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Chapter 8: Problem 19

Animal Propulsion. Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracting the cavity to force out the water through an opening. A \(6.50-\) kg squid (including the water in the cavity) at rest suddenly sees a dangerous predator. (a) If the squid has 1.75 \(\mathrm{kg}\) of water in its cavity, at what speed must it expel this water to suddenly achieve a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) to escape the predator? Neglect any drag effects of the surrounding water. (b) How much kinetic energy does the squid create by this maneuver?

Short Answer

Expert verified
(a) 6.79 m/s, (b) 14.84 J

Step by step solution

01

Identify essential information

The total mass of the squid including water is 6.50 kg. Out of this, the water mass is 1.75 kg, leaving the squid's own mass as 4.75 kg (6.50 kg - 1.75 kg). The desired final speed for the squid is 2.50 m/s. We assume conservation of momentum between the squid and the expelled water, as there is no external drag.
02

Apply the conservation of momentum

The conservation of momentum states that the total initial momentum of the system (squid + water) should equal the total final momentum of the system. Initially, both are at rest, so:\[m_\text{squid} \cdot v_\text{squid final} + m_\text{water} \cdot v_\text{water final} = 0\]Rearranging gives us:\[m_\text{squid} \cdot 2.50 \, \text{m/s} = m_\text{water} \cdot v_\text{water final}\]
03

Substitute known values into the momentum equation

Substituting the known values gives:\[4.75 \, \text{kg} \cdot 2.50 \, \text{m/s} = 1.75 \, \text{kg} \cdot v_\text{water final}\]Solving for \(v_\text{water final}\), we get:\[v_\text{water final} = \frac{4.75 \, \text{kg} \cdot 2.50 \, \text{m/s}}{1.75 \, \text{kg}} = 6.79 \, \text{m/s}\]
04

Calculate the kinetic energy of the squid

The kinetic energy (KE) of the squid is given by the formula:\[\text{KE} = \frac{1}{2} m_\text{squid} \cdot v_\text{squid final}^2\]Substituting the known values, we have:\[\text{KE} = \frac{1}{2} \cdot 4.75 \, \text{kg} \cdot (2.50 \, \text{m/s})^2 = 14.84 \, \text{J}\]

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy of motion. Whenever an object moves, it possesses kinetic energy. The amount of kinetic energy depends on two main factors: the mass of the object and its velocity. For example, in the case of the squid, part of its energy when escaping a predator is kinetic.

To calculate kinetic energy, we use the formula:
  • KE = \(\frac{1}{2} m v^2\)
where \(m\) is the mass of the object and \(v\) is its velocity.

In the squid's situation, where it reaches a speed of 2.50 m/s, its kinetic energy acts as a sudden burst of energy to enhance its chances of escape. Efficient conversion of stored potential energy into kinetic energy helps it propel forward rapidly, which is crucial for survival against predators.
Animal Propulsion
Animal propulsion is how creatures like squids and octopuses move or propel themselves. Unlike humans, they utilize a unique method of expelling water from their bodies to generate movement. This action results in a rapid burst of speed, allowing them to swiftly evade threats or catch prey.

For squids and octopuses, they store a certain amount of water inside their bodies. When threatened, they contract muscle cavities to force the water out through a narrow opening. This action propels them forward in a swift motion.
  • This is quite similar to how rockets work, as they expel gas to create thrust.
  • This type of propulsion is a stunning example of how simple mechanical strategies can be employed by marine life.
The squid, for instance, achieves significant motion by combining water ejection and body streamlining. Their ability to modulate water expulsion speed is vital for effective animal propulsion.
Momentum Equation
The momentum equation is a critical principle in physics that helps us understand how motion is transferred in a system of objects. Momentum itself is a measure of the quantity of motion an object

The conservation of momentum states that in a closed system, the total momentum before an event is equal to the total momentum after the event, provided no external forces are acting on it.
  • Initial Momentum = Final Momentum
In mathematical terms, it is often expressed as:
  • \(m_1 v_1 + m_2 v_2 = m_1 u_1 + m_2 u_2\)
For our squid, this means that the momentum of the expelled water is equal but opposite to the momentum of the squid, resulting in movement. This exquisite balance allows the squid to reach a desired speed effectively, helping it in defensive maneuvers against predators.

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Most popular questions from this chapter

(a) What is the magnitude of the momentum of a \(10,000-\mathrm{kg}\) truck whose speed is 12.0 \(\mathrm{m} / \mathrm{s} ?\) (b) What speed would a \(2000-\mathrm{kg}\) SUV have to attain in order to have (i) the same momentum? (ii) the same kinetic energy?

Two friends, Burt and Ernie, are standing at opposite ends of a uniform log that is floating in a lake. The log is 3.0 \(\mathrm{m}\) long and has mass 20.0 \(\mathrm{kg} .\) Burt has mass 30.0 \(\mathrm{kg}\) and Ernie has mass 40.0 \(\mathrm{kg} .\) Initially the log and the two friends are at rest relative to the shore. Burt then offers Ernie a cookie, and Ernie walks to Burt's end of the log to get it. Relative to the shore, what distance has the log moved by the time Ernie reaches Burt? Neglect any horizontal force that the water exerts on the log and assume that neither Burt nor Ernie falls off the log.

A Variable-Mass Raindrop. In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water droplets. Some of these small droplets adhere to the raindrop, thereby increasing its mass as it falls. The force on the raindrop is $$F_{\mathrm{ext}}=\frac{d p}{d t}=m \frac{d v}{d t}+v \frac{d m}{d t}$$ Suppose the mass of the raindrop depends on the distance \(x\) that it has fallen. Then \(m=k x,\) where \(k\) is a constant, and \(d m / d t=k v\) . This gives, since \(F_{\text { ext }}=m g\) , $$m g=m \frac{d v}{d t}+v(k v)$$ Or, dividing by \(k\) $$x g=x \frac{d v}{d t}+v^{2}$$ This is a differential equation that has a solution of the form \(v=\) at, where \(a\) is the acceleration and is constant. Take the initial velocity of the raindrop to be zero. (a) Using the proposed solution for \(v\) , find the acceleration \(a\) . (b) Find the distance the raindrop has fallen in \(t=3.00\) s. (c) Given that \(k=2.00 \mathrm{g} / \mathrm{m},\) find the mass of the raindrop at \(t=3.00 \mathrm{s} .\) (For many more intriguing aspects of this problem, see \(\mathrm{K} .\) S. Krane, American Journal of Physics, Vol. 49 \((1981),\) pp. \(113-117.)\)

A 1500 -kg blue convertible is traveling south, and a \(2000-\mathrm{kg}\) red SUV is traveling west. If the total momentum of the system consisting of the two cars is 7200 \(\mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\) directed at \(60.0^{\circ}\) west of south, what is the speed of each vehicle?

You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between your feet and the ice. A friend throws you a 0.400 -kg ball that is traveling horizontally at 10.0 \(\mathrm{m} / \mathrm{s}\) . Your mass is 70.0 \(\mathrm{kg}\) . (a) If you catch the ball, with what speed do you and the ball move afterward? (b) If the ball hits you and bounces off your chest, so afterward it is moving horizontally at 8.0 \(\mathrm{m} / \mathrm{s}\) in the opposite direction, what is your speed after the collision?
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