Law of Conservation of Momentum and Jumping Off a Boat with Constant Velocity

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In summary, the correct equation for conservation of momentum in this scenario is ##m_1v_{1f} + m_2v_{2f} = 0##, with the initial velocity of both the boat and person being zero.
  • #1
pierce15
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If a person jumps off of a boat with constant velocity, should the equation resulting from the law of conservation of momentum be (where 1 denotes the mass/velocity of the person, 2 denotes those of the boat, and f denotes a final velocity):

m1 * 0 + m2 * v2 = m2 * v2f + m1 * v1f

or

m1* 0 + m2 * v2 = (m2-m1) * v2f + m1* v1f
 
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piercebeatz said:
If a person jumps off of a boat with constant velocity, should the equation resulting from the law of conservation of momentum be (where 1 denotes the mass/velocity of the person, 2 denotes those of the boat, and f denotes a final velocity):

m1 * 0 + m2 * v2 = m2 * v2f + m1 * v1f

or

m1* 0 + m2 * v2 = (m2-m1) * v2f + m1* v1f

Neither. The left hand side is wrong because the initial momentum is ##(m_1+m_2)v_2## and that's what has to be equal to the final momentum ##m_1v_{1f}+m_2v_{2f}##.

These problems are easiest to work if you start with the total momentum equal to zero, which in this case means that the initial velocity of the boat and person is zero. You can make the problem turn out that way just by using the speeds measured by a second boat that is pacing the first boat (relative speed zero) until the person jumps. Working with that definition, it's easy: ##m_{1}v_{1f} + m_{2}v_{2f} = 0##. Then you just have to remember to add the speed of the observer back in if you want to get the speed of the boat and/or the person relative to the shore.
 

1. How does the Law of Conservation of Momentum apply to jumping off a boat with constant velocity?

The Law of Conservation of Momentum states that the total momentum of a closed system remains constant. In the case of jumping off a boat with constant velocity, the person and the boat are considered a closed system. This means that the total momentum of the person and the boat before the jump must be equal to the total momentum after the jump.

2. Does the momentum of the person change when they jump off a boat with constant velocity?

According to the Law of Conservation of Momentum, the momentum of the person does not change when they jump off a boat with constant velocity. This is because the total momentum of the system remains constant, so any change in the momentum of the person will be offset by an equal and opposite change in the momentum of the boat.

3. What factors affect the momentum of the person when jumping off a boat with constant velocity?

The momentum of the person when jumping off a boat with constant velocity is affected by their mass and velocity. A person with a higher mass will have a greater momentum, while a person with a higher velocity will also have a greater momentum. Additionally, the momentum of the boat and the direction in which it is moving also play a role in the momentum of the person.

4. Can the person change their momentum while jumping off a boat with constant velocity?

No, the person cannot change their momentum while jumping off a boat with constant velocity. This is because of the Law of Conservation of Momentum, which states that the total momentum of a closed system remains constant. Any change in the momentum of the person would result in a corresponding change in the momentum of the boat, thus keeping the total momentum of the system constant.

5. Are there any safety concerns when jumping off a boat with constant velocity?

Yes, there are safety concerns when jumping off a boat with constant velocity. The person must ensure that they have a clear understanding of the direction and speed of the boat before jumping off to avoid any collisions or accidents. Additionally, the person should also be aware of their own momentum and make sure to land safely in the water to avoid any injuries.

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