Today… Ch 6, Momentum and Collisions System of particles Elastic vs. inelastic collision Elastic collision in 1D Collision in 2D Center of mass Motion of system of particles (Motion of center of mass) 1 System of objects (particles) We consider system of “multiple” objects, which we can choose any way convenient for us. For example, The system: m1, m1 m3 m2 and m3; however m4 and m2 m5 are external objects! m4 m5 2 1 Internal vs. External Forces Internal forces: Forces between objects within the system External forces: Forces from outside the system m1 m3 m2 m4 m5 3 Net Momentum of System of Particles (or, Objects) G GG G G Pppnet =++=12... mvmv 1122 + + ... G G v1 v2 The system: m1, m1 m3 G m2 and m3; G v4 however m4 and m2 v3 m5 are external objects! m4 G m5 v5 4 2 Net external force, Net external impulse G GG Net external force : FFFnet,,1,2 ext= ext++ ext ... sum of all “external” forces acting on the system m1 system m3 m2 m4 G m5 G Net external impulse: IFtnet,, ext= net ext Δ 5 ImpulseImpulse--MomentumMomentum Theorem for System of Particles GGG Inet,,, ext=−PP net f net i NtNet exter na l impu lse is equal to ne t mo me ntum chan ge. Proof: use Newton’s 3rd law (see textbook) Note: Internal forces do not matter for net momentum change. 6 3 Conservation of momentum G G If I = 0 t is small net, ext , for example, Fnet, ext = 0 or Δ GG Æ Pnet,, f= P net i If the net external impulse is zero, for example, if the net external force is zero, the net momentum of a system is conserved. 7 Elastic vs. Inelastic Collisions In a collision within a system of particles Before After collision collision 11 Net kinetic energy: KKK= ++=... mvmv22 + + ... net 1222 1 2 If KKnet,, i= net f Æ elastic collision. If KKnet,, i≠ net f Æ inelastic collision. If KKnet,, i≠ net f and objects move together after collision Æ perfectly inelastic collision. 8 4 Elastic Collisions in 11--DimensionDimension Both total momentum & total kinetic energy are conserved. 11 1 1 mv+=+ mv mv mv & mv22+=+ mv mv 2 mv 2 1 1,iif 2 2, 1 1, 2 2, f2211iif 2 2 2 11 2 2 2 f v1i v2i Before collision m1 m2 +x direction v1f=? v2f=? After collision mm− 2 m 2mmm− vvv=+12 2 vvv=+121 1,f 1,ii 2, and 2,f 1,ii 2, mm12++ mm 12 mm12++ mm 12 (see text for proof) iClicker Quiz In 1D elastic collision, if m1=m2, vi1=+10 m/s, v2i=0, then, after collision v1f= _____ and v2f= _____. a) 5 m/s; 5 m/s b) -5 m/s; 5 m/s c) 0 m/s; 10 m/s d) 0 m/s; -10 m/s e) -10 m/s; 0 m/s V1i=10m/s V2i=0 Before collision m1 m2 +x direction v1f=? v2f=? After collision mm− 2 m 2mmm− vvv=+12 2 vvv= 121+ 1,f 1,ii 2, and 2,f 1,ii 2, mm12++ mm 12 mm12++ mm 12 5 Collisions in Two Dimensions If net external force is zero, net momentum is conserved. GG G G G Æ P = P , where net,, i net f Pmvmvnet = 11++ 2 2 ... Or y PPnet,, i x= net , f , x V1 PPnet,, i y= net , f , y If net kinetic energy is v1 conserved Æ Elastic collision x not conserved Æ Inelastic collision m1 v2 m2 If moving together after collision Æ Perfectly Inelastic collision V2 (Kinetic energy is not conserved.) Example 2: Collision at an intersection A 1500 kg car traveling east with a speed of 25 m/s collides at an intersection with a 2500 kg van traveling north at a speed of 20 m/s. Find the direc tio n and magnit u de of the velocity of the wreckage after the collision, assuming the vehicles stick together after the collision. 6 Example 1: 90 degree deflection rule in a game of pool G v2, f m G m v1,i G v1, f Assume that the collision is elastic, and the two balls have the same mass. Show that the angle between the outgoing balls is 90 degree. (No forward, back or side spin is in effect.) How could we analyze the motion of extended objects, or system of particles? 7 Concept of Center of Mass m1=3 kg m3=10 kg m2=6 kg For a system of particles or an extended object, “Cen ter of mass” is an “average” position for mass distdistibributi on. Definition of center of mass (com) in 1D m1 m2 m3 x3 x1 x2 In 1D, mx11+ mx 2 2+... xcom = mm12++... , where x is position of mass i mi 8 Definition of center of mass (com) in 2D mx11++ mx 2 2 ... xcom = mm12++... , where my11++ my 2 2 ... ycom = mm12++... (,xyii ) is the position of GG mi G mr11++ mr 2 2 ... rcom = mm12++... Example 3 Three particles of masses m1 = 1.1 kg, m2 = 2.5 kg, and m3 = 3.4 kg are located as shown in the figure: m1 is at (0,0), m2 is at (140 m,0), and m3 is at (70 m, 120 m). Find the coordinate of the center of mass. 9 iClicker Quiz A two-section piece, represented by the gray area on the figure, is cut from a metal plate of uniform thickness. The point that corresponds to the center of mass of this piece is closest to (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 Velocity of center of mass (com) Δxcom mv1,1xx+ mv 2,2+... vxcom, == Δ++tmm12... ()(mm12++...) vx ,com = mvmv 1,12,2 x + x +=... P net , x Similar for y component G G G mv11+ mv 2 2+... vcom = mm12++... G GG G where Pnet=+mv11 mv 2 2 +=... Mv com Mmm= 12++... 10 Example 1 A 2.0 kg particle has a velocity (2.0 m/s, -3.0 m/s), and a 3.0 kg particle has a velocity (1.0 m/s, 6.0 m/s). Find (()a) velocity of the center of mass and ()(b) the total momentum of the system. Acceleration of center of mass (com) Δ++vmamaxcom,1,12,2 x x ... axcom, == Δ++tmm12... Similar for y component. In vector form G G G ma11+ ma 2 2+... acom = mm12++... G GG ()(mm12++...) acom = mama 1122 + +... GGG Macom =+ m11 a m 2 a 2 +... 11 If we have only two objects in the system… system GGG m1 Mamamacom =+11 2 2 G G F1,ext G GGG F1,int ma ==+FFF F F+ F G 11 1,nettt 1,ext 1int1,int F G GGG 2,int G m2 ma2 2==+ F 2,net F 2, ext F 2,int F2,ext m3 GGGGGG ma1 1+=+++ ma 2 2 F 1,ext F 2, ext F 1,int F 2,int G G rd Newton’s 3 law for internal forces: FF1,int+ 2,int = 0 GGGG GG Macom=+ m11 a m 2 a 2 =+ F 1, ext F 2, ext = F net , ext 23 Newton’s secondG law for center of Gmass of a system of objects Fnet, ext= Ma com G : Sum of all external forces that act on the system Fnet, ext (Internal forces are not included) Mmm=++12... : Total mass of the system G acom : Acceleration of the center of mass Internal forces do NOT change the motion of C.O.M.!! Only external forces matter for the motion of C.O.M. !!!!!! 12 Motion of center of mass under gravity force G G Newton’s second law for C.O.M.: FManet, ext= com Under gravity, G Fnet,1 ext =−(0, m g ))(0 +− (0, m 2g ) +... =(0, −−−mg12 mg ...) =−++(0, (mm ...)g ) =− (0,Mg ) 12G ∴(0,−=Mg ) Macom G Æ ∴agcom =−(0, ) Usual projectile motion Center of mass moves like a particle of mass M under the net external force. Motion of COM is simple! 13 iClicker Quiz Two objects with unknown mass and velocity collide and stick together moving at 3 m/s along x direction. Assuming that net external force on the two objects is zero, what is the velocity of the center of mass before the collision? (a) 0 (b) 3 m/s along x (c) -3 m/s along x (d) Not enough information 14.