Man and Woman on Boat - Velocity of Boat after Dive

In summary, a 180-lb man and a 120-lb woman dive off a 300-lb boat with a velocity of 16-ft/s relative to the boat. The boat will move with a velocity of -9.20 ft/s to the left if the woman dives first and -9.37 ft/s to the left if the man dives first. The equations used to solve this problem include impulse-momentum equations, conservation of linear momentum, and the equation v_{person}=v_{boat}+16.
  • #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)

diving.jpg


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)?
 
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  • #2
Hi JJBladester! :smile:
JJBladester said:
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 :redface: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? :wink:)

so just put that into your conservation of momentum equations :smile:
 
  • #3
tiny-tim said:
Hi JJBladester! :smile:erm :redface: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? :wink:)

so just put that into your conservation of momentum equations :smile:

My conservation of momentum equations seem to get me to the same place:

Woman Jumps:

[tex]m_Bv_B+m_Mv_M+m_Wv_W=(m_B+m_M)v'_{(B+M)}+m_Wv_W[/tex]

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

Then Man Jumps:

[tex](m_B+m_M)v'_{(B+M)}=m_Bv''_B+m_Mv''_M[/tex]

[tex]v''_B=\frac{(m_B+m_M)v'_{(B+M)}-m_Mv''_M}{m_B}=-16ft/s[/tex]
 
  • #4
I'm confused …

where does the given v1 - v2 = 16 come in that? :confused:
 
  • #5
tiny-tim said:
where does the given v1 - v2 = 16 come in

[tex]v_{person}-v_{boat}=16[/tex]

[tex]v_{person}=v_{boat}+16[/tex]

Woman Jumps:

[tex](m_{boat}+m_{woman}+m_{man})v_{boat}=(m_{boat}+m_{man})v'_{boat}+m_{woman}(v_{boat}+16)[/tex]

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

Then Man Jumps:

[tex](m_{man}+m_{boat})v'_{boat}=m_{boat}v''_{boat}+m_{man}(v'_{boat}+16)[/tex]

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

Still not getting the desired answer of -9.20 ft/s... Any other pointers?
 
  • #6
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? :confused:
 
  • #7
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.

[tex]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)}[/tex]

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

Doing the above calculations did not yield the correct answer of -9.20 ft/s for [itex]v''_{boat}[/itex]
 
Last edited:
  • #8
(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)
 
  • #9
JJBladester said:
[tex]v_{person}-v_{boat}=16[/tex]

[tex]v_{person}=v_{boat}+16[/tex]

Woman Jumps:

[tex](m_{boat}+m_{woman}+m_{man})v_{boat}=(m_{boat}+m_{man})v'_{boat}+m_{woman}(v_{boat}+16)[/tex]

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

Then Man Jumps:

[tex](m_{man}+m_{boat})v'_{boat}=m_{boat}v''_{boat}+m_{man}(v'_{boat}+16)[/tex]

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

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


You've almost got it with this equation:

[tex](m_{boat}+m_{woman}+m_{man})v_{boat}=(m_{boat}+m_{man})v'_{boat}+m_{woman}(v_{boat}+16)[/tex]

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 [tex](v_{boat}+16)[/tex], it is [tex](v'_{boat}+16)[/tex]. 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.
 
  • #10
tiny-tim said:
(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.


Xerxes1986 said:
You've almost got it with this equation:

[tex](m_{boat}+m_{woman}+m_{man})v_{boat}=(m_{boat}+m_{man})v'_{boat}+m_{woman}(v_{boat}+16)[/tex]

...the velocity of the woman ISN'T [itex](v_{boat}+16)[/itex], it is [itex](v'_{boat}+16)[/itex]. 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 [itex]v'[/itex] for the woman's jump and [itex]v''[/itex] for the man's jump.

Case closed on this ambiguously written textbook problem. Whew!
 

1. What is the velocity of the boat after the dive?

The velocity of the boat after the dive depends on several factors such as the initial velocity of the boat, the mass of the boat, and the direction and force of the dive. It can be calculated using the laws of motion and is typically lower than the initial velocity due to the resistance of the water.

2. How does the velocity of the boat affect the man and woman on the boat?

The velocity of the boat can affect the man and woman on the boat in several ways. If the boat is moving at a high velocity, it can create a strong force on the man and woman, making it difficult for them to maintain balance and potentially causing them to fall. Additionally, if the boat is moving at a high velocity and then suddenly slows down after the dive, the man and woman may experience a sudden change in inertia, potentially causing them to be thrown off balance.

3. Why is the velocity of the boat important in this scenario?

The velocity of the boat is important in this scenario because it directly affects the safety and stability of the man and woman on the boat. If the boat is moving too fast or suddenly changes velocity, it can lead to accidents and injuries for the man and woman. It is also important in understanding the physics behind the dive and how the boat's motion affects the dive itself.

4. Is the velocity of the boat the only factor that determines the success of the dive?

No, the velocity of the boat is not the only factor that determines the success of the dive. Other important factors include the angle and force of the dive, the height of the dive, and the coordination and skill of the man and woman performing the dive. The velocity of the boat is just one piece of the puzzle in this scenario.

5. How can the velocity of the boat be controlled during the dive?

The velocity of the boat can be controlled during the dive by adjusting the speed and direction of the boat before the dive, as well as using techniques such as braking or steering to control the boat's motion during the dive. It is important for the person steering the boat to have good communication and coordination with the man and woman performing the dive to ensure a safe and successful dive.

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