Q24P - Fundamentals Of Physics

Question

Catapulting mushrooms. 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 the speed of 51.6 m/s in a $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{\mu }\mathit{m}$ launch; its speed is then reduced to zero in 1.0 mm by the air. Using that data and assuming constant accelerations, (a) find the accelerations in terms of g during the launch. (b)Find the accelerations in terms of g during the speed reduction.

Step-by-Step Solution

Answer

(a) The acceleration in terms of g during the launch is $2.6\times {10}^{4}g$ .

(b) The acceleration in terms of g during the speed reduction is $-1.3\times {10}^{2}g$ .

Step 1: Given Data

Launch distance,=$5.0\mu m$

The speed in launch is 1.6 m/s 

The distance for which speed reduces to zero is 1.0 mm

Step 2: Understanding the concept of acceleration

The value of acceleration or deceleration can be found using the kinematic equations. The acceleration would be positive in sign whereas deceleration would be negative in sign. However, one needs to be careful about the directions of initial velocity and final velocity as they can cause the signs of acceleration to change accordingly.

The formula is as given below,

$v_f^2=v_i^2+2ax$                                                                                                                                                  (i)

Here, $v_f$ is the final speed, $v_i$ is the initial speed, a is the acceleration and x is the distance.

Step 3: (a) Determination of the acceleration during the launch

Convert $5.0 \mathrm{\mu m}$ to meter.

$$\begin{gathered} 5.0 \mathrm{\mu m}=5.0 \mathrm{\mu m}\times \frac{1\times {10}^{-6}\mathrm{m}}{1.0\mathrm{\mu m}} \\ =5\times {10}^{-6}\mathrm{m} \end{gathered}$$

Putting values in equation (i),

$$\begin{gathered} 1.6^2=0^2+2\times a\times 5\times {10}^{-6} \\ a=\frac{2.56}{{10}^{-5}} \\ =2.56\times {10}^{5}m/s \end{gathered}$$

Since, the value of $\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2$ therefore, the acceleration during the launch is,

 $$\begin{gathered} a=2.56\times {10}^5\mathrm{m}/{\mathrm{s}}^2\times \frac{\mathrm{g}}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ =2.6\times {10}^4\mathrm{g} \end{gathered}$$

Thus, the value of acceleration during the launch in terms of g is $2.6\times {10}^4\mathrm{g}$ .

Step 4: (b) Determination of the acceleration during the speed reduction

For the speed reduction, the initial speed is 1.6m/s and final speed is zero, and the distance is 1.0 mm .

Convert 1.0 mm to meter.

 $$\begin{gathered} 1.0 \mathrm{mm}=1.0\mathrm{mm}\times \frac{{10}^{-3}\mathrm{m}}{1\mathrm{mm}} \\ =1.0\times {10}^{-3}\mathrm{m} \end{gathered}$$

Putting values in equation (i),

$$\begin{gathered} v_f^2=v_i^2+2ax \\ {0}^2=1.6^2+2a\times 1.0\times {10}^{-3} \\ a=-1.28\times {10}^3\mathrm{m}/\mathrm{s} \end{gathered}$$

Divide the acceleration by the value of g to get the acceleration in terms of g.

$$\begin{gathered} a=-1.28\times {10}^3\mathrm{m}/{\mathrm{s}}^2\times \frac{\mathrm{g}}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ =-1.30\times {10}^2\mathrm{g} \end{gathered}$$

Thus, the acceleration in terms of g during the speed reduction is $-1.30\times {10}^2\mathrm{g}$.