(1) Center of mass of a symmetric object always lies on an axis of symmetry. (2) Center of mass of an object does NOT need to be on the object.

Transcription

1

2 x com = 1 M N i=1 m ix i x com = 1 M xdm x com = 1 V xdv y com = 1 M N i=1 m iy i y com = 1 M ydm z com = 1 M N i=1 m iz i z com = 1 M zdm M = N i=1 m i ρ = dm dv = M V Here mass density replaces mass (1) Center of mass of a symmetric object always lies on an axis of symmetry. (2) Center of mass of an object does NOT need to be on the object.

3 F 1 m 2 F 2 x O z F 3 m 3 y Ma com = F net F F F net, x net, y net, z = = Ma Ma = Ma com, x com, y com, z

4 F 1 m 2 F 2 x O z F 3 m 3 y Ma com = F net F F F net, x net, y net, z = = = Ma Ma Ma com, x com, y com, z p = mv - Linear Momentum F net = dp dt F net = d P dt = d p 1 dt d p n dt Δ P = 0 - Conservation of Linear Momentum

5 P R,after Collision What changes momentum of each? P R,before Definition p f d p R = F L R (t)dt dp R = p i p f J t f t i Impulse Change in Momentum is equal to Impulse acting on it t f t i F L R (t)dt F net (t)dt p i = Δp = J Impulse Vector! Must satisfy for each direction! Impulse-momentum theorem

6 Conservation of Linear Momentum If F net = d P dt = d p 1 dt d p n dt Δ P = P f Pi = P1f P nf = 0 - Conservation of Linear Momentum P1i+...+ P ni = 0 If system is closed and isolated, the total linear momentum P cannot change. What about energy?

7 Collisions: Elastic vs. Inelastic Elastic collision : TOTAL KE is conserved (~ Conservative forces ) AND if system is closed and isolated, the total linear momentum P cannot change (whether the collision is elastic or inelastic!). ΔKE = K f K i = ( K 1f K nf ) ( K 1i K ni ) = 0 Δ P = P f Pi = P1f P nf P1i+...+ P ni = 0 Inelastic collision : KE is not conserved (~ thermal energy ) However, if system is closed and isolated, the total linear momentum P cannot change (whether the collision is elastic or inelastic!). ΔKE = K f K i = ( K 1f K nf ) ( K 1i K ni ) 0 Δ P = P f Pi = P1f P nf P1i+...+ P ni = 0

8 Concept Questions A 9-kg object is at rest. Suddenly, it explodes and breaks into two pieces. The mass of one piece is 6 kg and the other is a 3-kg piece. Which one of the following statements concerning these two pieces is correct? a) The speed of the 6-kg piece will be one eighth that of the 3-kg piece. b) The speed of the 3-kg piece will be one fourth that of the 6-kg piece. c) The speed of the 6-kg piece will be one forth that of the 3-kg piece. d) The speed of the 3-kg piece will be one half that of the 6-kg piece. e) The speed of the 6-kg piece will be one half that of the 3-kg piece.

9 Concept Questions A sled of mass m is coasting at a constant velocity on the ice covered surface of a lake. Three birds, with a combined mass 0.5m, gently land at the same time on the sled. The sled and birds continue sliding along the original direction of motion. How does the kinetic energy of the sled and birds compare with the initial kinetic energy of the sled before the birds landed? a) The final kinetic energy is one half of the initial kinetic energy. b) The final kinetic energy is two third of the initial kinetic energy. c) The final kinetic energy is one quarter of the initial kinetic energy. d) The final kinetic energy is one ninth of the initial kinetic energy. e) The final kinetic energy is equal to the initial kinetic energy.

10 Velocity of COM In a closed, isolated system the COM velocity ( ) of the system is CONSTANT. Why? v com F net = d P dt tot = 0 v com = P tot = Mv com = ( ) v com P tot = p 1i + p 2i p 1i + p 2i ( ) = p 1 f + p 2 f ( ) P tot ( ) = Constant!! Exploding things from Chapt. 9:

11 9.9 1D Inelastic Collisions Inelastic collision : KE is not conserved (~ thermal energy ) However, if system is closed and isolated, the total linear momentum P cannot change (whether the collision is elastic or inelastic!). P before = P after 1- D ( v 2i ) = ( v 1 f v ) 2 f Only COLM: Conservation of Linear Momentum Special Case: Completely Inelastic Collision (hit- n-stick) ( 0) = ( V V) V = target at rest ( )

12 Pure Inelastic Collision -- Hit- n-stick P before = P after 1- D ( ) ( v 2i ) = v 1 f v 2 f ( + 0) = V f ( ) v com = V = 1(1) ( + 3 ) V = v com = 1 4 m /s

13 Inelastic Collisions?

14 Inelastic Collision Ballistic pendulum: A bullet of mass m and initial velocity v 0 collides and sticks to a pendulum of mass M supported on a rope of length L. How high does the pendulum go before coming to rest? Conservation of momentum p i = mv 0 = p f = (m + M )v Kinetic Energy to Potential Energy (M + m)v 2 2 = ( M + m)gh v = mv 0 m + M mv 0 m + M 2g 2 = h

15 Inelastic Collision P. # 54: Ch. 9: A bullet of mass m is moving directly upward a tt a velocity v 0. It strikes a block of mass M initially at rest. It passes through the block but comes out with a velocity v 0 /4. How high does the block rise above its initial position?

16 D Elastic Collision: 2 particles 1 2 m 2 1 1v 1f + 2 m 2 1 2v 2 f = 2 m m 2 2v 2i v 1f v 2 f = v 2i v 1f = m 2 + 2m 2 v 2i v 2 f = 2 v 2i 6 variables and 2 Equations 2 mass & 4 velocity

17 1D Elastic Collision: 2 particles with v 2i =0 v 1f = m 2 v 2 f = 2 5 variables and 2 Equations 2 mass & 3 velocity Special Cases: 1) Equal masses If = m 2 then v f 1 = 0 and v f 2 = v o1 v 1f = 0 v 2 f =

18 1D Elastic Collision: 2 particles with v 2i =0 v 1f = m 2 v 2 f = 2 5 variables and 2 Equations 2 mass & 3 velocity Special Cases: 2) Massive target If << m 2 then f 1 v o1 and f 2 v 1f v 2 f 2 m 2 m 2 v o1

19 1D Elastic Collision: 2 particles with v 2i =0 v 1f = m 2 v 2 f = 2 5 variables and 2 Equations 2 mass & 3 velocity Special Cases: 3) Massive Projectile If >> m 2 then v 1f f 1 v o1 1i and v f2 f 2v 2 v o1 1i

20 Proble0-4 A ball of mass m is fastened to a cord of length L. The ball is released when cord is horizontal. At bottom of path, the ball elastically strikes block of mass M initially at rest on frictionless floor. a) What is the speed of the ball right after the collision? b) What is the speed of the block right after the collision? c) Use Conservation of Mechanical Energy to find the velocity of the ball right before collision: E mech = 0 ( KE i + U i ) = KE f + U f ( 0 + mgl) = 1 mv 2 bottom ( ) ( 2 + 0) v bottom = 2gL = v mi Because collision is elastic, use COLM+COKE to find the velocity of the ball right after collision: Eqn# 9-67 v mf = m M m + M v mi if M>m, negative? d) What is the speed of the block right after the collision? Because collision is elastic, use COLM+COKE to find the velocity of the block right after collision: Eqn# 9-68 v Mf = 2m m + M v mi

21 1-D Collisions If system is closed and isolated, the total linear momentum P cannot change Inelastic collision : KE is not conserved (~ thermal energy ) P before = P after 1- D 2 particles ( ) ( v 2i ) = v 1 f v 2 f Elastic collision : TOTAL KE is conserved (~ Conservative forces ) v 1 f = m 2 + 2m 2 v 2i (Eqn. 9-75) P before = P after 1- D KE before = KE after 2 particles v 2 f = 2 v 2i (Eqn. 9-76) 1 m 2 1v 2 1i + 1 m 2 ( 2 2v 2i ) = 1 m v m v 2 ( f ) f

22 Momentum and Kinetic Energy in Collisions

23 Examples of Elastic Collisions

24 1D Elastic Collision with m 2 =3 and v 2i =0 v 1 f = m 2 = = 1 2 v 2 f = 2 = = 1 2 1/2

25 Elastic Collision: Velocity of COM In a closed, isolated system the COM velocity ( ) of the system is CONSTANT. Why? v com F net = d P dt tot = 0 v com = P tot = Mv com = ( ) v com P tot = p 1i + p 2i p 1i + p 2i ( ) = p 1 f + p 2 f ( ) P tot ( ) = Constant!! Exploding things from Chapt. 9:

26 Summary: 1D Elastic Collisions Elastic collision : TOTAL KE is conserved (~ Conservative forces ) AND if system is closed and isolated, the total linear momentum P cannot change (whether the collision is elastic or inelastic!). For example: v 2i = 0 KE before = KE after ( 1 2 ) = 1 m v f 2 v 1 i ( m v 2 ) f COKE P before = P after During Collision: transfer KE and momentum between objects through conservative internal forces 1- D ( ) = ( v 1 f v ) 2 f COLM

27 Summary: Velocity & Position of COM (closed & isolated system) Special Case: Completely Inelastic Collision (hit- n-stick) ( ) = ( V V) v 2i V = target at rest ( ) = v com x-t plot position x v 2i slope in x-t plot gives velocity t 0 time t

28 Question Two bodies that form a closed, isolated system undergo an elastic collision in 1-D. Which of the three choices best represents the position-versus-time (x-t plot) of those bodies and their center of mass velocity (v com )?