# Man and Woman on Boat - Velocity of Boat after Dive

Gold Member
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.

(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 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):
$$m_{(B+M)}v_{(B+M)}+F_{W}t=m_{(B+M)}v'_{(B+M)}$$

Impulse-momentum equation for Woman (Eqn 2):
$$m_Wv_W-F_Wt=m_Wv'_W$$

Adding Eqn 1 and Eqn 2:
$$0=m_{(B+M)}v'_{(B+M)}+m_Wv'_W$$
$$v'_{(B+M)}=\frac{-m_Wv'_W}{m_{(B+M)}}=-4ft/s$$

Then the man jumps...:

Impulse-momentum equation for Boat (Eqn 3):
$$\left ( m_B \right )\left ( \frac{-m_Wv'_W}{m_{(B+M)}} \right )+F_Mt=m_Bv'_B$$

Impulse-momentum equation for Man (Eqn 4):
$$\left ( m_M \right )\left ( \frac{-m_Wv'_W}{m_{(B+M)}} \right )-F_Mt=m_Mv'_M$$

Adding Eqn 3 and Eqn 4:
$$V'_B=\frac{\left ( m_B+m_M \right )(-4)-(m_M)(16)}{m_B}=-16ft/s$$

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)?

#### tiny-tim

Homework Helper
Hi JJBladester! 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.

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.

Is there some kind of m1v1+m2v1=m1v2+m2v2 way of solving it (conservation of linear momentum)?
erm yes!

all the question means is that vperson - vboat = 16 ft/s (that's the x components … the y components won't affect the position of the boat, will they? )

so just put that into your conservation of momentum equations Gold Member
Hi JJBladester! erm yes!

all the question means is that vperson - vboat = 16 ft/s (that's the x components … the y components won't affect the position of the boat, will they? )

so just put that into your conservation of momentum equations My conservation of momentum equations seem to get me to the same place:

Woman Jumps:

$$m_Bv_B+m_Mv_M+m_Wv_W=(m_B+m_M)v'_{(B+M)}+m_Wv_W$$

$$v'_{(B+M)}=\frac{-m_Wv'_W}{m_B+m_M}=-4ft/s$$

Then Man Jumps:

$$(m_B+m_M)v'_{(B+M)}=m_Bv''_B+m_Mv''_M$$

$$v''_B=\frac{(m_B+m_M)v'_{(B+M)}-m_Mv''_M}{m_B}=-16ft/s$$

#### tiny-tim

Homework Helper
I'm confused …

where does the given v1 - v2 = 16 come in that? Gold Member
where does the given v1 - v2 = 16 come in
$$v_{person}-v_{boat}=16$$

$$v_{person}=v_{boat}+16$$

Woman Jumps:

$$(m_{boat}+m_{woman}+m_{man})v_{boat}=(m_{boat}+m_{man})v'_{boat}+m_{woman}(v_{boat}+16)$$

$$v'_{boat}=\frac{(m_{boat}+m_{woman}+m_{man})v_{boat}-m_{woman}(v_{boat}+16)}{m_{boat}+m_{man}}=-4ft/s$$

Then Man Jumps:

$$(m_{man}+m_{boat})v'_{boat}=m_{boat}v''_{boat}+m_{man}(v'_{boat}+16)$$

$$v''_{boat}=\frac{(m_{man}+m_{boat})v'_{boat}-m_{man}(v'_{boat}+16)}{m_{boat}}=-13.6ft/s$$

Still not getting the desired answer of -9.20 ft/s... Any other pointers?

#### tiny-tim

Homework Helper
It's very difficult to follow what you're doing without seeing any figures.

The initial speed of the boat is zero … have you used that? Gold Member
Yes, I did use that. Then, all of the masses (m_m, m_w, and m_b) were all just their weights divided by 32.2.

$$v'_{boat}=\frac{\left [(300/32.2)+(120/32.2)+(180/32.2) \right ]0-(120/32.2)(0+16)}{(300/32.2)+(180/32.2)}$$

$$v''_{boat}=\frac{\left [(180/32.2)+(300/32.2) \right ](-4)-(180/32.2)(-4+16)}{(300/32.2)}$$

Doing the above calculations did not yield the correct answer of -9.20 ft/s for $v''_{boat}$

Last edited:

#### tiny-tim

Homework Helper
(just got up :zzz: …)

No, your (0 + 16) should be (v'boat + 16) …

v'woman = v'boat + 16 …

if the boat was fixed, v'woman would be 16, so since the boat isn't fixed, v'woman will obviously be less.

(and there's no need to divide everything by g … the ∑mv equation works just as well with weights instead of masses)

#### Xerxes1986

$$v_{person}-v_{boat}=16$$

$$v_{person}=v_{boat}+16$$

Woman Jumps:

$$(m_{boat}+m_{woman}+m_{man})v_{boat}=(m_{boat}+m_{man})v'_{boat}+m_{woman}(v_{boat}+16)$$

$$v'_{boat}=\frac{(m_{boat}+m_{woman}+m_{man})v_{boat}-m_{woman}(v_{boat}+16)}{m_{boat}+m_{man}}=-4ft/s$$

Then Man Jumps:

$$(m_{man}+m_{boat})v'_{boat}=m_{boat}v''_{boat}+m_{man}(v'_{boat}+16)$$

$$v''_{boat}=\frac{(m_{man}+m_{boat})v'_{boat}-m_{man}(v'_{boat}+16)}{m_{boat}}=-13.6ft/s$$

Still not getting the desired answer of -9.20 ft/s... Any other pointers?

You've almost got it with this equation:

$$(m_{boat}+m_{woman}+m_{man})v_{boat}=(m_{boat}+m_{man})v'_{boat}+m_{woman}(v_{boat}+16)$$

But you have made one mistake....the initial velocity of the boat ( v_{boat} ) is zero. So the whole left side of the eqn goes away. Also the velocity of the woman ISN'T $$(v_{boat}+16)$$, it is $$(v'_{boat}+16)$$. Fix that and your equation should yield -3.2ft/s for the velocity of the boat after the woman jumps. Then setup a similar equation for the man and you should get the correct answer of -9.2ft/s.

Gold Member
(just got up :zzz: …)

No, your (0 + 16) should be (v'boat + 16) …

v'woman = v'boat + 16 …

if the boat was fixed, v'woman would be 16, so since the boat isn't fixed, v'woman will obviously be less.

(and there's no need to divide everything by g … the ∑mv equation works just as well with weights instead of masses)
tiny-tim, I appreciate your help! I also learned something valuable, that if gravity (g) is going to cancel out in the end, don't bother converting all of the weights to masses. It's an unnecessary step. Awesome.

You've almost got it with this equation:

$$(m_{boat}+m_{woman}+m_{man})v_{boat}=(m_{boat}+m_{man})v'_{boat}+m_{woman}(v_{boat}+16)$$

...the velocity of the woman ISN'T $(v_{boat}+16)$, it is $(v'_{boat}+16)$. Fix that and your equation should yield -3.2ft/s for the velocity of the boat after the woman jumps. Then setup a similar equation for the man and you should get the correct answer of -9.2ft/s.
Xerxes, thanks for the help. That was where I went wrong... And I obtained the correct answers after working it out using $v'$ for the woman's jump and $v''$ for the man's jump.

Case closed on this ambiguously written textbook problem. Whew!

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