Question

A typical raindrop is much more massive than a mosquito and falling much faster than a mosquito flies. How does a mosquito survive the impact? Recent research has found that the collision of a falling raindrop with a mosquito is a completely-inelastic collision. That is, the mosquito is "swept up" by the raindrop and ends up traveling along with the raindrop. Once the relative speed between the mosquito and the raindrop is zero, the mosquito is able to detach itself from the drop and fly away.
Part A

A hovering mosquito is hit by a raindrop that is 50 times as massive and falling at 8.1m/s , a typical raindrop speed. How fast is the raindrop, with the attached mosquito, falling immediately afterward if the collision is perfectly inelastic?

Part B

Because a raindrop is "soft" and deformable, the collision duration is a relatively long 8.0 ms. What is the mosquito's average acceleration, in g's, during the collision? The peak acceleration is roughly twice the value you found, but the mosquito's rigid exoskeleton allows it to survive accelerations of this magnitude. In contrast, humans cannot survive an acceleration of more than about 10 g.

Answer 1

Answer:

A: 7.94 m/s

B: 101.2 g

Explanation:

PART A

As stated in the question, this is a case of a completely-inelastic collision. This means that after the drop impacts the mosquito, both move together at the same speed.

By the law of conservation of energy, the sum of the individual kinetic energy of the raindrop and the mosquito must be equal before and after the collision.

Writting this as a formula:

$K_{f} = K_{r} + K_{m}\\m_{f}.v_{f} = m_{r}.v_{r} + m_{m}.v_{m}$

 

Where:

$K_{f} = Final\:\:kinetic\:\:energy\\K_{r} = Raindrop\:\:initial\:\:kinetic\:\:energy\\K_{m} = Mosquito\:\:initial\:\:kinetic\:\:energy\\m_{f},v_{f} = Final\:\:mass\:\:and\:\:speed \\m_{r},v_{r} = Raindrop\:\:mass\:\:and\:\:initial\:\:speed \\m_{m},v_{m} = Mosquito\:\:mass\:\:and\:\:initial\:\:speed$

And isolating $v_{f}$:

$v_{f} = \frac{m_{r}.v_{r} + m_{m}.v_{m}}{m_{f}}\\v_{f} = \frac{m_{r}.v_{r} + m_{m}.v_{m}}{m_{r}+m_{m}}$

Now, the problem states that the raindrop's speed is 8.1 m/s and its mass is 50 times greater than the mosquito's:

$m_{m} = \frac{m_{r}}{50}$

Replacing on the speed equation:

$v_{f} = \frac{m_{r}*v_{r} + m_{m}*v_{m}}{m_{r}+m_{m}}\\v_{f} = \frac{m_{r}*8.1 + m_{m}*0}{m_{r}+\frac{m_{r}}{50}}\\v_{f} = \frac{m_{r}*8.1}{\frac{51}{50}*m_{r}}\\v_{f} = \frac{50*8.1}{51}\\v_{f} = 7.94 \frac{m}{s}$

 

PART B

By definition, acceleration is the variation of speed by unit of time. In this case the mosquito initial state is hovering still (vertically), and reaching the previously calculated $v_{f}$ speed in 8.0 miliseconds (0.008 s).

Writting this as a formula:

$a = \frac{\Delta v}{\Delta t}\\a = \frac{v_{f}-v_{m}}{8*10^{-3}}\\a = \frac{7.94-0}{0.008}\\a = \frac{7.94}{0.008}\\a = 992,5 \frac{m}{s^{2}}$

Knowing that 9.8 m/s^2 is equivalent to 1g acceleration:

$a = 992,5 \frac{m}{s^{2}} = 101.2 g$

Answer 2

Answer:

Part a)

$v = 7.94 m/s$

Part b)

$a = 992.6 m/s^2$

Explanation:

Part a)

As we know that we can use momentum conservation for this

so we will have

$m_1v_1 = (m_1 + m_2)v$

$(50m)8.1 = (50m + m)v$

$v = 7.94 m/s$

Part b)

As we know that acceleration is rate of change in velocity

so we have

$a = \frac{v_f - v_i}{t}$

so we have

$a = \frac{7.94 - 0}{8 \times 10^{-3}}$

$a = 992.6 m/s^2$