UNIVERSITY PHYSICS I CHAPTER 9 Impulse, Momentum and Collision 99.1 Momentum and Newton's second law of motion Collisions: All around us(cars, buses) Grand astronomical scale (stars, galexies) Nuclear, atomic and molecular physics The impulse-momentum theorem relates the change in the momentum of a system to the total force on it and the time interval during which this total forces acts
1 Collisions: All around us (cars, buses) Grand astronomical scale (stars, galexies) Nuclear, atomic and molecular physics The impulse-momentum theorem relates the change in the momentum of a system to the total force on it and the time interval during which this total forces acts. §9.1 Momentum and Newton’s second law of motion
89.1 Momentum and Newton 's second law of motion 1. momentum (1)a single particle of mass m Define:p=mν=m (2 a group of particles of mass m1,m2,…,m,… ∑i P1=∑mv Position vector of center of mass 2. the center of mass Total The total momentum of a system of particle: P- p=∑n=∑ dr d M dt 89.1 Momentum and Newtons second law of motion (1)The position vector of mass center(weighted average) Define: =∑ M 元1+22+ M That is IcM= m1F1+m2F2+…+mN m1+m2+…+mN Total mass of the particle system M=n n1+n,+…+n1 12
2 1. momentum (1)a single particle of mass m Define: t r p mv m d d r r r = = (2)a group of particles of mass m1, m2, …, mi , … = ∑ = ∑ i i i i i p p m v r r r 2. the center of mass ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ =∑ =∑ = ∑ i i i i i i i i M m r t M t r p p m r r r r d d d d The total momentum of a system of particle： t r p M d d CM r r = ? Position vector of center of mass Total mass §9.1 Momentum and Newton’s second law of motion x y z 1r r Nr r m1 m2 mN O 2 r r (1)The position vector of mass center (weighted average) N N N m m m m r m r m r r + + + + + + = L r L r r r 1 2 1 1 2 2 CM That is N N i i i r M m r M m r M m M m r r r L r r r r = + + + = ∑ 2 2 1 1 CM Define: M = mtotal = m1 + m2 +L+ mN Total mass of the particle system: C c r r §9.1 Momentum and Newton’s second law of motion
89.1 Momentum and Newtons second law of motIon The components in the Cartesian coordinate system: For the system composed of single particles ∑ md x CM S i=l M J CM Ⅵ ∑ miz CM M 89.1 Momentum and Newton's second law of motion For the common objects as essentially continuous distribution of matter: rd t dm(x, CM S M rd CM M ycMs火J M zd M
3 The components in the Cartesian coordinate system: M m z z M m y y M m x x N i i i N i i i N i i i ∑ ∑ ∑ = = = = = = 1 CM 1 CM 1 CM x y z 1r r Nr r m1 m2 mN O C c r r 2 r r For the system composed of single particles M m r r N i ∑ i i = = 1 CM r r §9.1 Momentum and Newton’s second law of motion M z m z M y m y M x m x ∫ ∫ ∫ = = = d d d CM CM CM o x z y M dm( ) x, y,z r r For the common objects as essentially continuous distribution of matter: M r m r ∫ = d CM r r §9.1 Momentum and Newton’s second law of motion
89.1 Momentum and Newtons second law of motion Example 1: P420 56(b) y 3m Soluti 2 mx0+2ml+3ml/2 7l CM 6m 12 m×0+2m×0+3msin60°3√31 J CM 6m 12 89.1 Momentum and Newton's second law of motion Example 2: a thin strip of material is bent into the shape of a semicircle of radius r Find its center of mass Solution: dm since the symmetry, we obtained R C 0 w=∫om= M (Rsin)-dφ R 2R SIn =0.637R 元 元
4 Example 1: P420 56(b) l l l m 2m 3m x y 12 3 3 6 0 2 0 3 sin60 12 7 6 0 2 3 / 2 l m m m ml y l m m ml ml x CM CM = × + × + = = × + + = o Solution: §9.1 Momentum and Newton’s second law of motion Example 2: A thin strip of material is bent into the shape of a semicircle of radius R. Find its center of mass. Solution: since the symmetry, we obtained R R R M R M y m M y x CM CM 0.637 2 sin d ( sin ) d 1 d 1 0 0 0 = = = = = = ∫ ∫ ∫ π π π φ φ π φ π φ x y R φ dm dφ §9.1 Momentum and Newton’s second law of motion
89.1 Momentum and Newtons second law of motion (2)Momentum of system of particles ∑D=∑m dr M M M Ptotal M CM My CM (3)Kinetic energy of system of particles 十 =Vo+ 89.1 Momentum and Newton's second law of motion The total Kinetic energy of the whole system KEHI 722m2= ∑,m(x+的) (om +v 2 mv ∑ 2 v/+2vcM vi ∑m"+∑m+m:∑m可 KE=-MI 2 2 CM 2 v
5 (2) Momentum of system of particles CM CM total d d Mv t r p M r r r = = t r M M m r t M t r p p m i i i i i i i i d d d d d d CM r r r r r =⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ =∑ =∑ = ∑ (3) Kinetic energy of system of particles i i i i v v v r r r = + ′ = + ′ r r r r r r CM CM §9.1 Momentum and Newton’s second law of motion x y z mi O C rCM r i r r i r ′ r The total Kinetic energy of the whole system i i i i i i i i i i i i i i i i i i i i i i i m v m v v m v m v v v v m v v v v KE m v m v v = + ′ + ⋅ ′ = + ′ + ⋅ ′ = + ′ ⋅ + ′ = = ⋅ ∑ ∑ ∑ ∑ ∑ ∑ ∑ r r r r r r r r r r CM 2 2 CM CM 2 2 CM CM CM 2 total 2 1 2 1 ( 2 ) 2 1 ( ) ( ) 2 1 2 1 2 1 ∑ ′ = 0 i i i m v r 2 2 total CM 2 1 2 1 i i i KE = Mv +∑ m v′ §9.1 Momentum and Newton’s second law of motion