Calculating Boat Speed with Rainfall Accumulation

[ProBasket]

A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 3.00 m/s when it starts to rain. The rain is falling vertically, and it accumulates in the boat at the rate of 10.0 kg/hr.

using this formula: mb*vb/(mb+k0+t1)

i got an answer of 2.86, but it's incorrect. it tells me I am close, but wrong. do i need to convert the units or something?

 

[whozum]

What is the question asking you? Whats k_0? t_1?

 

[ProBasket]

oops, so sorry. i forgot to post the question.

What is the speed of the boat after time 2.00 hr has passed? Assume that the water resistance is negligible.

T_1 is just time(2hr) and K_0 is 10.0 kg/hr

 

[whozum]

Is this calc based or algebra based?
And how did you derive your eq?

 

[xanthym]

A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 3.00 m/s when it starts to rain. The rain is falling vertically, and it accumulates in the boat at the rate of 10.0 kg/hr.

using this formula: mb*vb/(mb+k0+t1)

i got an answer of 2.86, but it's incorrect. it tells me I am close, but wrong. do i need to convert the units or something?

You seem to correctly be trying to use conservation of horizontal momentum:
v_b(t=2 hr) = m_b(t=0)*v_b(t=0)/m_b(t=2 hr)
Recalculate your value. (Should be about 2.778 m/sec).


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[dextercioby]

Hold on,a second.U can't add things with different units.A flow (Kg/hr) with a mass (Kg) with a time (s)...

What's the initial momentum of the boat?What's the boat's momentum after 2hrs??

Daniel.

 

[ProBasket]

thanks a lot, 2.778 was correct and i know how you got it, but i don't UNDERSTAND how it works, can someone help?

"What's the initial momentum of the boat?What's the boat's momentum after 2hrs??"
p=mv
initial momentum = 250*3
after 2hr = 250*(10*2*3) (for this one, do you use the momentum formula also?)
cause what would be v?

 

[whozum]

Well this is basically an inelastic collision right? You were going a certain speed and you gained mass. What is your new speed?

$$m_iv_i = m_fv_f$$

Your final momentum is $$m_fv_f$$. Your final mass is not 250(10*2*3), that would mean you gained about 50 times your mass. In fact your final mass is $$mass_{boat} + mass_{rain}$$. How much mass did the rain add to you?

using $$p_f = m_fv_f$$ you can find your final momentum.

Do you see how this works?

 

[ProBasket]

ahh... thanks a lot!

 

[alexialight]

I got stuck on the 3rd part of this question:

Assume that the boat is subject to a drag force F due to water resistanceThe drag is proportional to the square of the speed of the boat, in the form F=0.5v^2. What is the acceleration of the boat just after the rain starts?

I don't even know how to go about approaching this part

 

[dextercioby]

Does that mean we can neglect varying mass & momentum of the boat due to the water falling down?

Daniel.