To solve this problem, we can use the principle of conservation of momentum. The total momentum before the jug is tossed is equal to the total momentum after the jug is tossed.
Let's denote the initial momentum of the boy and skateboard as \( p_{\text{initial}} \) and the final momentum of the boy and skateboard as \( p_{\text{final}} \).
The initial momentum is:
\[ p_{\text{initial}} = 0 \, \text{kg m/s} \]
The final momentum can be calculated as:
\[ p_{\text{final}} = (M_{\text{boy}} + M_{\text{skateboard}}) \cdot V_{\text{boy/skateboard}} + M_{\text{jug}} \cdot V_{\text{jug}} \]
Where:
- \( M_{\text{boy}} \) is the mass of the boy
- \( M_{\text{skateboard}} \) is the mass of the skateboard
- \( M_{\text{jug}} \) is the mass of the jug
- \( V_{\text{boy/skateboard}} \) is the speed at which the boy and skateboard are moving after the jug is tossed
- \( V_{\text{jug}} \) is the speed at which the jug is moving after being tossed
We are given that:
- \( M_{\text{skateboard}} = 2.0 \, \text{kg} \)
- \( M_{\text{jug}} = 8.0 \, \text{kg} \)
- \( V_{\text{boy/skateboard}} = -0.60 \, \text{m/s} \) (opposite direction to the jug)
- \( V_{\text{jug}} = 3.0 \, \text{m/s} \)
Now we can substitute these values into the equation for the final momentum and solve for the mass of the boy:
\[ p_{\text{final}} = (M_{\text{boy}} + 2.0) \cdot (-0.60) + 8.0 \cdot 3.0 \]
\[ 0 = (-0.60M_{\text{boy}} - 1.20) + 24.0 \]
\[ 0 = -0.60M_{\text{boy}} + 22.80 \]
\[ 0.60M_{\text{boy}} = 22.80 \]
\[ M_{\text{boy}} = \frac{22.80}{0.60} \]
\[ M_{\text{boy}} = 38 \, \text{kg} \]
Therefore, the mass of the boy is 38 kg.
4. 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.
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