/*! 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 62 Shows a bird catching a fish. As... [FREE SOLUTION] | 91Ó°ÊÓ

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Shows a bird catching a fish. Assume that initially the fish jumps up and that the bird coasts horizontally and does not touch the water with its feet or flap its wings. (a) Is this kind of collision (1) elastic, (2) inelastic, or (3) completely inelastic? Why? (b) If the mass of the bird is \(5.0 \mathrm{~kg}\), the mass of the fish is \(0.80 \mathrm{~kg}\), and the bird coasts with a speed of \(6.5 \mathrm{~m} / \mathrm{s}\) before grabbing, what is the speed of the bird after grabbing the fish?

Short Answer

Expert verified
(a) The collision is (3) completely inelastic. (b) The speed of the bird after grabbing the fish is approximately 5.6 m/s.

Step by step solution

01

Analyze the Type of Collision

This scenario involves a bird catching a fish in the air. In this event, the bird and the fish stick together after the interaction. In collisions where the objects stick together, the collision is known as completely inelastic. Thus, this collision is (3) completely inelastic.
02

Setup Conservation of Momentum Equation

In completely inelastic collisions, the mass after collision is the sum of the masses of the two objects, and only momentum is conserved. The momentum before the collision is \( m_b v_b + m_f v_f \), and after the collision is \( (m_b + m_f) v' \), where \( v' \) is the final velocity of the bird-fish system, \( m_b \) is the mass of the bird, and \( m_f \) is the mass of the fish.
03

Determine Initial Momentum

Before the collision, the fish's initial vertical speed does not contribute to horizontal momentum of the bird-fish system, so \( v_f = 0 \) m/s. The initial total horizontal momentum is thus \( m_b v_b = 5.0 \times 6.5 = 32.5 \text{ kg m/s} \).
04

Calculate Final Velocity of Bird and Fish System

Since the momentum is conserved, set the initial horizontal momentum equal to the final horizontal momentum: \( 32.5 = (5.0 + 0.80) v' \). Solve for \( v' \): \( v' = \frac{32.5}{5.8} \approx 5.6 \text{ m/s} \).

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Key Concepts

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

Conservation of Momentum
In physics, momentum is a fundamental concept that embodies the motion of objects. The conservation of momentum principle is crucial in understanding interactions like collisions. Conservation of momentum means that in a closed system with no external forces, the total momentum remains constant before and after a collision.

Imagine a scenario where a bird catches a fish. From the problem, we assume no external horizontal forces act on the bird-fish system during the interaction. This makes it an excellent example of momentum conservation.
  • Before the collision, the bird is flying horizontally with momentum calculated as the product of its mass and velocity, given by \( m_b v_b \).
  • The fish leaps into the air and has no initial horizontal momentum, \( v_f = 0 \) m/s.
After they collide and stick together, momentum conservation allows us to solve for the final velocity of the joint bird-fish system. The principle is expressed as:
\[ m_b v_b + m_f v_f = (m_b + m_f) v' \]Where:
\( m_b \) and \( m_f \) are the masses of the bird and fish respectively,
\( v_b \) and \( v_f \) are their respective velocities before the collision,
\( v' \) is the shared velocity after the collision.
Collision Types
Collisions in physics can be broadly categorized based on how energy and momentum are conserved during the interaction.

These categories include elastic, inelastic, and completely inelastic collisions. Specifically, a completely inelastic collision is backed by unique characteristics:
  • In an **elastic collision**, both momentum and kinetic energy are conserved.
  • In an **inelastic collision**, only momentum is conserved, not kinetic energy.
  • In a **completely inelastic collision**, as seen in the problem, momentum is conserved but objects stick together post-collision.
In our bird-fish example, after the bird catches the fish, they move as one entity. This sticking together indicates a completely inelastic collision. Despite some kinetic energy being transformed into other forms, such as sound or heat, the total system momentum remains constant. This is an important principle in solving for their shared velocity.
Momentum Calculation
Calculating the momentum in systems like a bird catching a fish helps understand the velocities involved in collisions. Initially, we focus on the momentum of each object before they interact:

  • The bird's initial momentum is computed as \( m_b \times v_b \).
  • For the fish, since it leaps upward, its horizontal velocity is \( v_f = 0 \), giving it zero initial horizontal momentum.
Using the problem's values, calculate:
\[ 5.0 \times 6.5 = 32.5 \text{ kg m/s} \]
This is the total initial momentum, entirely contributed by the bird's flight. Post-collision, the final momentum has to match this initial value since momentum is conserved:
\[ (m_b + m_f) \times v' = 32.5 \text{ kg m/s} \]
Where:
  • \( m_b = 5.0 \text{ kg} \) is the mass of the bird,
  • \( m_f = 0.80 \text{ kg} \) is the mass of the fish,
  • \( v' \) is the velocity to find.
Solving the equation for \( v' \):
\[ v' = \frac{32.5}{5.8} \approx 5.6 \text{ m/s} \]
This result gives the speed of both the bird and fish after they collide and move together, highlighting the concept of momentum conservation in action.

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Most popular questions from this chapter

In a laboratory setup, two frictionless carts are placed on a horizontal surface. Cart A has a mass of \(500 \mathrm{~g}\) and cart B's mass is \(1000 \mathrm{~g}\). Between them is placed an ideal (very light) spring and they are squeezed together carefully, thereby compressing the spring by \(5.50 \mathrm{~cm} .\) Both carts are then released and \(\mathrm{B}^{\prime}\) s recoil speed is measured to be \(0.55 \mathrm{~m} / \mathrm{s}\). (a) Will cart A's speed be (1) greater than, (2) less than, or (3) the same as B's speed? Explain. (b) Determine B's recoil speed to see if your conjecture in (a) was correct. (c) Determine the spring constant of the spring.

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