Question

question 8
a 73.6 kg person jumps along a dock at 5.92 m/s and jumps into a 700 kg rowboat. what is the final velocity of the boat and its new occupant as they drift away from the dock?
1 point
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Explanation:

Step1: Apply the law of conservation of momentum

The initial momentum of the person is $p_{i}=m_{1}v_{1}$, where $m_{1} = 73.6\ kg$ and $v_{1}=5.92\ m/s$. The initial momentum of the boat is $p_{i - boat}=0$ (assuming it is initially at rest). According to the law of conservation of momentum $p_{i}=p_{f}$, and $p_{f}=(m_{1}+m_{2})v_{f}$, where $m_{2}=700\ kg$ is the mass of the boat.

Step2: Solve for the final velocity $v_{f}$

We have $m_{1}v_{1}=(m_{1}+m_{2})v_{f}$. So, $v_{f}=\frac{m_{1}v_{1}}{m_{1}+m_{2}}$. Substitute $m_{1} = 73.6\ kg$, $v_{1}=5.92\ m/s$ and $m_{2}=700\ kg$ into the formula:
\[v_{f}=\frac{73.6\times5.92}{73.6 + 700}\]
\[v_{f}=\frac{435.712}{773.6}\]
\[v_{f}\approx0.563\ m/s\]

Answer:

$0.563\ m/s$