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2 days ago

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In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water d

roplets. Some of these small droplets adhere to the raindrop, thereby increasing its mass as it falls. The force on the raindrop is
Fext=dp/dt=m dv/dt+v dm/dt
Suppose the mass of the raindrop depends on the distance x that it has fallen. Then m = kx, where k is a constant, and dm/dt=kv
dm/dt=kv This gives, since Fext=mg
Fext=mg,
mg=m dv/dt+v(kv)
Or, dividing by k,
xg=x dv/dt+v2
This is a differential equation that has a solution of the form v = at, where a is the acceleration and is constant. Take the initial velocity of the raindrop to be zero.
(a) Using the proposed solution for v find the acceleration a.
(b) Find the distance the raindrop has fallen in t = 3.00 s.
(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.

1 answer:

Keith_Richards [2.8K]2 days ago

6 0

Answer:

a) a = g / 3

b) x (3.0) = 14.7 m

c) m (3.0) = 29.4 g

Explanation:

Provided:-

- The differential equation related to (x) the distance traveled by a raindrop takes the following form:

$x*g = x * \frac{dv}{dt} + v^2$

- Where, v = Velocity of the raindrop

- Proposed solution for the provided ODE:

v = a*t

Where, a = raindrop's acceleration

Find:-

(a) Use the proposed solution for v to find the acceleration a.

(b) Find the distance the raindrop falls in t = 3.00 s.

(c) Given that k = 2.00 g/m, determine the mass of the raindrop at t = 3.00 s.

Solution:-

- We're aware that acceleration (a) corresponds to the first derivative of velocity (v):

a = dv / dt... Eq 1

- Similarly, velocity (v) is the first derivative of displacement (x):

v = dx / dt, v = a*t... proposed solution (Eq 2)

v.dt = dx = a*t. dt

- Integrate both sides:

∫a*t. dt = ∫dt

x = 0.5*a*t^2... Eq 3

- Substitute Eq1, 2, 3 into the original ODE:

0.5*a*t^2*g = 0.5*a^2 t^2 + a^2 t^2

= 1.5 a^2 t^2

a = g / 3

- Using the derived acceleration of the raindrop (a) and t = 3.00 seconds, plug into Eq 3:

x (t) = 0.5*a*t^2 x (t = 3.0) = 0.5*9.81*3^2 / 3

x (3.0) = 14.7 m

- Utilizing the relationship for mass with k = 2.00 g/m, find the mass of the raindrop at t = 3.0 s:

m (t) = k*x (t)

m (3.0) = 2.00*x(3.0) m (3.0) = 2.00*14.7

m (3.0) = 29.4 g

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At the outset, the raindrop possesses only gravitational potential energy:

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Work by air = work done by air resistance.

The kinetic energy at ground level is computed as follows:

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Now, we can find the work done by air resistance:

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