A Momentum Problem

  • #1
Hyperreality
202
0
A peron on a boat with a total mass of M are stationary in the river, the man throws his shoes mass m at a velocity v successive, find the expression for the boat's final velocity.

Here is what I did.

First let u be the velocit of the boat
After the first shoes:
(M + m)u = mv
u = mv(1/(M +m))

After the second shoes:
Change in momentum of the boat = Change in momentum of the shoe

So, Mu'- (M + m)u = mv, but (M + m)u = mv

Therefore Mu' = 2mv and u' = 2mv/M

u + u' = mv(2/M + 1/(M + m)) is the final velocity of the boat.

But the correct answer is mv(1/(M + m) + 1/(M + 2m)), can anyone please tell me where I've got it wrong?
 

Answers and Replies

  • #2
arcnets
508
0
Originally posted by Hyperreality

Change in momentum of the boat = Change in momentum of the shoe

So, Mu'- (M + m)u = mv
The 2nd shoe ist not at v relative to the water, since it's thrown from a moving boat.
 
  • #3
Hyperreality
202
0
So this is the correct expression?

Mu'- (M + m) = m(u + v)? Because I still can't find the correct expression...
 
  • #4
arcnets
508
0
Originally posted by Hyperreality
Because I still can't find the correct expression...

Nor can I reproduce that answer.
There's also something unclear in the problem: It says 'total mass M'. Does that include the shoes, or not?

Let u be the boat velocity after 1st throw.
Let u' be the boat velocity after 2nd throw.

If we assume that M includes the shoes, then

I. (M-m)u = mv
II. (M-2m)u' - (M-m)u = m(v-u)

This leads to
u' = mv(1/(M - m) + 1/(M - 2m)).

Now if M doesn't include the shoes, this will sure mean the transformation
M -> M+2m, yielding
u' = mv(1/(M + m) + 1/M).

I can't see how to get closer to the 'correct' answer.
 

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