. Initially, the skateboard and the person are at rest.

Calculate the final speed of the person and the skateboard relative to the ground if the person throws the bricks one at a time. Assume that each brick is thrown with a speed of 16.7m/s relative to the person.

## Homework Equations

m1v1+m2v2 = m1v1'+m2v2'

## The Attempt at a Solution

0 = m1v1' + m2v2' + m3v3'

0 = (59.8+2.14)(v1') + (.902)(16.7)+(.902)(16.7)

v1' = 0.486 m/s

Why is this incorrect? The initial momentum is zero because they are at rest, then if I consider the system as the skateboard, bricks, and the person, the final momentum has to equal zero too. So the momentum of the two bricks, each moving at 16.7 m/s, and the momentum of the skateboard and the person should equal zero.