At African Lion Safari an Ape has a head on elastic collision with a Flamingo. The Ape has a mass of 190 kg and an initial velocity of 4.0 m/s [right]. It hits the Flamingo which has a mass of 20 kg and a velocity of 0.50 m/s [left]. Determine the final velocities.
At African Lion Safari an Ape has a head on elastic collision with a Flamingo. The Ape has a mass of 190 kg and an initial velocity of 4.0 m/s [right]. It hits the Flamingo which has a mass of 20 kg and a velocity of 0.50 m/s [left]. Determine the final velocities.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Kindly provide the solution to the following question using the GRASS method. Elastic and Inelastic Collisions - Ape vs Flamingo question, (Unit: Energy and Momentum).
At African Lion Safari an Ape has a head on elastic collision with a Flamingo. The Ape has a mass of 190 kg and an initial velocity of 4.0 m/s [right]. It hits the Flamingo which has a mass of 20 kg and a velocity of 0.50 m/s [left]. Determine the final velocities.
The image attached are the formulas for this unit. Please make sure to show all your work using the GRASS method and using formulas from this unit (Energy and Momentum).
![Energy
W = F d
1
Eg == k
EE
- lxx²
2
Momentum
→
Px = Px
1
E₁ = mgh_E₁₂ = mv²
Eg
Ex
2
T=2
Py = Py
Unit 2 Energy & Momentum
m
V k
Px
px = mix
Em = Ep + Ek
1 k
2 Vm
==
Py = mvy
F
spring
E₁ = E + Ek
EFAT = Ap
= kx Eth=Fd
Fnet = 4P
At
p = p²
Ek = Ek'
Ек](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7b519efd-481f-44c9-b15a-1bd08236bdce%2F2437eba8-77fe-4761-9c01-6f39d6a29921%2Fep16cz9_processed.png&w=3840&q=75)
Transcribed Image Text:Energy
W = F d
1
Eg == k
EE
- lxx²
2
Momentum
→
Px = Px
1
E₁ = mgh_E₁₂ = mv²
Eg
Ex
2
T=2
Py = Py
Unit 2 Energy & Momentum
m
V k
Px
px = mix
Em = Ep + Ek
1 k
2 Vm
==
Py = mvy
F
spring
E₁ = E + Ek
EFAT = Ap
= kx Eth=Fd
Fnet = 4P
At
p = p²
Ek = Ek'
Ек
![Gravitational constant: 6.67 × 10-¹¹ Nm²/kg²
Mass of Earth: 5.98 × 10²4 kg
Radius of Earth: 6.38 × 106 m
RIGHT TRIANGLE
sin 0
Oppl Hyp
cos 0
Adjl Hyp
tan 0 = Oppl Adj
=
Hypotenuse
Side adjacent to
Side opposite to 8
Universal Constants
SINE LAW
a
sin A
Miscellaneous
b
sin B
Mass of Sun: 1.99 × 10³⁰ kg
Mass of Moon: 7.36 × 10²² kg
Radius of Moon: 1.74 × 10⁰ m
C
sin C
COSINE LAW
2 = b² + c² - 2bccosA
=
- 2 accos B
+ c²
a
a² + b² - 2abcos C
62
c² =
C
a](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7b519efd-481f-44c9-b15a-1bd08236bdce%2F2437eba8-77fe-4761-9c01-6f39d6a29921%2F0jjtvj_processed.png&w=3840&q=75)
Transcribed Image Text:Gravitational constant: 6.67 × 10-¹¹ Nm²/kg²
Mass of Earth: 5.98 × 10²4 kg
Radius of Earth: 6.38 × 106 m
RIGHT TRIANGLE
sin 0
Oppl Hyp
cos 0
Adjl Hyp
tan 0 = Oppl Adj
=
Hypotenuse
Side adjacent to
Side opposite to 8
Universal Constants
SINE LAW
a
sin A
Miscellaneous
b
sin B
Mass of Sun: 1.99 × 10³⁰ kg
Mass of Moon: 7.36 × 10²² kg
Radius of Moon: 1.74 × 10⁰ m
C
sin C
COSINE LAW
2 = b² + c² - 2bccosA
=
- 2 accos B
+ c²
a
a² + b² - 2abcos C
62
c² =
C
a
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