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
89.1 Momentum and Newtons second law of motion (I)dynamics of a single partcle d law 3. Momentum and Newton's seco F dpd(mν) n dtdt (2)dynamics of a system of particles (a)The velocity of the center of mass: dr d ∑ mi ' va v M CM dt dt ∑=∑ M dt 或 (b The acceleration of the center of mass: dvcs= d ∑ m1. adm CM CM or dt dt M M 89.1 Momentum and Newton's second law of motion (c)Momentum of a system of particles Ptotal =M-CM M CM (d)Newton's second law for the system of particles admt) v F total CM total Ma dt The longer the total force acts on a system, the longer the system is subjected to an acceleration and the greater the change in its velocity. 6
6 3. Momentum and Newton’s second law (1)dynamics of a single particle ma t mv t p F r r r r = = = d d( ) d d total §9.1 Momentum and Newton’s second law of motion (2)dynamics of a system of particles M v m M m v t r M m M m r t t r v i i i i i i CM i i CM ∫ ∑ = = ∑ =∑ = d d d d d d d r r r r r r 或 (a)The velocity of the center of mass: M a m M m a t r t v a i i i CM CM CM ∫ ∑ = = = d or d d d d 2 2 r r r r r (b)The acceleration of the center of mass: (d) Newton’s second law for the system of particles §9.1 Momentum and Newton’s second law of motion CM i i i Ma t v M t m v t p F r r r r r = = = = ∑ d d d d( ) d d total CM total The longer the total force acts on a system, the longer the system is subjected to an acceleration and the greater the change in its velocity. CM CM Mv t r p M r r r = = d d total (c)Momentum of a system of particles
89.1 Momentum and Newtons second law of motion (e) Internal and external forces The forces from other particles within the system of particles is called internal force. The forces from sources outside the system of particles is called external force. The total force of the system: F1+F2+…+F2+… Flin t Fle+ F2in t Fret =F1++F2+…+F+… 89.1 Momentum and Newtons second law of motion According Newton’ s third law Fu F2r F23 ∑Fm=0 F3n F13 se f= F+ F 1 in dt =F.+F= dt PN N F、+F N dt
7 The forces from other particles within the system of particles is called internal force. The forces from sources outside the system of particles is called external force. The total force of the system: (e) Internal and external forces L r L r r L r r r r r L r L r r r = + + + + + = + + + + = + + + + + 1ex 2ex i ex 1in 1ex 2in 2ex total 1 2 F F F F F F F F F F Fi FN §9.1 Momentum and Newton’s second law of motion §9.1 Momentum and Newton’s second law of motion m1 m2 m3 F12 r F21 r F13 r F31 r F32 r F23 r F1ex r F3ex r F2ex r According to Newton’s third law in = ∑ in ≡ 0 i F Fi r r t p F F F t p F F F t p F F F N N N N d d d d d d ex in 2 2 2 ex 2 in 1 1 1ex 1in r r r r LLL r r r r r r r r = + = = + = because = + =
89.1 Momentum and Newtons second law of motIon Consider F. ∑ F,≡0 The sum of the equations all above is d le F,+∵+ (P1+P PN) dy total total M M dt dt 9.2 Impulse-momentum theorem 1. Impulse-momentum theorem of a single particle because Fota dt=dp let dI= Ftotal dt -differential impulse The impulse of the total force on the particle 7-d=Fm(0=∫"=,-=4 8
8 Consider The sum of the equations all above is ( ) d d 1ex 2ex Nex p1 p2 p N t F F F r L r r r L r r + + + = + + + 0 in = ∑ in ≡ i F Fi r r §9.1 Momentum and Newton’s second law of motion MaCM t v M t p F r r r r = = = d d d d total CM total §9.2 Impulse-momentum theorem 1. Impulse-momentum theorem of a single particle F t p r r total totald = d d d F t p r r because = dI F dt total r r let = --differential impulse The impulse of the total force on the particle I I F t p p p p f i p p t t f i f i r r r r r r r r = = = r = − = ∆ ∫ ∫ ∫ d total(t)d d
89.2 Impulse-momentum theorem 2. The Cartesian components of the impulse F dt=4 px dt Apy tavE y total dt= Ap 3. Impulse and average impulse force i=F total dt= F.at Ix= favdt=FAr I =FaA I=faAr 9.2 Impulse-momentum theorem 4. Impulse-momentum theorem of a system of rticle For a particle system dpo -ft f dt ps ap vota total Components in the Cartesian coordinate system notice x total =k Fr tota dt =4p rtotal y total Fmad=4rmum=∑Fm=0 F total 4 totalis total n tota
9 z t t z z y t t y y x t t x x I F t p I F t p I F t p f i f i f i ∆ ∆ ∆ = = = = = = ∫ ∫ ∫ d d d total total total I F t F t I F t I F t y y z z t t x = x = x ∆ = ∆ = ∆ ∫ ave ave ave ave 2 1 d O t Fxtotal Fxave i t f t I F t F t t t total ave∆ 2 1 d r r r = = ∫ 2. The Cartesian components of the impulse 3. Impulse and average impulse force §9.2 Impulse-momentum theorem total total total Itotal F dt dp p f i f i p p t t r r r r r = = r = ∆ ∫ ∫ total total total total total total total total total d d d z t t z z y t t y y x t t x x I F t p I F t p I F t p f i f i f i ∆ ∆ ∆ = = = = = = ∫ ∫ ∫ 4. Impulse-momentum theorem of a system of particles total total d d F t p r r For a particle system = Components in the Cartesian coordinate system: 0 1 in total = ∑ in ≡ = N i F Fi r r in total = in totald ≡ 0 ∫ I F t f i t t r r notice： §9.2 Impulse-momentum theorem
89.2 Impulse-momentum theorem Example 1: throw a ball of mass 0. 40kg against a brick wall. Find (a)the impulse of the force on the ball; (b )the average horizontal force exerted on the ball. if the ball is in contact with the wall for 0.01s Solution: (a)I=Px -pxi v1=-30m/s Px=mv1=0.40×(-30) v2=20m/ =-12kg·m/s x=m2=0.40×20 8. 0kg. m/s 1=Px -Pxi=8.0-(12)=20 kg.m/s 9.2 Impulse-momentum theorem Fdt= F.4t 20 =2000N t0.01 Example 2: a base ball of mass 0. 14kg is moving horizontally at speed of 42m/s when it is truck by the bat. It leaves the bat in a direction at an angle 35 above its incident path and with a speed of 50m/.(a)Find the impulse of the force exerted on the ball.(b) assuming the collision last for 1.5 ms, what is the average force? (c)Find the change in the momentum of the bat. 10
10 Example 1: throw a ball of mass 0.40kg against a brick wall. Find (a)the impulse of the force on the ball;(b)the average horizontal force exerted on the ball, if the ball is in contact with the wall for 0.01s. 30m/s v1 = − 20m/s v2 = x Solution: (a) x pxf pxi I = − 8.0 (12) 20 kg m/s 8.0kg m/s 0.40 20 12kg m/s 0.40 ( 30) 2 1 = − = − = ⋅ = ⋅ = = × = − ⋅ = = × − x xf xi xf xi I p p p mv p mv §9.2 Impulse-momentum theorem (b) I F t F t x t t x x f i = d = ave∆ ∫ 2000N 0.01 20 ave = = = t I F x x ∆ §9.2 Impulse-momentum theorem Example 2: a base ball of mass 0.14kg is moving horizontally at speed of 42m/s when it is truck by the bat. It leaves the bat in a direction at an angle φ=35º above its incident path and with a speed of 50m/s. (a)Find the impulse of the force exerted on the ball. (b) Assuming the collision last for 1.5 ms, what is the average force? (c)Find the change in the momentum of the bat. o 35
