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A \(15.0 \mathrm{~kg}\) fish swimming at \(1.10 \mathrm{~m} / \mathrm{s}\) suddenly gobbles up a \(4.50 \mathrm{~kg}\) fish that is initially stationary, Ignore any drag effects of the Water. (a) Find the speed of the large fish just after it eats the small one. (b) How much total mechanical energy was dissipated during this meal?

Short Answer

Expert verified
The speed of the large fish just after it eats the small one is approximately 0.83 m/s. The total mechanical energy dissipated during this meal is around 1.43 J.

Step by step solution

01

Understand the Principle of Conservation of Momentum

The problem involves two bodies moving together after a collision. This is a perfect situation to apply the Principle of Conservation of Momentum. Given that before the 'collision' we have only the momentum of the large fish moving and after the 'collision' we have both fishes moving together, we can create the equation for this conservation principle: \(Momentum_{before}=Momentum_{after}\). Since Momentum = mass x velocity, then we will have: \((15.0 kg)*(1.10 m/s)= (15.0 kg + 4.50 kg)*v\) where v is the velocity of both fishes after the gobbling.
02

Solve for the Velocity

In this step, we will solve for the velocity 'v'. Re-arranging the equation from step 1, we will get: \(v = \frac{(15.0 kg)*(1.10 m/s)}{(15.0 kg + 4.50 kg)}\). After doing the calculation we will find the resulting speed of the large fish just after it eats the small one.
03

Calculate the Total Mechanical Energy Dissipated

The total mechanical energy (kinetic in this case) before and after the gobbling is given by \(E_{initial} - E_{final}\) = ΔE, where \(E_{initial}\) is the kinetic energy of the large fish initially (given by \(\frac{1}{2}mv_{1}^{2}\)) and \(E_{final}\) is the total kinetic energy of both fishes after the event (given by \(\frac{1}{2}Mv_{2}^{2}\) where M = \(m_{1} + m_{2}\) is the total mass and \(v_{2}\) is the final speed from Step 2). The result of this calculation will show how much total mechanical energy was dissipated during this meal event.

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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 a type of energy that a body possesses due to its motion. It is calculated using the formula \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. In this exercise, we initially have a large fish moving at a certain speed while the smaller fish is stationary.
The initial kinetic energy is only due to the large fish, as the stationary fish has zero kinetic energy. When the two fishes become one system after gobbling, the total kinetic energy needs to be recalculated considering the new combined mass and the reduced velocity.

Though the total momentum stays the same due to the principle of conservation of momentum, the kinetic energy typically changes during inelastic collisions, leading to some energy dissipation. This results from the conversion of kinetic energy into other forms such as thermal energy, sound, and, in reality, sometimes due to water resistance too, though it's ignored here.
Collision Mechanics
Collision mechanics uses the principles of physics to analyze the interactions happening between colliding bodies. In the problem scenario, two fishes collide - one actively gobbles up the other. This situation perfectly illustrates the momentum conservation principle, a fundamental aspect of collision mechanics.
When the large fish collides with the small fish, the principle of conservation of momentum applies. This law states that within a closed system without external forces, the total momentum before the collision is equal to the total momentum after the collision. As such, no momentum is gained or lost.

Collision mechanics is crucial in predicting outcomes of impacts in car crashes, sports, and various engineering applications. It involves calculating variables like velocity and understanding kinetic exchanges during the impact, which is why it’s essential to differentiate between elastic and inelastic collisions situations.
Inelastic Collision
An inelastic collision is a type of collision where the colliding objects stick together afterwards, resulting in a combined mass moving with a common velocity. In the given problem, the large fish eating the small fish becomes an inelastic collision, with both masses then moving as one.
  • Momentum is conserved in every inelastic collision.
  • Total kinetic energy is not conserved, as some is turned into other forms of energy.
This kind of collision is contrasted with an elastic collision where objects collide and rebound, maintaining both kinetic energy and momentum. In cases of inelastic collisions, you often see energy lost in the system. Here, the energy loss is determined by subtracting the final kinetic energy from the initial kinetic energy.
Through such analyses, better understanding can be gained concerning how real-world interactions involve energy dissipations often neglected in ideal conditions.

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Most popular questions from this chapter

You are standing on a large shect of frictionless ice and holding a large rock. In orver to get off the ice. you throw the rock so it has velocity \(12.0 \mathrm{~m} / \mathrm{s}\) relutive to the eurth at an angle of \(35.0^{\circ}\) above the horizontal. If your mass is \(70.0 \mathrm{~kg}\) and the rock's mass is \(3.00 \mathrm{~kg}\), what is your speed after you throw the rock? (See Discussion Question \(Q 8.7 .)\)

Two figure skaters, one weighing \(625 \mathrm{~N}\) and the other \(725 \mathrm{~N}\) push off against cach other on frictionless ice. (a) If the heavicr skater travels at \(1.50 \mathrm{~m} / \mathrm{s}\), how fast will the lighter one travel? (b) How much kinetic energy is "created" during the skaters' maneuver, and where does this cnergy come from?

A packing crute with mass \(80.0 \mathrm{~kg}\) is at rest on a horicontal. frictionlcss surface. At \(t=0\) a net horizontal force in the \(+x\) -dircction is applied to the crate. The force has a constant value of \(80.0 \mathrm{~N}\) for \(12.0 \mathrm{~s}\) and then decrcases lincarly with time so it becomes \(7 \mathrm{cro}\) after an ad. Jitional \(6.00 \mathrm{~s}\). What is the final speed of the crate. \(18.0 \mathrm{~s}\) after the force was first applied?

Starting at \(t=0\). a horizontal net externul force \(F=(0.280 \mathrm{~N} / \mathrm{s}) t \hat{\imath}+\left(-0.450 \mathrm{~N} / \mathrm{s}^{2}\right) r^{2} \hat{\jmath}\) is applied to a box that has an initial momentum \(\vec{p}=(-3.00 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}) \hat{\imath}+(4.00 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}) \hat{\jmath} .\) What is the momentum of the box at \(t=2.00 \mathrm{~s} ?\)

A \(1200 \mathrm{~kg}\) SUV is moving along a straight highway at \(12.0 \mathrm{~m} / \mathrm{s}\). Another car, with mass \(1800 \mathrm{~kg}\) and speed \(20.0 \mathrm{~m} / \mathrm{s}\), has its center of mass \(40.0 \mathrm{~m}\) ahead of the center of mass of the SUV (Fig. E8.54). Find (a) the position of the center of mass of the system consisting of the two cars; (b) the magnitude of the system's total momentum, by using the given data: (c) the speed of the system's center of mass; (d) the system's total momentum, by using the speed of the center of mass. Compare your result with that of part (b).

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