Jumping off the front of a boat

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A passenger jumps off the front of a moving boat at a velocity of 4.0 m/s relative to the boat, which is traveling at 0.5 m/s. The discussion revolves around calculating the total velocity of the passenger relative to the ground, with some participants asserting the answer should be 5.0 m/s based on momentum transfer. However, others argue that the maximum possible speed should be 4.5 m/s due to the mass ratio of the passenger and the boat. The passenger's mass is 72 kg and the boat's mass is 1000 kg, leading to confusion regarding the correct application of momentum conservation principles. The thread highlights the need for clarity in the problem's parameters and calculations.
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Homework Statement


A boat is moving at .5 m/s relative to the ground with a passenger riding inside of the boat. If the passenger jumps off of the front of the boat at a velocity of 4.0 m/s (relative to the boat), what is the total velocity of the passenger relative the the ground/water?


Homework Equations





The Attempt at a Solution


I know the answer should be 5.0 m/s, but I am having trouble explaining this. I believe it comes from the momentum of the passenger along with the momentum of the boat being transferred to the passenger when he jumps giving him a velocity of 1.0 m/s + the 4.0 giving a total of 5.0 m/s. Can someone better explain this?
 
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Does this problem give the mass of the passenger and the mass of the boat?

I don't think it's possible for the speed of the passenger, relative to the ground, to be greater than 4.5 m/s.
If the mass of the boat M is much larger than the mass of the passenger m, m << M, than the greatest speed possible for the passenger, relative to the ground, for M->Infinity, should be 4.5 m/s.
 
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That is what I was thinking too, but the answer key to the book says 5.0m/s, but it doesn't explain why. Yeah, the mass of the passenger is 72 kg. The mass of the boat is 1000 kg.
 
fish399 said:
That is what I was thinking too, but the answer key to the book says 5.0m/s, but it doesn't explain why. Yeah, the mass of the passenger is 72 kg. The mass of the boat is 1000 kg.

Good Morning,

I just happened to be logged on...

Here's the picture I've drawn for the problem:

http://img15.imageshack.us/img15/4843/boat1st.jpg

In this picture, \rm v_i = .5 m/s, v = 4 m/s, m = 72 kg, and M = 1000 kg.

I get 4.23 m/s for the final speed of the passenger relative to the ground (a frame at rest), which corresponds to \rm v_f + v in my picture.

I messed up a similar problem last week, so I might be making an error here. Perhaps someone else will take a look.
 
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fishy welcome to pf!

hi fish399! welcome to pf! :smile:
fish399 said:
That is what I was thinking too, but the answer key to the book says 5.0m/s, but it doesn't explain why. Yeah, the mass of the passenger is 72 kg. The mass of the boat is 1000 kg.

use conservation of momentum, together with vp - vb = 4.0 :wink:

(but I don't see how it can be more than 4.5 :confused: … are you sure all the numbers in the question are correct?)
 
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