Mechanics 1 Flashcards

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1
Q

Definition of Momenum

A

p=mv

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2
Q

Definition of Kinetic Energy

A

K=1/2(m)(v)^2

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3
Q

Conservation of Momentum

True for every interaction between objects

A

∑Pi=∑Pf

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4
Q

Conservation of Kinetic Energy

Only true for elastic collisions

A

∑Ki=∑Kf

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5
Q

Conservation of Momentum for a 1D, 2 body interaction

A

(Ma)(Vai)+(Mb)(Vbi)=(Ma)(Vaf)+(Mb)(Vbf)

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6
Q

Conservation of Momentum and Kinetic Energy for a 1D, 2 body elastic collision

A

(Ma)(Vai)+(Mb)(Vbi)=(Ma)(Vaf)+(Mb)(Vbf)

1/2(Ma)(Vai)^2+1/2(Mb)(Vbi)^2=1/2(Ma)(Vaf)^2+1/2(Mb)(Vbf)^2

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7
Q

Speed of Approach = Speed of Separation

Consequence of 1D, 2 body elastic collisions

A

Vai-Vbi=Vbf-Vaf

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8
Q

Conservation of Momentum and loss of Kinetic Energy for a 1D, 2 body inelastic collision

A

(Ma)(Vai)+(Mb)(Vbi)=(Ma)(Vaf)+(Mb)(Vbf)

1/2(Ma)(Vai)^2+1/2(Mb)(Vbi)^2 > 1/2(Ma)(Vaf)^2+1/2(Mb)(Vbf)^2

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9
Q

1D, 2 body, perfectly inelastic collision. Objects stick together. Greatest possible loss of KE while retaining momentum

A

(Ma)(Vai)+(Mb)(Vbi)=(Ma+Mb)Vf

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10
Q

1D, 2 Body, “explosion”

Energy is added during the interaction

A

(Ma)(Vai)+(Mb)(Vbi)=(Ma)(Vaf)+(Mb)(Vbf)

1/2(Ma)(Vai)^2+1/2(Mb)(Vbi)^2 < 1/2(Ma)(Vaf)^2+1/2(Mb)(Vbf)^2

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11
Q

Conservation of Momentum for a 2D, 2 Body interaction

A

(Ma)(Vaxi)+(Mb)(Vbxi)=(Ma)(Vaxf)+(Mb)(Vbxf)

Ma)(Vayi)+(Mb)(Vbyi)=(Ma)(Vayf)+(Mb)(Vbyf

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12
Q

2D, 2 body, elastic collision, kinetic energy is conserved and the Pythagorean Theorem can be used to break the V’s into x and y componenets

A

1/2(Ma)(Vaxi+Vayi)^2+1/2(Mb)(Vbxi+Vbyi)^2=1/2(Ma)(Vaxf+Vayf)^2+1/2(Mb)(Vbxf+Vbyf)^2

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13
Q

Newtons Second Law

Must be customized for each new situation

A

∑F=ma

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14
Q

Newtons Third Law

A

Fab=-Fba

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15
Q

On an incline, we rotate to coordinate system and force due to gravity can be broken into an x and y component

A

Fgx=Fgll=Fg:downhill=mgsinθ

Fgy=Fg⊥=Fg;into=mgcosθ

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16
Q

Kinetic (Sliding) Friction, always opposite to relative motion of objects, typically 0<u></u>

A

Fk=UkN

17
Q

Static Friction, typically 0<u></u>

A

Fs

18
Q

Magnitude of torque

A

T=rFsinθ

19
Q

First condition of equilibrium

A

∑F=0

20
Q

Second condition of equilibrium

A

∑T=0