Question

Pilobolus is a genus of fungi commonly found on dung and known for launching its spores a large distance for a sporangiophore only 1 cm tall. To achieve this, Pilobolus accelerates its spores (m = 10−8 kg) to 7.0 m/s in 2.0 μs. For calculations, use g = 10 m/s2.
How much work is done by the Pilobolus head when it ejects the spore?

If the spores are shot out horizontally at the maximum speed, how far away from the fungi do they land? Neglect air resistance.

A. 5.0 m
B. 1.0 m
C. 10.0 m
D. 0 m

Answer 1

1) The work done on the spores is $2.45\cdot 10^{-7}J$

2) The spores land 0.32 m away

Explanation:

1)

According to the work-energy theorem, the work done on the ejected spores is equal to the change in kinetic energy of the spores:

$W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$

where

W is the work done

m is the mass of one spore

v is the final velocity

u is the initial velocity

For the spores in this problem, we have:

$m=10^{-8} kg$ is hte mass

u = 0 is the initial velocity (they start from rest)

v = 7.0 m/s is the final velocity

Substituting, we find the work done:

$W=\frac{1}{2}(10^{-8})(7.0)^2-0=2.45\cdot 10^{-7}J$

2)

The motion of the spores is a projectile motion, so it follows a parabolic path which consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction  

We start by considering the vertical motion: this is a free fall motion, so we can use the suvat equation  

$s=ut+\frac{1}{2}at^2$  

where  

s = 1 cm = 0.01 m is the vertical displacement (the height of the fungi)

u = 0 is the initial vertical velocity of the spore (it is ejected horizontally)

t is the time of flight

$a=g=10 m/s^2$ is the acceleration of gravity  

Solving for t, we find:

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(0.01)}{10}}=0.045 s$  

Now we can consider the horizontal uniform motion. Since the velocity is constant, the horizontal distance travelled is given by :

$d=v_x t$  

where:

$v_x = 7.0 m/s$ is the horizontal velocity, which is constant

t = 0.045 s is the time of flight  

Solving for d, we find:

$d=(7.0)(0.045)=0.32 m$

So, the spores land 0.32 m away.

Learn more about work:

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Learn more about projectile motion:

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