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 1.6 m/s in a 5.0 \mu m launch; its speed is then reduced to zero in 1.0 mm by the air. Using those data and assuming constant accelerations, find the acceleration in terms of g during (a) the launch and (b) the speed reduction.
Solution:
Let us use the following equation to solve this question.
v^{2}=v_0^2+2a(x-x_{0})
(This is a fundamental equation that describes the motion of a particle with constant acceleration)
a) To find the acceleration of the spores, we will isolate for a and then substitute the values given to us in the problem.
a=\frac{v^{2}-v_0^2}{2x}
a=\frac{(1.6m/s)^2}{2(5.0\times 10^-6 m)}
a=2.56\times 10^5 m/s^{2}
and to express this in terms of g, we simply divide 2.56\times 10^5 m/s^{2} by 9.8 m/s^2 which gives us 2.6\times 10^4 g.
b) When the speed is in the reduction stage, the acceleration is equal to:
a=\frac{v^{2}-v_0^2}{2x}
a=\frac{0-(1.6m/s)^2}{2(1.0\times 10^-3 m)}
a=-1.28\times 10^{3} m/s^{2}
and in terms of g, this is -1.3\times 10^2 g.
This question can be found in Fundamentals of Physics, 10th edition, chapter 2, question 24.
