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A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 1.00 m/s when it starts to rain with incredible intensity. The rain is falling vertically, and it accumulates in the boat at the rate of 100 kg/hr . What is the speed of the boat after time 0.500 hr has passed7 Assume that the water resistance is negligible. Express your answer in meters per second. Now assume that the boat is subject to a drag force Fd due to water resistance. Is the component of the total momentum of the system parallel to the direction of motion still conserved? The boat is subject to an external force, the drag force due to water resistance, and therefore its momentum is not conserved. The drag is proportional to the square of the speed of the boat, in the form Fd = bv2 where b = 0.5 N. s2/m2. What is the acceleration of the boat just after the rain starts'7 Take the positive x axis along the direction of motion. Express your answer in meters per second per second.

Respuesta :

mavila18

Answer:

Explanation:

1. We use the conservation of momentum for before the raining and after. And also we take into account that in 0.5h the accumulated water is

100kg/h*0.5h = 50kg

[tex]p_{b}=p_{a}\\M_{b}v_{b}=(M_{b}+m_{r})v_{a}\\v_{a}=\frac{M_{b}v_{b}}{M_{b}+m_{r}}=\frac{250kg*1\frac{m}{s}}{250kg+50kg}=0.83\frac{m}{s}[/tex]

2. the momentum does not conserve because the drag force of water makes that the boat loses velocity

3. If we assume that the force of the boat before the raining is

[tex]F=ma=m\frac{v-v_{0}}{t-t_{0}}=250kg\frac{1m}{s^{2}}=250N[/tex]

where we have assumed that the acceleration of the boat is 1m/s{2} just before the rain starts

And if we take the net force as

[tex]F_{net}=M_{b}a_{net}=F-F_{d}=250N-bv^{2}\\F_{net}=250N-0.5N\frac{s^{2}}{m^{2}}(1\frac{m}{s})^{2}=249.5N\\a_{net}=\frac{249.5N}{M_{b}}=\frac{249.5N}{250kg}=0.99\frac{m}{s^{2}}[/tex]

where we take v=1m/s because we are taking into account tha velocity just after the rain stars.

I hope this is useful for you

regards

snehashish65

(1) The speed of the boat after 0.500 hr is 0.83 m/s.

(2) The momentum will not remain conserved when system is subjected to drag force.

(3)  The acceleration of the boat is  [tex]9.79 \;\rm m/s^{2}[/tex].

Given data:

The mass of boat is, m = 250 kg.

The speed of boat is, u = 1.00 m/s.

The rate of accumulation of rain water is, r = 100 kg/hr.

The time elapse is, t = 0.500 hr.

The given problem is divided into three parts. So let us answer in part (1), (2) and (3) respectively.

(1)

Mass of water accumulated in 0.5 h is,

[tex]m'=\dfrac{r}{t}\\\\m'=\dfrac{100}{0.5}\\\\m'=50 \;\rm kg[/tex]

We use the conservation of momentum for before the raining and after the rain. Then,

momentum before rain = momentum after rain

[tex]m \times u=(m+m') v\\\\v = \dfrac{m \times u}{ m +m'} \\\\v = \dfrac{250 \times 1.00}{ 250 +50} \\\\v = 0.83 \;\rm m/s[/tex]

Thus, we can conclude that the speed of the boat after 0.500 hr is 0.83 m/s.

(2)

When the boat is subject to some drag force due to water resistance, then the component of the total momentum of the system parallel to the direction of motion is not conserved because the drag force of water makes that the boat loses velocity.

(3)

The net force acting on the boat is,

[tex]F_{net} = W - F_{d}[/tex]

Here, W is the weight of boat. And [tex]F_{d}[/tex] is the drag force and its value is,

[tex]F_{d}=b v^{2}\\\\F_{d} =0.5 \times 0.83^{2}\\\\F_{d}=0.344 \;\rm N[/tex]

The net force is also expressed by the Newton's second law as,

[tex]F_{net} = m \times a[/tex]

Here, a is the acceleration of boat.

Solving as,

[tex]W - F_{d} = m \times a\\\\(mg) - F_{d} = m \times a\\\\(250 \times 9.8) - 0.344 = 250 \times a\\\\a =\dfrac{(250 \times 9.8) - 0.344}{250} \\\\a=9.79 \;\rm m/s^{2}[/tex]

Thus, the acceleration of the boat is  [tex]9.79 \;\rm m/s^{2}[/tex].

Learn more about the Newton's law of motion here:

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