Jumping off the front of a boat

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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?


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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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  • #3
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.
 
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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, [itex] \rm v_i[/itex] = .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 [itex] \rm v_f + v[/itex] 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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