/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 Catapulting mushrooms. Certain m... [FREE SOLUTION] | 91Ó°ÊÓ

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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 a speed of \(1.6 \mathrm{~m} / \mathrm{s}\) in a \(5.0 \mu \mathrm{m}\) launch; its speed is then reduced to zero in \(1.0 \mathrm{~mm}\) 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.

Short Answer

Expert verified
Launch acceleration is \(2.61 \times 10^5 \, \text{m/s}^2 \) or \(26122g\). Speed reduction acceleration is \(-1280 \, \text{m/s}^2 \) or \(-130.61g\).

Step by step solution

01

Understanding the Problem

To solve this problem, we need to find the accelerations during two phases: the launch and the speed reduction. We're given the spore's final speed during launch, the distance it travels during launch, and how quickly it decelerates to a halt. We will assume constant acceleration for both scenarios.
02

Calculating Acceleration During Launch

Use the kinematic equation \(v^2 = u^2 + 2as\). Here, final speed \(v = 1.6 \, \text{m/s}\), initial speed \(u = 0\), and distance \(s = 5.0 \, \mu\text{m} = 5.0 \times 10^{-6} \, \text{m}\). Rearrange for acceleration \(a\): \[a = \frac{v^2 - u^2}{2s} = \frac{(1.6)^2 - 0^2}{2 \times 5.0 \times 10^{-6}} = \frac{2.56}{1.0 \times 10^{-5}} = 2.56 \times 10^5 \, \text{m/s}^2\]
03

Expressing Launch Acceleration in Terms of g

Given that \(g = 9.8 \, \text{m/s}^2\), express the acceleration in terms of \(g\): \[a = \frac{2.56 \times 10^5}{9.8} \approx 26122 \times g\]
04

Calculating Acceleration During Speed Reduction

Use the same kinematic equation \(v^2 = u^2 + 2as\) for the deceleration. Here, final speed \(v = 0\), initial speed \(u = 1.6 \, \text{m/s}\), and distance \(s = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m}\). So, \[a = \frac{0 - (1.6)^2}{2 \times 1.0 \times 10^{-3}} = \frac{-2.56}{2 \times 10^{-3}} = -1280 \, \text{m/s}^2\]
05

Expressing Speed Reduction Acceleration in Terms of g

Again, express the acceleration in terms of \(g\): \[a_{reduction} = \frac{-1280}{9.8} \approx -130.61 \times g\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematic Equations
In projectile motion, kinematic equations are vital to determining quantities such as distance, speed, and acceleration. For this exercise with the catapulting mushrooms, we utilize one of these key equations:
  • \(v^2 = u^2 + 2as\)
Here, \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the distance covered.
Using this equation, we can calculate acceleration during both the launch and the deceleration phase. It's crucial to re-arrange the equation to solve for the acceleration, \(a\).
For the launch: the final speed was \(1.6 \text{ m/s}\), and the initial speed \(0\). The distance \(5.0 \mu\text{m}\). Plug these into the equation to find the acceleration.
Kinematic equations allow us to perfectly link the motion of an object in a straight line, under constant acceleration conditions. They help decode the varying speeds and the distances covered.
Acceleration
Acceleration is the rate at which an object changes its velocity. It is a vector quantity, meaning it has both magnitude and direction.
During the launch of the mushroom’s spore, acceleration is calculated using the kinematic equation. Since the spore moves from rest to a specific velocity, its acceleration is determined by:
  • The change in velocity
  • The distance over which this change occurs
For the spore’s launch, acceleration was found to be a staggering \(2.56 \times 10^5 \text{ m/s}^2\). This number reflects the rapidity of the spore's velocity change over the very short launch distance.
In daily situations, acceleration due to gravity \(g\) is used as a reference point for understanding. In our Earth's environment, \(g\) is approximately \(9.8 \text{ m/s}^2\). Therefore, acceleration in terms of \(g\) for the launch was about \(26122 \times g\), indicating an extraordinary rate compared to regular human experiences.
Deceleration
Deceleration is simply acceleration that causes an object to slow down, and it is often referred to in negative terms since it opposes the direction of motion. In this context, deceleration of the spore occurs once launched, as air resistance rapidly brings it to a halt.
Using the same kinematic equation, but setting the final velocity to zero, we are able to compute the deceleration rate:
  • Initial speed: \(1.6 \text{ m/s}\)
  • Distance for deceleration: \(1.0 \text{ mm}\)
The calculated deceleration was \(-1280 \text{ m/s}^2\), indicating how swiftly the spore comes to rest. Expressing this in terms of \(g\), it equates to \(-130.61 \times g\).
Understanding deceleration is key in projectile motion for predicting how and where an object will stop after being projected.
Newton's Laws of Motion
Newton’s Laws are foundational to understanding motion. They are critical when analyzing projectile motion like the mushroom’s spore launch.
  • First Law: An object in motion stays in motion unless acted upon by an external force. For the spore, air resistance is the force that decelerates it to a stop.
  • Second Law: The net force on an object is equal to the mass of the object multiplied by its acceleration \( (F = ma) \). This is directly applied in the problem as we calculate the acceleration and deceleration, revealing the forces acting upon the spore.
  • Third Law: For every action, there is an equal and opposite reaction. The mechanism that catapults the spore demonstrates this principle: as the water film merges with the drop, the spore is pushed upward with force.
Understanding these laws helps to clarify how projectiles behave in motion and allows us to calculate and predict their paths and final positions.

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