Physics 2210 Fall smartphysics 10 Center-of-Mass 11 Conservation of Momentum 10/21/2015

Transcription

1 Physics 2210 Fall 2015 smartphysics 10 Center-of-Mass 11 Conservation of Momentum 10/21/2015

2 Collective Motion and Center-of-Mass Take a group of particles, each with mass m i, position r i and velocity v i (both r i and v i are functions of time) for i = 1,2,3,,. The net force on the i tt particle can be written as F i = F jj j i + F i (ext) where F jj is the (internal) force exerted by another (j th ) particle in the group, and F i (ext) is the net external force (vector sum of forces on the i th particle, not exerted by another particle in the group). We now sum over the group: Remember we generally are not allowed to do this but here we are treating the group as a single object!!! (Definition) The net force on the group is given by F = F i = F jj j i + F i (ext) ow: By ewton s 3 rd Law (units are all about the consequences of 3L) Each F jj in the sum j i F jj is cancelled by an equal and opposite F ii

3 i.e. Collective Motion (continued) 3L F jj for any group of partcles. So the net force on the group is always equal to just the sum over the external forces on the individual particles in the group. F = F i j i 0 = F i (ext) ow By ewton s 2 nd Law, the net force on the i tt particle is related to its acceleration by a i dv i dd d2 r i dt 2 = 1 m i F i ote THIS DOES OT MEA a i = 1 m i F i (ext) Because the (ext) desgnation here means outside of the group, but for an individual particle you have to count the external (to the particle) force exerted by other particles in the group!!!

4 ewton s 2 nd Law for the Collective Multiplying a i by m i and summing over the group, interchanging the order of summation and differentiation (derivatives are linear ) ote that And so: d 2 r i m i dt 2 d 2 = d2 dt 2 m i r i m i r i = M = F i 1 M m ir i = F i (ext) = MR CC dt 2 MR CC = M d2 R CC dt 2 M dv CC MA dd CC = F Which looks just like ewton s 2 nd Law for a particle of mass M = located at the center of mass of the group. = F m i, We have implicitly defined the velocity and acceleration of the center-of-mass by V CC dr CC dd, A CC dv CC dd d2 R CC dt 2

5 p mv [ Units: kg m/s ] Unit 11

6 p mv [ Units: kg m/s ] Unit 11

7 Momentum ewton did not actually formulate his laws in terms of acceleration. Instead he used a quantity called momentum Definition of momentum: velocity multiplied by mass p mv [ Units: kg m/s ] ewton s Second Law: (time) rate of change of the momentum vector is equal to the net force (vector sum of all external forces) on a body. dp dd = F These are vector relations: Definition of momentum: p x mv x m dd dd, ewton s Second Law: dp x dd = F x, p y mv y m dy dd, dp y dd = F y, dp z dd = F z p z mv z m dz dd

8 Total Momentum of a System of Particles Take a group of particles, each with mass m i, position r i and velocity v i (both r i and v i are functions of time) for i = 1,2,3,,. Definition of Total Momentum: vector sum of individual momenta P p i m i dr i v i m i dd = MV CC dx i dy i P x p ix m i v ii m i, P dd y p iy m i v iy m i, dd Differentiating the total momentum w.r.t. time: d dd P = d dd p i = F i = F i (ext) = F Which is the alternate form of ewton s Second Law for the group. This leads to the Law of Conservation of Momentum: If the net (external) force on the group of particles is zero, the total momentum is conserved (this is true for each component independetly) F x = 0 d dd P x d dd p ii = 0, F y = 0 d dd P y d dd p iy = 0,

9 Poll Suppose you are on a cart, initially at rest, which rides on a frictionless horizontal track. If you throw a ball off the cart towards the left, will the cart be put into motion (neglect friction between cart and ground)? A. Yes, and it moves to the right. B. Yes, and it moves to the left. C. o, it remains in place

10 Poll Suppose you are on a cart, initially at rest, which rides on a frictionless horizontal track. You throw a ball at a vertical surface that is firmly attached to the cart. If the ball bounces straight back as shown in the picture, will the cart be put into motion after the ball bounces back from the surface? A. Yes, and it moves to the right. B. Yes, and it moves to the left. C. o, it remains in place

11 Poll Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface. The ball hitting box 1 bounces back, while the ball hitting box 2 gets stuck. Which box ends up moving faster? A. Box 1 B. Box 2 C. Same

12 Example 11-1 (1/3) A 12 kg block is at rest on a level floor. A 419 g glob of putty is thrown at the block such that it travels horizontally, hits the block, and sticks to it. The block and putty slide 15 cm along the floor. If the coefficient of sliding friction is 0.40, what is the initial speed of the putty? (%i1) /* Completely inelastic collision initially Let putty be m1=0.419kg, block m2=12 kg Total momentum: */ P: m1*v1 + m2*v2; (%o1) m2 v2 + m1 v1 (%i2) Pi: P, v1=v0, v2=0; (%o2) m1 v0 (%i3) Pf: P, v1=vf, v2=vf; (%o3) m2 vf + m1 vf (%i4) soln1: solve(pi=pf, vf); m1 v0 (%o4) [vf = ] m2 + m1 (%i5) vf: rhs(soln1[1]); m1 v0 (%o5) m2 + m1... continued

13 Example 11-1 (2/3) A 12 kg block is at rest on a level floor. A 419 g glob of putty is thrown at the block such that it travels horizontally, hits the block, and sticks to it. The block and putty slide 15 cm along the floor. If the coefficient of sliding friction is 0.40, what is the initial speed of the putty? (%i1) (%i6) /* second part: friction force does work Wf on block+putty */ M = m1 + m2; (%o6) M = m2 + m1 (%i7) KEi: 0.5*M*vf^2; m1 v0 M (%o7) (m2 + m1) (%i8) KEf: 0; (%o8) 0 (%i9) /* normal force is equal to weight in this problem */ : M*g; (%o9) g M (%i10) /* friction force is in the -x direction */ Ff: -mu_k*; (%o10) - g mu_k M... continued

14 Example 11-1 (3/3) A 12 kg block is at rest on a level floor. A 419 g glob of putty is thrown at the block such that it travels horizontally, hits the block, and sticks to it. The block and putty slide 15 cm along the floor. If the coefficient of sliding friction is 0.40, what is the initial speed of the putty? (%i11) /* work done by friction force */ Wf: Ff*Dx; (%o11) - Dx g mu_k M (%i12) /* KEi + Wf = KEf by work-energy theorem: solve for v0 */ soln2: solve(kei+wf=kef, v0); (sqrt(2) m2 + sqrt(2) m1) sqrt(dx g mu_k) (%o12) [v0 = , m1 (sqrt(2) m2 + sqrt(2) m1) sqrt(dx g mu_k) v0 = ] m1 (%i13) /* take positive root */ v0: rhs(soln2[2]); (sqrt(2) m2 + sqrt(2) m1) sqrt(dx g mu_k) (%o13) m1 (%i14) v0, m1=0.419, m2=12.0, Dx=0.15, g=9.81, mu_k=0.40, numer; (%o14) Answer: v0 = 32.2 m/s

15 Conservation of Momentum in a Collision Collision Experiment: A cart of mass m 1 is traveling at speed v i in the +x direction towards a second cart of mass m 2, which is at rest. They collide and stick together. What is their (common) speed v f after the collision? Case 1: m 2 = m 1 = 1.0 SC This is called a totally inelastic collision From

16 Conservation of Momentum in a Collision Theory: Total momentum is conserved in the x direction because no external forces with non-zero x-components act on the group (they interact but the internal forces must cancel because of 3L) P ii = m 1 v i P fx = m 1 + m 2 v f Setting P ii = P ff we get m 1 v i = m 1 + m 2 v f So the final speed of the conjoined carts is given by m 1 v f = v m 1 + m i 2 Or: the ratio v f /v i is given by v f m 1 = v i m 1 + m 2 Predictions for three cases (1) m 1 = 1.0 SC, m 2 = 1.0 SC: v f /v i = 1/2 = (2) m 1 = 2.0 SC, m 2 = 1.0 SC: v f /v i = 2/3 = (3) m 1 = 1.0 SC, m 2 = 2.0 SC: v f /v i = 1/3 = 0.333

17 Totally Inelastic Collision: m2 = m1 Case 1: m1 = m2 This window dump missed the first digitized point

18 Totally Inelastic Collision : m2 = m1 t (s) x (pix) We have m 1 = 1.0 SC m 2 = 1.0 SC Before the collision: p 1i = (1.0 SCU)*(342 pix/s) ı = 342 SC pix/s ı p 2i = (1.0 SCU)*(0.0 pix/s) SC pix/s ı = 0 ı P i = p 1i + p 2i = 342 SC pix/s ı After the collision: P f = ( ) SCU * (168 pix/s) ı = 336 SC pix/s ı Within a few % of P i!!! Predicted v f /v i = 1/2 = Measured: v f /v i = 168/342 = Within a few % of pppppppppp!!!

19 Unit 12

20 Unit 12

21 Conservation of Momentum in Elastic(?) Collisions Collision Experiment: A cart of mass m 1 is traveling with velocity +v i (in the +x direction) towards a second cart of mass m 2, which is at rest. They collide elastically. What are their velocities after the collision? m 1 = 2.0 SC m 2 = 1.0 SC From

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23 (Almost) Elastic Collision Case 1: m2 = 0.5 m1 t (s) x1 (pix) x2 (pix) xcm (pix) Digitized x 1 and x 2 every 3 frames (0.10 s) Since m 2 = 2 (SC) and 0.5 m 1 = 1.0 (SC) then X CC = m 1x 1 + m 2 x 2 = 2x 1 + x 2 m 1 + m 2 3 Measurement of V1, V2: (from slope near time of collision) V1i = Vi = 340 pix/s V1f=133 pix/s V2f = 421 pix/s

24 0.00s 0.10s 0.20s 0.30s 0.40s 0.50s 0.60s 0.70s 0.80s The collision as seen in the center of mass frame 0.90s 1.00s 1.10s 1.20s 1.30s 1.40s

25 Comparison of Prediction and Measurement From (1) Conservation of Momentum and (2) Conservation of Energy v 1f = m 1 m 2 v m 1 + m i, v 2f = 2m 1 v 2 m 1 + m i 2 The experimental Results gave us V1i = Vi = 340 pix/s V1f=133 pix/s V2f = 421 pix/s 1. V1f/Vi: predicted: v 1f v i = m 1 m 2 m 1 + m 2 = = Measured: V1f/Vi = 133/340 = V1f/Vi: predicted: v 2f v i = 2m 1 m 1 + m 2 = = Measured: V2f/Vi = 421/340 = Conclusion: the Collision was OT completely elastic