A boy on a 1.9 kg skateboard initially at rest tosses a(n) 8.6 kg jug of water in the forward direction. If the jug has a speed of 3.1 m/s relative to the ground and the boy and skateboard move in the opposite direction at 0.61 m/s, find the boy's mass. Answer in units of kg.

Since the initial momentum is equal to the final momentum (Pi=Pf) then the equation used would be MaVa+MbVb = MaVa+MbVb. But, this problem has 3 masses so i don't know what to do. Can any one help me?

## Answers (1)

You do it as if the skateboard and the boy are one item, and then you can subtract from it afterwards.

a = jug

b = skateboard + boy

initial momentum = final momentum

MaVa + MbVb = MaVa + MbVb

initial momentum = 0 kg·m/s (because everything is at rest)

0 = MaVa + MbVb

0 = (8.6 kg)(3.1 m/s) + (Mb)(-0.61 m/s) <-- negative because in opposite direction

0 = 26.66 kg·m/s + Mb(-0.61 m/s)

-26.66 kg·m/s = (-0.61 m/s)Mb

Mb = 43.7 kg

Mb = mass of skateboard + mass of boy

43.7 kg = 1.9 kg + mass of boy

mass of boy = 41.8 kg = 42 kg