Velocity of boat after two people have jumped off

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The discussion focuses on calculating the final velocity of a boat after two individuals, each with mass m, jump off with a horizontal velocity u relative to the boat's initial velocity. The participants analyzed two scenarios: (a) both individuals jumping together and (b) jumping one after the other. The consensus is that the final velocity of the boat is greater when both jump together due to the conservation of momentum principles. The calculations revealed that the final speed in case (b) is lower than in case (a), which aligns with the expected outcomes based on momentum transfer dynamics.

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Original question: Two people (each of mass m) are standing on the stern of a boat of mass M. Ignoring friction, find the boat’s velocity v after they jump out with a horizontal velocity u relative to the boat’s initial velocity
(a) if they jump together
(b) if they jump one after another
(c) which velocity is greater?

I followed the path of using equal total momentum before and after to solve the problem, and I just but for some reason, my (a) velocity expression is greater than my (b) velocity expression. Is this supposed to happen? I assumed this is like a multi-stage system, where final velocity should be higher if there are more stages.

My set up is (M+2m)vi = 2m(vi - u) + M(vf) where vi is initial velocity of boat and vf is final (what I am looking for)
(vi - u) would be velocity of the two people relative to water.. right?
 
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QuanH said:
Original question: Two people (each of mass m) are standing on the stern of a boat of mass M. Ignoring friction, find the boat’s velocity v after they jump out with a horizontal velocity u relative to the boat’s initial velocity
(a) if they jump together
(b) if they jump one after another
(c) which velocity is greater?
.

My set up is (M+2m)vi = 2m(vi - u) + M(vf) where vi is initial velocity of boat and vf is final (what I am looking for)
(vi - u) would be velocity of the two people relative to water.. right?
It is true when the two people jump together. But in case b, the second man jumps with relative velocity u with respect to the new velocity of the boat.

ehild
 
yes i took that into account too. i divided (b) into two parts, 1st jump and 2nd jump, where the 2nd jump's initial velocity is the final velocity of the 1st jump
 
QuanH said:
yes i took that into account too. i divided (b) into two parts, 1st jump and 2nd jump, where the 2nd jump's initial velocity is the final velocity of the 1st jump

Show your work, please. The final speed must be higher in case b.

ehild
 
ehild said:
Show your work, please. The final speed must be higher in case b.

ehild

This looks counter intuitive to me. If they jump together, they will both carry away momentum -mu in the initial boat rest frame. In case b only the first will while the second carries momentum m(v1-u) where v1 is the velocity of the boat after the first guy jumps. This is of course assuming that they still jump with the same velocity relative to the initial boat velocity before their jump which may not be physically reasonable.
 
Your answer sounds right. When the first person jumps, he is adding velocity to both the boat and the second person.
 
(Editing doesn't seem to be working yet.)

This is different than a two stage rocket problem because when people jump off they impart energy to the boat after they remove their weight from it.
 
ahh that makes sense now. thank you all.
 
this is my work if any is still interested to see
a.jpg
b.jpg
 
  • #10
Orodruin said:
This looks counter intuitive to me. If they jump together, they will both carry away momentum -mu in the initial boat rest frame. In case b only the first will while the second carries momentum m(v1-u) where v1 is the velocity of the boat after the first guy jumps. This is of course assuming that they still jump with the same velocity relative to the initial boat velocity before their jump which may not be physically reasonable.

It is right, I made some silly mistake, thespeed is bigger in case a).

ehild
 

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