Conservation of Momentum and Relative Velocity

AI Thread Summary
A 55kg student on a 4.6 kg surfboard is moving at 2.0 m/s east while walking at 1.9 m/s east relative to the board. The resultant velocity of the surfboard relative to the water needs to be calculated using the conservation of momentum. The initial approach mistakenly assumed the board's velocity remained constant while the student walked, which is incorrect. The correct equation incorporates the board's changing velocity as the student walks. The final answer for the surfboard's velocity relative to the water is 0.25 m/s east.
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Homework Statement



A 55kg student stands on a 4.6 kg surfboard moving at 2.0 m/s [E]. The student then walks with a velocity of 1.9 m/s[E] relative to the surfboard. Determine the resultant velocity of the surfboard, relative to the water. Neglect Friction.
Answer: 0.25 m/s[E]

Homework Equations



Conservation of momentum (P = P')

The Attempt at a Solution



I plugged in the values given (relative to the water) into the equation, but I'm not getting the right answer ?

vGoBS.jpg


ms = mass of student
mb = mass of board
vs = 2.0m/s
vb = 2.0m/s
vs' = 1.9m/s + 2.0m/s
vb' = ?
 
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whoareyou said:

Homework Statement



A 55kg student stands on a 4.6 kg surfboard moving at 2.0 m/s [E]. The student then walks with a velocity of 1.9 m/s[E] relative to the surfboard. Determine the resultant velocity of the surfboard, relative to the water. Neglect Friction.
Answer: 0.25 m/s[E]

Homework Equations



Conservation of momentum (P = P')

The Attempt at a Solution



I plugged in the values given (relative to the water) into the equation, but I'm not getting the right answer ?

vGoBS.jpg
Your top equation is correct. But before solving for v'B, you'll need to correct a mistake.

(The bottom equation is technically correct too, but it's too early to solve for v'B. Solving for it at this point doesn't make the problem easier.)
ms = mass of student
mb = mass of board
vs = 2.0m/s
vb = 2.0m/s
vs' = 1.9m/s + 2.0m/s
vb' = ?

The error is your
vs' = 1.9m/s + 2.0m/s​
equation. That's not right.

The way you've expressed the equation, the board maintains its 2.0 m/s velocity even after the surfer starts walking. But that's not right. The board has a different velocity when the surfer is walking. You need to replace that 2.0 m/s with with the new velocity of the board. :wink: (Hint: the new velocity of the board is a variable that you haven't solved for just yet.)
 
But isn't the new velocity of the board what we're trying to solve for?
 
whoareyou said:
But isn't the new velocity of the board what we're trying to solve for?
Yes, that's right! :approve:

(Another hint: you have a little more algebra to do. :wink:)
 
So then if 2.0m/s is not the final velocity of the surfboard, then it would be 1.9m/s + vb' ?
 
whoareyou said:
So then if 2.0m/s is not the final velocity of the surfboard, then it would be 1.9m/s + vb' ?
No not quite.

But the student is walking at 1.9 m/s relative to the board.

So that means the velocity of the student is v's = 1.9 m/s + v'b.

[Edit: I was calling the student a 'surfer'. Changed it back to 'student' in this post.]
 
It works out! Thanks a lot! :D

o9zAl.jpg
 
There ya' go. :approve:
 
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