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Chapter 6: Many-body Interactions Chapter 6 Goals: To appreciate the ubiquity of action-reaction force pairs • To understand and utilize Newton’s Third (and final) law • To define a system of massive particles • To identify the critical importance of the center-of-mass point in a system • To derive general principles for conservation of energy, momentum, and angular momentum, for a system • To define and utilize the notion of the impulse of an interaction: the change in momentum • To apply these principles to two-body collisions Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Newton’s Third law • ‘to every action, there is an equal and opposite [sic] reaction’ • by ‘action’ we mean a force (and therefore a momentum change) • the actions occur on/to DIFFERENT bodies: the bodies are said to interact • How? strings… contact… springs… The 2 forces on the monitor, and their COLLISIONS Newton’s-third-law counterparts • FBA = ─ FAB Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example The pair of blocks accelerates at 5 m/s2. Find all of the forces on both blocks. F 5 kg N1 F 15 kg N2 R21 R 12 W1 W2 Note that R12 = ─ R21 FBD 1 N1 = m1g =50 N F ─ R21 = m1a = 25 N FBD 2 N2 = m2g = 150 N R12 = m2a = 75 N But R12 = R21 = 75N and so from above, F = 100N Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 5 kg Example 15 kg N1 FK T21 W1 T12 The pair of blocks accelerate. The coefficient of kinetic friction is 0.40. Find a and T. String and pulley are ideal. Note that T12 = ─ T21 FBD 1 N1 = m1g = 50 N ─ FK + T21 = m1a and note FK = mKN1 = 20 N so T21 = 20 N + 5 kg a FBD 2 ─ T12 + m2g = m2a So T12 = 150 N ─ 15 kg a But T12 = T21 so we get 20 N + 5 kg a = 150 N ─ 15 kg a W2 a = 130 N/ 20 kg = 6.5 m/s2 T = 20 N + (5 kg)( 6.5 m/s2) = 52.5 N Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A system of masses interacting with one another Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A system of masses and their felt forces • a system is a collection of N masses {mi} i = 1, 2, 3,…, N • each one feels both internal and external forces N dpi the net force on mass i is Fi (Fi ) ext f j i dt N N j i define system mass M : mi and system momentum P : pi i 1 • Take the time derivative of P, using sum rule dP : dt N dpi dt i 1 N N N Fi ext f ji i 1 i 1 Messy double sum... i 1 j i • first term: total external force on system N Fi ext : FEXT i 1 • second term: messy double sum that contains all pairs of internal forces added up , like f12 + f21 +… Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Newton’s action-reaction pairs for two bodies • the forces may be central (like gravity or electric) • but they don’t have to be (like magnetism) • they may be gigantic and short-lived (like collisions) • for this situation we will introduce the useful notion of the impulse of a force • there are exceptions in advanced electricity due to the finite speed of light Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A system of masses and the forces on it • by N3, f12 = ─ f21 : the entire double sum adds to zero!! • in English: the internal forces in a system add to zero and thereforce cannot affect the momentum of the system!! dP we are left with N2 for a system : FEXT dt • important special case: if FEXT = 0 (either because there are no external forces, of because they happen to add to 0) dP we get that 0 and so system momentum is conserved!! dt • an extremely powerful conservation law that relies on the truth of N3 and permits one to not worry too much about the internal doings of a system or extended body Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The center of mass of a system M = ∑mi • Let the positions of the bodies be given by {ri} dri d N P pi mi v i mi mi ri (constant masses) dt dt i 1 i 1 i 1 i 1 1 N define the center ofmass position R C : mi ri M i 1 N N N N or, equivalent ly, miri : MR C i 1 dR C dR C d N P MVC where VC : mi ri M dt i 1 dt dt • very cool!! The system momentum is the simply the system mass times the velocity of center of mass!! Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Further implications, and where is it? dVC dVC dP N Fi ext FEXT M MA C where A C : dt i 1 dt dt • very cool!! The rate of change of system momentum is the simply the system mass times the acceleration of the center of mass!! • And if FEXT = 0, it’s N1 time!! No matter what the system is doing (spinning, writhing, exploding… the center of mass just cruises along without accelerating • but.. where is the center of mass?? • example: two bodies. Where is c.o.m.? • answer: along the line joining the 1 N mi ri bodies, between them, and closer to the R C : M i 1 more massive body Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Where is it for the two-body problem? 1 R C : M N m r i 1 i i define r '1 and r '2 via ri r 'i R C • the primed positions have the c.o.m. as their origin r '1 r1 R C so Mr '1 Mr1 MR C m1 m2 r1 m1r1 m2r2 Mr '1 m2r1 m2r2 m2 r1 r2 ; easy to get Mr '2 m1 r2 r1 • both primed positions are vectors parallel to the vector that starts on one body and ends on the other • they are both shorter than that vector and proportional in length to the other mass/M • they are in opposite directions • Thus, the c.o.m. lies where we said it did • For N > 2, the c.o.m. is a ‘weighted’ position average and visualizable at least: see figure next slide Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. m2 r1 r2 r '1 M m1 r2 r1 r '2 M r’1 Three bodies much more complicated r’2 r1 C r1–r2 r1 C r2 r3 r2 O O {show Active Figure 09_14} Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Not in book: the idea of the impulse • The impulse J delivered to any body in an interaction is the change in momentum of that body Dp • it is the net force Fnet(t) that one graphs • interaction duration Dt • useful for very rapid highforce situations Fnet, x dpx Dp x : J x dt t f Fnet, x (t ) dt t i Fnet(t) graph • The impulse is the area under Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. More about the impulse • if the graph is a simple shape (triangle, rectangle) one gets the area easily • if a triangle, J = bh/2 = (Dt)(Fnet,max)/2 so we see Fnet,max that Fnet,max = 2J/ Dt • for any shape graph, the average net force is given by < Fnet> J/ Dt from the rectangle’s area Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example Assume the collision lasts for .25 s. The car’s mass is 800 kg and Fnet(t) has a triangular shape. The net force is essentially the collision force since it is dominant. Find the impulse delivered to the car, and <F> and Fmax . J Dp D(mv ) mDv 800 (2.6 (15)) 14,100 kg m/s F > J 14,080 Dt .25 56,400 N (wow!!) Fmax 2 F > 113,000 N (WOW!!!) Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The reduced mass of a two-body system • a way to replace the motions of the pair of bodies, as seen from a fixed origin, with the motion of one of the bodies (here, #1), as seen relative to a moving origin located at the other body (here, #2) • of course, the moving origin is probably accelerating so it is not an inertial frame • the kinematics (motion) is straightforward • the dynamics (forces) may require fictitious forces to explain the kinematics • we start with the origin at the c.o.m. and we define r to be the vector from #2 to #1: r := r1 ─ r2 • thus, r is the position of r1 relative to r2 • recall that the primed positions are parallel to r Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Some remarks about the notation Relative to the fixed origin O • positions relative to O are ri • velocities relative to O are vi := dri/dt • accelerations relative to O are ai := dvi/dt Relative to the c.o.m. origin • positions relative to c.o.m. are r’i := ri – RC • velocities relative to c.o.m. are ui := vi – VC • accelerations relative to c.o.m. are a’i := ai – AC For the two-body problem • position of m1 relative to m2 is r = r1 – r2 [= r’1 – r’2 ] • velocity of m1 relative to m2 is u = u1 – u2 [= v1 – v2] • acceleration of m1 relative to m2 is a = a1 – a2 [=a’1 – a’2] Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The reduced mass II • let the internal net forces be F21 = ─ F12 := F • a1 = F/m1 ; a2 = ─ F/m2 a = a1 ─ a2 = F/m1 + F/m2 • common denominator: a = F(m1+m2)/m1m2 2 m1m2 m1m2 d r so F a M M dt 2 m1m2 and we get F m a where m M • m is called reduced mass; m < m1 and m < m2 • the motion of m1 as seen from an origin on m2 is controlled by F, but it is as if there is only one body moving, and that body has the reduced mass!! Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The reduced mass III • rephrase K1: when 2 bodies orbit, they both move on (different) ellipses, with c.om. at common focus • here, the two stars have comparable masses so the c.om. is near the center, and the reduced mass is about half of either mass taken alone • in general, m is less than and closer to the smaller mass Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The reduced mass IV • obviously, VC = dRC/dt ; again vi = dri /dt • then u = dr/dt = v1 ─ v2 since r = r1 ─ r2 • now consider VC ─ vi , which is the velocity of c.o.m. as seen from body i (we’ll do #1 and #2) m1v1 m2 v 2 m1v1 m2 v 2 m1 m2 v1 VC v1 v1 M M m2 v 2 m2 v1 m2u m2 v1 VC u M M M m1 similarly, by processing VC v 2 v 2 VC u M • you can get the same stuff just by taking d/dt of our previous r1 and r2 equations much more easily • note that we have re-expressed velocities in terms of c.o.m velocity VC and relative velocity u Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. What happens to system kinetic energy? • definition is straightforward; the interpretation less so N N 1 1 K : Ki mi vi2 mi VC u i 2 2 i 1 i 1 i 1 N N mi 2 VC 2VC u i ui2 2 i 1 N MVC2 VC mi u i 2 i 1 N 2 miVC2 2 i 1 [used u i : v i VC ] N N 1 mi VC u i mi ui2 2 i 1 i 1 N 1 mi ui2 2 i 1 • term 2 has velocity of c.o.m. relative to c.o.m. = 0 two terms for K: kinetic energy of the MVC2 N 1 2 K mi ui center of mass (as if system were a 2 2 i 1 point mass M moving at c.o.m. velocity VC) plus kinetic energy relative to the c.o.m. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. What happens to two-body kinetic energy? • look at the second term and re-express in terms of u m 2 M recall r '1 N 1 2 m u 2 i i i 1 r and r '2 m 1 2 m 2 1 M 2 m M1 r u 2 m u1 M2 u and u 2 2 m 1 1 u 2 m 2 2 M m M2 m1m22 m2m12 2 m1m2 2 1 2 u u 2 μu 2 2 M 2M K 1 2 MVC2 1 2 μu 2 • two terms for K: kinetic energy a body of mass M moving at VC, plus kinetic energy of a body of mass m moving at u: the two-body problem is effectively now a one-body problem, if you observe from the c.o.m. • recall system momentum: much simpler P = MVC Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. u What happens to the system potential energy? What about the mechanical energy conservation? • If definable, U is a function of position only •U = U(r1 , r2) could include internal and external forces • If the forces are only internal and central and vary only with separation distance (common!) we say more: • U(r1 , r2) = U(|r1 ─ r2|) = U(|r|) = U(r) so we have MVC2 m u 2 E : K U U (r ) 2 2 • but there is no external force: the first term can’t change MVC2 EC : a constant (N1situati on for M ) 2 E m u2 2 U (r ) a constant (N2 situation for m ) Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. What happens to system angular momentum? • proceed as we have been doing, writing positions ri as c.o.m. position plus positions referred to c.o.m [and for vi] L : N N N i 1 N i 1 i 1 Li miri vi mi R C r'i VC ui N N N i 1 i 1 i 1 mi R C VC mi R C ui mir'i VC mir'i ui i 1 N M R C VC R C mi u i mi r 'i VC i 1 i 1 N N mir'i ui i 1 • term 2 contains velocity of c.o.m. as seen from c.o.m = 0 • term 3contains position of c.o.m. as seen from c.o.m = 0 L : M R C VC N mir'i ui = ang mom of c.o.m + ang i 1 mom relative to c.o.m Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. How does system angular momentum change? • take the time derivative of the system ang mom dL : dt N dL i dt i 1 N d ri pi dt i 1 N dr dv i mi dt v i ri dti i 1 N mi v i v i ri a i ri Fi ext f j i i 1 i 1 j i N N N N M i ext ri f j i : M EXT ri f j i i 1 j i i 1 j i N N • term 2: assume N3 so there are terms like (ri ─ ri) x fji = vector from j to i crossed into fji • further assume central forces, so those two vectors are parallel: entire double sum is zero!! Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. How does system angular momentum change? dL : M EXT N2 for angular momentum for system dt • if there are no external torques (or if they add to zero) then angular momentum of the system is conserved Process L to express in terms of reduced mass • look at the second term and re-express in terms of u N m m m m 2 2 mi r 'i u i m1 r u m2 1 r 1 u M M M M i 1 m2 2 m 2 m m2 m m2 m1 m2 1 r u 1 2 2 2 1 r u M M M mm 1 2 r u μr u M L M R C VC μr u R C P μr u Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Summary of the system quantities • system momentum is P = MVC • it can only change if (F)EXT ≠ 0 N 2 • system kinetic energy K MVC 1 mi ui2 2 2 i 1 N • system angular momentum L : R P m r' u C K i i i i 1 •it can only change if (M)EXT ≠ 0 • two-body kinetic energy 1 2 MVC2 1 2 μu 2 • two-body angular momentum L R C P μr u • if c.o.m. obeys N1 there is an inertial frame, called the center-of-mass frame, from which the physics is particularly simple (since the c.o.m. is stationary!!) Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Important scenario: the two-body collision in 1d • m1 and m2 collide and all motion is in 1d • velocities before collision are u1 and u2 (known) • velocities after collision are v1 and v2 (not known) • momenta before collision are p1 and p2 • momenta after collision are q1 and q2 • during collision, all important forces are internal • therefore system momentum is conserved! u1 ; p1 u2 ; p2 v1 ; q1 v2 ; q2 P Q p1 p2 q1 q2 m1 (v1 u1 ) m2 (u2 v2 ) u 2 v2 m 1 m 2 (v1 u1 ) [statement about velo city changes] Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The elastic collision in 1d • if DE = 0 DK = 0 too (DU = 0 too because there is no height change) elastic collision • conventional approach is to derive expressions for the final velocities v1 and v2 in terms of the masses m1 and m2 and the initial velocities u1 and u2 • there are two conditions to impose: system momentum conservation, and kinetic energy conservation two equations in two unknowns • trouble is, one of them is ‘quadratic’ statement so the arithmetic is very tricky • Watch carefully… we will make use of u2 v2 m 1 m 2 (v1 u1 ) Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The final velocities for the elastic 1d collision 1 2 m u 2 1 1 1 1 1 2 m2u 22 2 m1v12 2 m2 v22 m2 u 22 v22 m1 v12 u12 m2 u 2 v2 u 2 v2 m1 v1 u1 v1 u1 but u 2 v2 m 1 m 2 (v1 u1 ) m1 m2 v1 u1 u 2 v2 m1 v1 u1 v1 u1 m2 u 2 v2 v1 u1 add this to the above m1 2u 2 u1 1 m v1 1 2 1 u1 m2 m1 v1 m2 m1 m2 m1 m2 2m2 2m2u 2 u1 m2 m1 v1M v1 u1 u2 M M 2m m m1 and we also get by solving for v2 v2 1 u1 2 u2 M M m 1 m 2 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The totally inelastic collision in 1d • if maximum energy is lost totally inelastic collision • the bodies stick together, so velocities v1 = v2 := v • there is only one condition to impose: system momentum conservation conservation one equation since u 2 v m1 m2 v m m 1 2 m 1 m 2 m m 1 (v u1 ) v1 m u 2 m1 u1 2 2 m u m u 2 2 11 m 1m2 v m u m u 2 2 11 M • proving that maximum energy is lost if the bodies stick together is not so easy unless one ‘goes to’ the center of mass frame of reference • this is alluded to in the book… we will illustrate! Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The totally inelastic collision in 1d as expressed relative to the c.o.m 1 m1u1 m2u2 and now define primed velocitie s : VC : m1 m2 u '1 : u1 in VC u '2 : u 2 VC ' relative to c.o.m.' velocitie s this ' frame of reference'P' Q' m1u '1 m2u '2 Mv' m1 u1 VC m2 u2 VC Mv' m m1u1 m2u2 m2u2 m2 M1 m1u1 m2u2 Mv' m1u1 Mm1u1 m12u1 m1m2u2 Mm2u2 m1m2u1 m22u 2 M 2 v' m 1 M everythingon left cancels so v' 0 • result is obvious: c.o.m. is motionless in c.o.m frame • there is NO kinetic energy left as seen in this frame Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Important case: initially stationary m2 (u2 = 0) • elastic collision is worked out in example 6.3 (p 135) 2m1 m1 m2 u1 u1 v2 v1 m1 m2 m1 m2 2r r 1 u1 u1 v2 v1 m2 r 1 r 1 some cases for r are interestin g (book has entire table!!0 m now book defines r 1 r 1 (puck collides with bus) v1 u1 v2 0 r 1 (puck collides with puck) v1 0 v2 u1 r >> 1 (bus collides with puck) v1 u1 v2 2u1 • totally inelastic collision is simplicity itself m1 1 v u1 u1 {show active figures 09_05-07} m1 m2 r 1 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example Problem 6.3 Two crates (masses 300 kg and 100 kg), joined by a light rope, are being pulled along a factory floor by a constant force exerted on the heavier crate at an angle of 25° to the horizontal. The coefficient of kinetic friction between the heavier crate and the floor is 0.11 and that between the lighter crate and the floor is 0.18. What should the magnitude of the applied force be to move the crates at constant speed? What applied force would be required to give the system an acceleration of 0.10 m/s2? What would the tension in the rope be in both cases? 100 kg 300 kg Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example 6.6 A glider of mass 180 g travelling on a horizontal airtrack at a speed of 3.0 m/s catches up with another glider of mass 360 g moving in the same direction at 0.9 m/s. If 10% of the energy is lost in the collision what are the speeds of the gliders after the collision? P m1u1 m2u2 (.18)(3.0) (.36)(.90) .540 .324 .864 Q .864 m1v1 m2v2 .180 v1 .360 v2 one equation v2 2.400 .500 v1 solved for v2 K init 12 m1u12 12 m2u22 12 (.18)(3.0) 2 12 (.36)(.90) 2 .810 .146 .956 K fin (.90)(.956) .860 12 m1v12 12 m2v22 second equation Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example cont’d this becomes, when squaring the above expression for v2 .860 .090 v12 (.180 ) 5.76 2.40v1 .250 v12 combining terms .135v12 .432 v1 .177 0 v1 .432 .432 2 4.135 .177 .432 .187 .096 2.135 2.135 .432 .302 2.72 m/s not sure which to choose... .270 .481 m/s Plug both of these into the v2 equation: 2.40 .50(2.72) 1.04 m/s makes sense v2 2.40 .50(.481) 2.16 m/s makes sense Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example cont’d P .864 Q (.180 ) (2.72) (.360)(2.16) 1.27 Q (.180 ) (.481) (.360)(2.16) .864 yes!! The other answer would have come from the 360 g mass moving in the opposite direction initially. This would have slowed down the 180 g mass a lot more, but would have corresponded to the same kinetic energy intially. So the quadratic nature of the kinetic energy introduces subtleties! Example 5.17 A 200 g block is constrained to move on a smooth horizontal table by a string that passes downward through a hole in the centre of the table. Its speed is 0.5 m/s and radius is 0.4 m. The string is pulled to reduce the radius to 0.3 m. What is new speed? How much work is done by the puller? Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example 5.17 cont’d • find the net torque on the body as seen TOP VIEW from the hole as origin • MT = r x T MT = rT sin 180° = 0 MW = r x W MW = rmg sin 90° = rmg r • direction is anti-parallel to p T • clearly, N = – W so MN = rmg sin 90° too O • but direction is parallel to p • net torque Mnet = 0 L is conserved • L = r x p L = mvr sin 90° = mvr; direction is ‘down’ L init L fin Linit L fin mvinit rinit mv fin r fin v fin rinit vinit r .54 3 0.67 m/s fin DK tension force's work 0 0 K fin K init .20 2 .50 2 .019 J . 67 2 Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson DK m 2 v 2 fin 2 vinit p Example 5.17 cont’d • find the net torque on the body as seen from the hole as origin • MT = r x T MT = rT sin 180° = 0 • direction is immaterial! • MW = r x W MW = rmg sin 90° = rmg • direction is anti-parallel to p • clearly, N = – W so MN = rmg sin 90° too • but direction is parallel to p • net torque Mnet = 0 L is conserved • L = r x p L = mvr sin 90° = mvr • direction is into table: ‘down’ L init L fin Linit L fin mvinit rinit mv fin r fin v fin rinit vinit r .54 3 6.7 m/s fin Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example of a two-body system I A mass m1 = 500 g moves at 4.0 m/s to the right, and collides elastically with a mass m2 = 1500 g moving to the left at 2.0 m/s. a) Find system P and K in this frame b) Find the c.o.m velocity and the reduced mass c) Find the quantities in (a) using M and m d) Find the relative-to-c.o.m velocities and the same 4 quantities relative to (“in”) the c.o.m. frame e) confirm that P and K are conserved in the original frame P m1u1 m2u 2 (.50)(4.0) (1.50)(2.0) 1.0 kg - m/s 1 1 2 K 2 m1u1 2 m2u 22 4.0 3.0 7.0 J 1 P 1.0 kg-m/s VC M pi M 2.0 kg 0.50 m/s m mm 1 2 m m 1 2 .75 kg2 2.0 kg .375 kg (less than either mass) Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example of a two-body system II c) Find the quantities in (a) using M and m d) Find the relative-to-c.o.m velocities and the same 4 quantities relative to (“in”) the c.o.m. frame u '1 u1 VC 4.0 (0.50) 4.5 m/s u '2 u2 VC 2.0 (0.50) 1.5 m/s P' m1u '1 m2u '2 (.50)(4.5) (1.50)(1.5) 0.0 kg - m/s 1 1 K ' 2 m1u '12 2 m2u '22 5.06 1.69 6.75 J V 'C 1 M p'i P' M 0.0 m/s P MVC (2.0)(0.50) 1.0 kg - m/s u u1 u2 4.0 (2.0) 6.0 m/s K 1 2 MVC 2 1 2 mu 2 0.25 6.75 7.0 J Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Example of a two-body system III e) confirm that P and K are conserved in the original frame v1 m m 1 2u 1 M 2m 2 M u2 .51.5 2(1.5) (4.0) 2.0 (2.0) 2.0 2.0 3.0 5.0 m/s v2 2m 1u M 1 m m 2 1u 2 M 2(.5) 1.5.5 ( 4 . 0 ) ( 2 . 0 ) 2.0 2.0 2.0 1.0 1.0 m/s Q m1v1 m2v2 (.50)(5.0) (1.50)(1.0) 1.0 kg - m/s 1 1 K 2 m1v12 2 m2v22 6.25 0.75 7.0 J one can confirm that P and K are conserved in the c.o.m. frame….. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.