A boy on a 2.0 kg skateboard initially at rest tosses an 8.0 kg jug of water in the forward direction. If the jug has a speed of 3.0 m/s relative to the ground and the boy and the skateboard move in the opposite direction at 0.60 m/s, find the boy's mass.

Answer 1

First, we need to calculate the total momentum before and after the jug is thrown.

Initially, the total momentum is:
0 = 2.0 kg * 0 m/s + 8.0 kg * 0 m/s

After the jug is thrown, the total momentum is:
(2.0 kg + boy's mass) * (-0.60 m/s) + 8.0 kg * 3.0 m/s

Setting the two equations equal to each other, we get:
(2.0 kg + m) * (-0.60 m/s) + 8.0 kg * 3.0 m/s = 0

Expanding and solving for the boy's mass, we get:
-1.2 kg - 0.60m + 24 kg = 0
23.4 kg = 0.60m
m = 39 kg

Therefore, the boy's mass is 39 kg.