What Is the Bird's Speed After Swallowing the Insect?

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Homework Help Overview

The discussion revolves around a physics problem involving conservation of momentum, specifically analyzing the speed of a bird after it swallows an insect. The scenario includes a bird of mass 300 g flying at 6.0 m/s and an insect of mass 10 g approaching at 30 m/s.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different methods to apply conservation of momentum, with some focusing on vector components and others on scalar values. There are attempts to calculate the final speed of the bird after the interaction, with varying interpretations of directionality and mass considerations.

Discussion Status

The discussion is active, with multiple participants providing different calculations and reasoning. Some participants question the treatment of momentum as vectors, while others suggest simplifying the problem by treating it as a linear interaction. There is no explicit consensus on the final speed, but various approaches are being examined.

Contextual Notes

Participants note the importance of direction in the momentum calculations, with some indicating that the problem does not specify a reference frame, leading to different interpretations of the final velocity's direction.

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A 300 g bird flying along at 6.0 m/s sees a 10 g insect heading straight toward it with a speed of 30m/s. The bird opens its mouth wide and enjoys a nice lunch. What is the bird’s speed immediately after swallowing?



Relevant equations
Jx = m(Vx)f - m(Vx)i
Jy = m(Vy)f - m(Vy)i

Given
m(b) = 0.3 kg
v(b) = 6.0 m/s
m(i) = 0.01 kg
v(i) = 30 m/s



The attempt at a solution

Jx = (0.01)(30) - (0.3)(6.0)
= 0.12kg.m/s

0.12kg.m/s * 1/0.3kg = 0.4 m/s

?
 
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.3 * 6 - .01 * 30 = (.3 + .01) * v

conservation of momentum

p1 + p2 = p1' + p2'

forget the components in this problem. the objects are traveling in a linear path, so a negative value and a positive value denote the direction. You can assign what each equal, though, just as long as you keep it consistent throughout the problem.
 
Last edited:
P1 = m1*v1
= (0.3)(6.0)
= 1.8

P2= m2*v2
= (0.01)(30)
=0.3

P1 +P2 = P3
1.8 +0.3 = P3
P3 = 2.1

P3 = m3v3
v3 = p3/m3
= 2.1/0.31
= 6.7741935
= 6.77 m/s?
 
jeeves_17 said:
P1 = m1*v1
= (0.3)(6.0)
= 1.8

P2= m2*v2
= (0.01)(30)
=0.3

P1 +P2 = P3
1.8 +0.3 = P3
P3 = 2.1

P3 = m3v3
v3 = p3/m3
= 2.1/0.31
= 6.7741935
= 6.77 m/s?

No. p1 and p2 are vectors and in this problem they point exactly opposite of each other

Think about it this way: this interaction happens ONLY on the x axis, so only use the x axis.p1 = .3 * 6 * cos(0)
p1 = .3 * 6 * 1
p1 = 1.8
p2 = .01 * 30 * cos(180)
p2 = .01 * 30 * -1
p2 = NEGATIVE .3

p1 + p2 = p1' + p2'

1.8 + -.3 = (.01 + .3) * v(the v after the eating happens)

if v is negative, then the answer will be at 180. if positive, 0. (remember, only because i defined 0 degrees as positive velocity and 180 degrees as negative velocity)

The answer is 4.84 m/s, 0 degrees or "in the direction the bird was flying"The problem doesn't define a perspective, so you could say the bird was flying from right to left and all that would happen is the answer will be 180 degrees instead of 0. I would just include the degrees and the words to explain your perspective. Just think about it. You could watch this collision from any perspective. The result won't change, only the way to express it.
 
Last edited:
just another way to do this is:


m1*(vf)1 + m2*(vf)2 = m1*(vi)1 + m2*(vi)2

because the bird swallows the insect (vf)1 and (vf)2 are going to be the same

(m1 + m2)*vf = m1*(vi)1 + m2*(vi)2

then solve for vf

vf= m1*(vi)1 + m2*(vi)2 divided by (m1 + m2)

to make things clear let's say the bird is flying in the positive direction, and since the insect is flying is the opposite direction of the bird making its velocity negative


sooo..

vf = (0.03kg * 6.0m/s) + (0.01kg * -30.0m/s) divided by (0.31kg)
vf= 4.8m/s
 

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