Final answer:
In this case,the joint velocity of the person and the boat after the person jumps off the dock and lands on the stationary boat is approximately 1.13 m/s.
Step-by-step explanation:
To determine the joint velocity of the person and the boat after the person jumps off the dock and lands on the stationary boat, we can apply the law of conservation of momentum.
1) Law of conservation of momentum: According to this law, the total momentum before an event is equal to the total momentum after the event, assuming there are no external forces acting on the system.
2) Momentum: Momentum is a vector quantity that describes the motion of an object. It is the product of an object's mass and velocity.
3) Calculation: To calculate the joint velocity, we can use the equation:
- Total initial momentum = Total final momentum
- Initial momentum of the person = mass of the person * velocity of the person
- Final momentum of the person and the boat = (mass of the person + mass of the boat) * joint velocity
So, we have:
(68.0 kg * 2.79 m/s) = (68.0 kg + 131 kg) * joint velocity
4) Solving for joint velocity: Rearranging the equation, we find:
joint velocity = (68.0 kg * 2.79 m/s) / (68.0 kg + 131 kg)
joint velocity ≈ 1.13 m/s
Therefore, the joint velocity of the person and the boat after the person jumps off the dock and lands on the stationary boat is approximately 1.13 m/s.