Two identical bathyspheres float in suspension, the first at depth $2d$, the second at depth $d$ ($d = 1 km$). At the initial moment of time the first bathysphere discharges ballast and starts to float. When it floats to depth d, the second bathysphere bathysphere discharges ballast and begins to float. The first bathysphere appeared on the water surface at $T = 10 s$ earlier than the second one. It is known that the first bathysphere moved with almost constant velocity $v_{0} = 1 m/s$ for the second half of its path. Find the ejective force. The mass of the bathysphere without ballast is $m$. (The force of resistance to the motion of the bathysphere from the water side can be considered directly proportional to the speed of the bathysphere.
Decision:
The force of resistance to the motion of the bathysphere $F = - kv$, where $v$ is its velocity relative to the water; $k$ is a constant coefficient (the minus sign indicates that the drag force is directed against the velocity). The equation of motion of the bathysphere without ballast is:
$ma = -kv + F_{A} - mg$. (1)
At steady motion with velocity $v_{0}$, equation (1) takes the form of
$0= -kv_{0} + F_{A}-mg$. (2)
We can now rewrite equation (1) as follows:
$ma = -k(v - v_{0})$. (3)
Let us apply this equation to the motion of the second bathysphere. In parentheses in the right side of this equation is the relative velocity of the two bathyspheres at the second stage of motion. Multiply equality (3) by $\Delta t \||| T$ and consider that $a \Delta t = \Delta v$ is the change in velocity of the second bathysphere, and $(v_{0}- v) \Delta t = \Delta S$ is the change in the distance between the bathyspheres. Their relationship follows from equation (3):
$m \Delta v = k \Delta S$. (4)
At the moment of surfacing of the first bathysphere, the second one will be at a depth $S \approx v_{0}T = 10 m$ - much less than $d$. This means that our approximation is sufficiently accurate and at this moment the second bathysphere will be moving with steady-state velocity $v_{0}$ (as well as the first one at the end of the first half of the path).
Passing in equation (4) from infinitesimal to finite changes of velocity and distance (i.e., adding up equations (4) for the whole time of ascent of the two bathyspheres), we obtain a relation of the type $mv_{0} = kS =kv_{0}T$, or $k = \frac{m}{T}$. Given this value of the coefficient $k$, we obtain from equation (2) the answer:
$F_{a} = mg + \frac{mv_{0}}{T}=mg \left ( 1+\frac{v_{0}}{gT} \right ) \approx 1.01mg$
Decision:
The force of resistance to the motion of the bathysphere $F = - kv$, where $v$ is its velocity relative to the water; $k$ is a constant coefficient (the minus sign indicates that the drag force is directed against the velocity). The equation of motion of the bathysphere without ballast is:
$ma = -kv + F_{A} - mg$. (1)
At steady motion with velocity $v_{0}$, equation (1) takes the form of
$0= -kv_{0} + F_{A}-mg$. (2)
We can now rewrite equation (1) as follows:
$ma = -k(v - v_{0})$. (3)
Let us apply this equation to the motion of the second bathysphere. In parentheses in the right side of this equation is the relative velocity of the two bathyspheres at the second stage of motion. Multiply equality (3) by $\Delta t \||| T$ and consider that $a \Delta t = \Delta v$ is the change in velocity of the second bathysphere, and $(v_{0}- v) \Delta t = \Delta S$ is the change in the distance between the bathyspheres. Their relationship follows from equation (3):
$m \Delta v = k \Delta S$. (4)
At the moment of surfacing of the first bathysphere, the second one will be at a depth $S \approx v_{0}T = 10 m$ - much less than $d$. This means that our approximation is sufficiently accurate and at this moment the second bathysphere will be moving with steady-state velocity $v_{0}$ (as well as the first one at the end of the first half of the path).
Passing in equation (4) from infinitesimal to finite changes of velocity and distance (i.e., adding up equations (4) for the whole time of ascent of the two bathyspheres), we obtain a relation of the type $mv_{0} = kS =kv_{0}T$, or $k = \frac{m}{T}$. Given this value of the coefficient $k$, we obtain from equation (2) the answer:
$F_{a} = mg + \frac{mv_{0}}{T}=mg \left ( 1+\frac{v_{0}}{gT} \right ) \approx 1.01mg$