#### JJBladester

Gold Member

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**1. The problem statement, all variables and given/known data**

A 180-lb man and a 120-lb woman stand side by side at the same end of a 300-lb boat, ready to dive, each with a 16-ft/s velocity relative to the boat. Determine the velocity of the boat after they have both dived, if

*(a)*the woman dives first,

*(b)*the man dives first.

__Answers:__

(a) 9.20 ft/s (to the left)

(b) 9.37 ft/s (to the left)

**2. Relevant equations**

F=ma

**3. The attempt at a solution**

What does "...each with a velocity of 16-ft/s relative to the boat..." mean? If the man and woman are both standing on the boat, wouldn't their velocities relative to the boat be 0-ft/s?

Is 16-ft/s the velocities of each of them w.r.t. the boat afteir their respective dives?

Do they dive straight out or with x- and y- components (as in a ballistics eqn)?

Do we assume the motion (diving, boat's reaction) is all in the x-direction?

Aside from the question being highly vague, I've attempted (and failed) to solve the problem using impulse-momentum equations below.

____________________________________________________________

**(Woman jumps first, creating an action-reaction pair of forces F**

_{w}and -F_{w}. So, she jumps and pushes off with a force equal to F_{w}. The boat feels -F_{w}.)Impulse-momentum equation for Boat + Man (Eqn 1):

[tex]m_{(B+M)}v_{(B+M)}+F_{W}t=m_{(B+M)}v'_{(B+M)}[/tex]

Impulse-momentum equation for Woman (Eqn 2):

[tex]m_Wv_W-F_Wt=m_Wv'_W[/tex]

Adding Eqn 1 and Eqn 2:

[tex]0=m_{(B+M)}v'_{(B+M)}+m_Wv'_W[/tex]

[tex]v'_{(B+M)}=\frac{-m_Wv'_W}{m_{(B+M)}}=-4ft/s[/tex]

Then the man jumps...:

Impulse-momentum equation for Boat (Eqn 3):

[tex]\left ( m_B \right )\left ( \frac{-m_Wv'_W}{m_{(B+M)}} \right )+F_Mt=m_Bv'_B[/tex]

Impulse-momentum equation for Man (Eqn 4):

[tex]\left ( m_M \right )\left ( \frac{-m_Wv'_W}{m_{(B+M)}} \right )-F_Mt=m_Mv'_M[/tex]

Adding Eqn 3 and Eqn 4:

[tex]V'_B=\frac{\left ( m_B+m_M \right )(-4)-(m_M)(16)}{m_B}=-16ft/s[/tex]

So, I found that the boat moves at -16 ft/s after both the man and woman jump. That's not the correct answer of -9.20 ft/s. Am I overcomplicating this whole matter? Is there some kind of m

_{1}v

_{1}+m

_{2}v

_{1}=m

_{1}v

_{2}+m

_{2}v

_{2}way of solving it (conservation of linear momentum)?