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Chapter 8: Problem 70

Animal propulsion. Squids and octopuses propel them- selves by expelling water. They do this by taking the water into a cavity and then suddenly contracting the cavity, forcing the water to shoot out of an opening. A 6.50 \(\mathrm{kg}\) squid (including the water in its cavity) that is at rest suddenly sees a dangerous predator. (a) If this squid has 1.75 \(\mathrm{kg}\) of water in its cavity, at what speed must it expel the 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 for this escape maneuver?

Short Answer

Expert verified
(a) 6.79 m/s; (b) 55.15 J.

Step by step solution

01

Understand the Problem

We have a 6.50 kg squid, including 1.75 kg of water in its cavity. Initially at rest, the squid must expel water to reach a speed of 2.50 m/s. We need to calculate the speed at which the squid must expel the water to achieve this and find the kinetic energy created for its escape.
02

Apply Conservation of Momentum

Since the squid is initially at rest, the system's total momentum is zero. By conservation of momentum, the momentum of the expelled water must equal the momentum gained by the squid:\[m_s \cdot v_s = m_w \cdot v_w\]where:\(m_s = 6.50 - 1.75 = 4.75 \) kg (mass of squid without water),\(v_s = 2.50\) m/s (velocity of squid), and\(m_w = 1.75\) kg (mass of water).We need to find \(v_w\), the velocity of the expelled water.
03

Solve for Velocity of Expelled Water

Rearrange the conservation of momentum equation to solve for \(v_w\):\[v_w = \frac{m_s \cdot v_s}{m_w}\]Substitute the values:\[v_w = \frac{4.75\, \text{kg} \times 2.50\, \text{m/s}}{1.75\, \text{kg}}\]\[v_w = 6.79\, \text{m/s}\]
04

Calculate Kinetic Energy of Squid

The kinetic energy \(KE\) of the squid is given by the formula:\[KE = \frac{1}{2} m_s v_s^2\]Substitute the known values:\[KE = \frac{1}{2} \times 4.75\, \text{kg} \times (2.50\, \text{m/s})^2\]\[KE = \frac{1}{2} \times 4.75 \times 6.25\, \text{J}\]\[KE = 14.84\, \text{J}\]
05

Calculate Total Kinetic Energy of Water

Similarly, calculate the kinetic energy of the expelled water:\[KE_w = \frac{1}{2} m_w v_w^2\]Substitute the values:\[KE_w = \frac{1}{2} \times 1.75\, \text{kg} \times (6.79\, \text{m/s})^2\]\[KE_w = \frac{1}{2} \times 1.75 \times 46.06\, \text{J}\]\[KE_w = 40.31\, \text{J}\]
06

Determine How Much Kinetic Energy the Squid Creates

Total kinetic energy is the sum of the kinetic energy of the squid and the expelled water:\[KE_{total} = KE + KE_w\]\[KE_{total} = 14.84\, \text{J} + 40.31\, \text{J}\]\[KE_{total} = 55.15\, \text{J}\]

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Key Concepts

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

Kinetic Energy Calculation
Kinetic energy is the energy that an object possesses due to its motion. Calculating kinetic energy involves understanding the relationship between mass and velocity. This concept plays a significant role in solving physics problems related to motion.In the context of our squid propulsion example, we calculated the kinetic energy that the squid generated during its escape maneuver. The formula to calculate kinetic energy is: \[ KE = \frac{1}{2} mv^2 \]where
  • \( m \) is the mass of the moving object.
  • \( v \) is the velocity of the object.
To determine the kinetic energy of the squid, we use the mass excluding the expelled water: 4.75 kg, and the desired escaping velocity of 2.50 m/s. By plugging these values into the kinetic energy formula, we find that the squid creates approximately 14.84 Joules of kinetic energy in its escape.Understanding how to calculate kinetic energy helps us analyze the energy dynamics of moving objects, and this becomes essential in various physics scenarios, from simple motions to more complex systems involving collisions and propulsion mechanics.
Propulsion Mechanics
Propulsion mechanics involves the study of forces that propel objects through different media. In this case, we are examining how a squid uses a jet propulsion mechanism to move rapidly through water, a common method seen in many aquatic creatures. The squid's propulsion method relies on the conservation of momentum. The squid first fills a cavity with water and then expels it forcefully. When the squid expels water at a certain velocity, an equal and opposite reaction propels the squid in the opposite direction. This phenomenon is based on Newton's third law of motion. Let's break it down:
  • The squid's mass without the expelled water is 4.75 kg.
  • The expelled mass of water is 1.75 kg.
  • The squid escapes at a velocity of 2.50 m/s.
  • The velocity of the expelled water must balance the momentum to zero since initially, the system (squid plus water) is at rest.
By calculating this using conservation of momentum, we determined that the water must be expelled at a velocity of 6.79 m/s. This intricate interaction between the forces shows how effectively the squid can utilize its physical capabilities for propulsion.
Physics Problem Solving
Solving physics problems often involves applying core principles like the conservation of momentum or energy equations to analyze and predict outcomes. To tackle physics problems efficiently, understanding the initial conditions and the laws governing the system is crucial.Our squid escape scenario is an excellent example of physics problem solving. Initially at rest, the momentum of the entire system must remain zero once the squid starts moving. The known values鈥攁 mass of 6.50 kg including water, a required speed of 2.50 m/s for escape, and a water mass of 1.75 kg鈥攁re key inputs.A structured approach involves:
  • Identifying the goal, such as calculating velocity or energy.
  • Applying appropriate formulas, such as momentum conservation \( m_s \cdot v_s = m_w \cdot v_w \).
  • Substituting known values into these formulas.
  • Calculating the unknowns step by step.
This methodical process allows us to successfully find the solutions鈥攖he water鈥檚 expulsion speed and the total kinetic energy. By practicing these strategies, students develop a better understanding and capability for addressing various physics challenges, enhancing their problem-solving skills in this domain.

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Most popular questions from this chapter

\(\bullet\) Tennis, anyone? Tennis players sometimes leap into the air to return a volley. (a) If a 57 g tennis ball is traveling horizontally at 72 \(\mathrm{m} / \mathrm{s}\) (which does occur), and a 61 kg tennis player leaps vertically upward and hits the ball, causing it to travel at 45 \(\mathrm{m} / \mathrm{s}\) in the reverse direction, how fast will her center of mass be moving horizontally just after hitting the ball? (b) If, as is reasonable, her racket is in contact with the ball for \(30.0 \mathrm{ms},\) what force does her racket exert on the ball? What force does the ball exert on the racket?

\(\bullet\) On a frictionless, horizontal air table, puck \(A\) (with mass 0.250 \(\mathrm{kg}\) ) is moving to the right toward puck \(B\) (with mass \(0.350 \mathrm{kg} ),\) which is initially at rest. After the collision, puck \(A\) has a velocity of 0.120 \(\mathrm{m} / \mathrm{s}\) to the left, and puck \(B\) has a veloc- ity of 0.650 \(\mathrm{m} / \mathrm{s}\) to the right. (a) What was the speed of puck \(A\) before the collision? (b) Calculate the change in the total kinetic energy of the system that occurs during the collision.

\(\bullet\) Jonathan and Jane are sitting in a sleigh that is at rest on frictionless ice. Jonathan's weight is \(800 \mathrm{N},\) Jane's weight is \(600 \mathrm{N},\) and that of the sleigh is 1000 \(\mathrm{N} .\) They see a poisonous spider on the floor of the sleigh and immediately jump off. Jonathan jumps to the left with a velocity (relative to the ice) of 5.00 \(\mathrm{m} / \mathrm{s}\) at \(30.0^{\circ}\) above the horizontal, and Jane jumps to the right at 7.00 \(\mathrm{m} / \mathrm{s}\) at \(36.9^{\circ}\) above the horizontal (relative to the ice). Calculate the sleigh's horizontal velocity (magnitude and direction) after they jump out.

\(\bullet\) On a frictionless air track, a 0.150 \(\mathrm{kg}\) glider moving at 1.20 \(\mathrm{m} / \mathrm{s}\) to the right collides with and sticks to a stationary 0.250 \(\mathrm{kg}\) glider. (a) What is the net momentum of this two- glider system before the collision? (b) What must be the net momentum of this system after the collision? Why? (c) Use your answers in parts (a) and (b) to find the speed of the glid- ers after the collision. (d) Is kinetic energy conserved during the collision?

\(\bullet\) On a cold winter day, a penny (mass 2.50 g) and a nickel (mass 5.00 g) are lying on the smooth (frictionless) surface of a frozen lake. With your finger, you flick the penny toward the nickel with a speed of 2.20 \(\mathrm{m} / \mathrm{s}\) . The coins collide elastically; calculate both their final velocities (speed and direction).
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