Mechanics 1 Flashcards
Definition of Momenum
p=mv
Definition of Kinetic Energy
K=1/2(m)(v)^2
Conservation of Momentum
True for every interaction between objects
∑Pi=∑Pf
Conservation of Kinetic Energy
Only true for elastic collisions
∑Ki=∑Kf
Conservation of Momentum for a 1D, 2 body interaction
(Ma)(Vai)+(Mb)(Vbi)=(Ma)(Vaf)+(Mb)(Vbf)
Conservation of Momentum and Kinetic Energy for a 1D, 2 body elastic collision
(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
Speed of Approach = Speed of Separation
Consequence of 1D, 2 body elastic collisions
Vai-Vbi=Vbf-Vaf
Conservation of Momentum and loss of Kinetic Energy for a 1D, 2 body inelastic collision
(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
1D, 2 body, perfectly inelastic collision. Objects stick together. Greatest possible loss of KE while retaining momentum
(Ma)(Vai)+(Mb)(Vbi)=(Ma+Mb)Vf
1D, 2 Body, “explosion”
Energy is added during the interaction
(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
Conservation of Momentum for a 2D, 2 Body interaction
(Ma)(Vaxi)+(Mb)(Vbxi)=(Ma)(Vaxf)+(Mb)(Vbxf)
Ma)(Vayi)+(Mb)(Vbyi)=(Ma)(Vayf)+(Mb)(Vbyf
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
1/2(Ma)(Vaxi+Vayi)^2+1/2(Mb)(Vbxi+Vbyi)^2=1/2(Ma)(Vaxf+Vayf)^2+1/2(Mb)(Vbxf+Vbyf)^2
Newtons Second Law
Must be customized for each new situation
∑F=ma
Newtons Third Law
Fab=-Fba
On an incline, we rotate to coordinate system and force due to gravity can be broken into an x and y component
Fgx=Fgll=Fg:downhill=mgsinθ
Fgy=Fg⊥=Fg;into=mgcosθ
