Question

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.

Answer 1

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

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

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

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

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

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

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