- #1
JJBladester
Gold Member
- 286
- 2
Homework Statement
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)
Homework Equations
F=ma
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 Fwand -Fw. So, she jumps and pushes off with a force equal to Fw. The boat feels -Fw.)
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 m1v1+m2v1=m1v2+m2v2 way of solving it (conservation of linear momentum)?