Question

Certain mushrooms launch their spores by a catapult mechanism. As water condenses from the air onto a spore that is attached to the mushroom, a drop grows on one side of the spore and a film grows on the other side. The spore is bent over by the drop's weight, but when the film reaches the drop, the drop's water suddenly spreads into the film and the spore springs upward so rapidly that it is slung off into the air. Typically, the spore reaches a speed of 1.60 m/s in a 5.30 μm launch; its speed is then reduced to zero in 1.30 mm by the air. Using that data and assuming constant accelerations, find the acceleration in terms of g during
(a) the launch and
(b) the speed reduction.

Answer 1

Answer:

Part a)

$a = 2.46 \times 10^4 g$

Part b)

$a = 100.37 g$

Explanation:

Part a)

During the launch

speed will increase from rest to final speed of 1.60 m/s

the distance moved by it is given as

$d = 5.30 \mu m$

now we have

$v_f^2 - v_i^2 = 2ad$

$1.60^2 - 0 = 2(a)(5.30 \times 10^{-6})$

so we have

$a = \frac{1.60^2}{2(5.30 \times 10^{-6})}$

$a = 2.4151 \times 10^5 m/s^2$

now in terms of g it is given as

$a = 2.46 \times 10^4 g$

Part b)

During  speed reduction

we know that final speed will be zero by air in distance 1.30 mm

so we have

$v_f^2 - v_i^2 = 2 a d$

$0 - 1.60^2 = 2(a)(1.30 \times 10^{-3})/tex]

[tex]a = 984.6 m/s^2$

Now in terms of g it is

$a = 100.37 g$