* Über 80% neue Produkte zum Festpreis; Das ist das neue eBay*. Finde Center Mass! Riesenauswahl an Markenqualität. Folge Deiner Leidenschaft bei eBay A special case of the center-of-momentum frame is the center-of-mass frame: an inertial frame in which the center of mass (which is a physical point) remains at the origin. In all COM frames, the center of mass is at rest, but it is not necessarily at the origin of the coordinate system. In special relativity, the COM frame is necessarily unique only when the system is isolated. Properties. Conservation of four-momentum gives pCμ = pAμ + pBμ, while the mass M of the heavier particle is given by −PC ⋅ PC = M2c2. By measuring the energies and three-momenta of the daughter particles, one can reconstruct the invariant mass of the two-particle system, which must be equal to M For the center of momentum frame it does not matter where we place the origin as only the motion of the CM-frame matters. Don't confuse this with the center of mass system in which the origin is at. Boosting to the center of momentum system - the general case As mentioned, calculations get particularly easy in the CM-system. We may In the center of momentum system the total spatial momentum is.

- Remember we said that if momentum is conserved, the center of mass velocity of the system is also. As the collision is taking place, it doesn't alter the motion of the center of mass a bit. It just plods along at a constant velocity. If we were coasting along on a bike at this center of mass velocity, watching the collision, what would we see? Well in this reference frame, the center of mass.
- Question: In The Center Of Mass Frame, By Definition, The Total 4-momentum Of A System Plotar = (Excm,0,0,0) Where Ecm Is The Center-of-mass Energy. An Easy Way To Figure Out Ecm Is To Use Em = C Plotal 1. A Pion With A Mass Of 2.4 X 10-25 Kg Is Traveling With Speed 0.5c In The +x Direction
- This 4-momentum squared value is an invariant (a constant) in every possible reference frame, even though the total 4-momentum vector varies from frame to frame. In the CoM frame, p t o t a l 2 = 0. So E 2 = m t o t a l 2
- With 4-vectors, finding this frame first is more work than necessary. You have a direct way to get the center-of-mass energy. binbagsss said: So by choosing a frame in which the final particles are at rest, are we able to evaluate the four-momentum in different frames before or after the collision? You can do that. binbagsss said:.

It does suggest however, that a natural frame to analyze reactions is the center of mass (CM) frame. Often we shall analyze a process in this frame, and use a Lorentz transformation to get information about processes in the laboratory frame. Since almost all processes involve the scattering (deflection) of one particle by another (or a number of others), this is natural example for such a. Lecture 6 - 4{momentum transfer and the kinematics of two body scattering E. Daw March 26, 2012 1 Review of Lecture 5 Last time we gured out the physical meaning of the square of the total 4 momentum in two particles about to collide. Where the 4{momentum is writ-ten p T = (p 1 + p 2), the dot product of p T with itself, p T p T is equal to 2M T c2. Here M T is the mass of the heaviest. Center Of Mass Frame 4 Momentum; masuzi. Leave a Comment Cancel reply. Save my name, email, and website in this browser for the next time I comment. Recent Posts. How To Paint Door Frames With Gloss; Broken Picture Frame Glass Meaning; A3 Glass Photo Frame; A3 Glass Picture Frame; A3 Glass Clip Frame; Recent Comments. Archives. September 2020 ; August 2020; July 2020; June 2020; May 2020. Chapter 9 - Center of mass and linear momentum I. The center of mass - System of particles / - Solid body II. Newton's Second law for a system of particles III. Linear Momentum - System of particles / - Conservation IV. Collision and impulse - Single collision / - Series of collisions V. Momentum and kinetic energy in collisions VI. Inelastic collisions in 1D-Completely inelastic collision. In fact, for every other frame of reference that is not the original center of mass / center of momentum frame there will be a net non-zero momentum. That net non-zero momentum can be transformed away by accounting for every chunk and moving the observer to the center of mass / center of momentum frame of reference. Dec 10, 2015 #5 PeroK. Science Advisor. Homework Helper. Insights Author. Gold.

- the so-called reduced mass. Note that the total momentum of the system can be written (445) The fact that the Hamiltonian is In other words, in the center of mass frame, two particles of mass and , moving in the potential , are equivalent to a single particle of mass , moving in the potential , where . This is a familiar result from classical dynamics. Next: Identical Particles Up: Multi.
- 1. 4momentum and the energymomentum invariant in the center‐of ‐mass (CM) frame, defined by the condition p 1CM + p 2CM = 0 (the total 3‐momentum is zero): sc2 = (E 1CM + E 2CM)2. (43) This is the square of the total CM energy of the system. 5.1 The total CM energy and the production of new particles The quantity c√s is the energy available for the production of new particles.
- Clearly the total momentum in the center of mass frame is zero 4 (as it should be), both before and after a collision, and is thus conserved. To find out what happens with the relative velocity in an elastic collision, we invoke conservation of kinetic energy, which we calculate using
- e an isolated system of particles from the frame of reference where the total momentum is zero. Since the system is isolated, the total momentum is constant and in this Center of Momentum frame remains zero as the particles interact with one another. Consider.
- a of negligible thickness, radius R and mass M distributed uniformly over the area. We know that Center of Mass is more equivalent to average, and this shape is symmetrical about the x axis. So by simple logic we can see that the x co-ordinate of Center of Mass is 0
- Momentum of a system from its center of mass frame is zero because, P=mv, where 'v' is the relative velocity of the centre of mass of the system with respect to the reference frame and velocity of center of mass with respect the center of mass is.
- The kinetic energy of the center of mass (i.e. modeling the system as a point particle with all of its mass concentrated at its center of mass) is called translational kinetic energy. K trans = 1 2 Mv2 cm = p2 2M (15) If you imagine that the center of mass is at rest (this is called the center of mass reference frame), the

particle in the center-of-mass frame is then given by The last term is zero due to the fact that the momentum of the system in the center of mass reference frame is zero (Eq. (15.2.11)). Therefore Eq. (15.2.16) becomes = 1 1 1 . v. 2 1 ′ 2 + v. 2 2 + (m. 1 + m. 2. K) v . (15.2.17) S . m. 1. m. 2 2 2 . 2. cm . The first two terms correspond to the kinetic energy in the center of mass. Center Of Mass Frame Momentum. masuzi January 4, 2020 Uncategorized 0. Tering in the center of mass frame tering in the laboratory frame elastic collisions center of mass physics mechanics momentum 5 of 9. Tering In The Center Of Mass Frame Tering In The Laboratory Frame Elastic Collisions Center Of Mass Reference Frame You Physics Mechanics Momentum 5 Of 9 Collision And Center Mass What Is. Other 4-vectors, such as the space-time coordinates of events, of course transform in the same way. The scalar product of two 4-momenta p1 ·p 2= E1E2 −p1 ·p is invariant (frame independent). 43.2. Center-of-mass energy andmomentum In the collision of two particles of masses m1 and m2 the total center-of-mass energ As before, the first particle is of mass , and is located at position vector , whereas the second particle is of mass , and is located at . By definition, there is zero net linear momentum in the center of mass frame at all times. Hence, if the first particle approaches the collision point with momentum then the second must approach with momentum Visit http://ilectureonline.com for more math and science lectures! In this video I will show how the center of mass changes as 2 objects moves towards each.

In this frame the momentum of the system is zero. This frame is also called the center-of-mass frame. The center of mass is a coordinate that is a mass-weighted average of the positions of the objects that make up the system. In a two-object system the center of mass is always somewhere in between the two objects 6.3.2 Velocity in the Centre of **Mass** **frame** 6.3.3 **Momentum** in the CM **frame** 6.3.4 Motion of CM under external forces 6.4 Kinetic energy and the CM 2. 6.1 Lab & CM **frames** **of** reference From hereon we will deal with 2 inertial **frames**: I The Laboratory **frame**: this is the **frame** where measurements are actually made I The centre of **mass** **frame**: this is the **frame** where the centre of **mass** **of** the system is. That is, it is the mass of an object in its rest frame. Sometimes γ We also can define new units for mass and momentum. For example, the mass of the electron can be expressed m. In other words, if you multiply the mass by c e =0 511 /c. MeV 2 2, you get the rest energy in electron-volts. Similarly, we know that pc has units of energy, so momentum can be expressed in units like MeV / c. In. Sample Problems in 104 Problem Set for Momentum: 1, 3. Center of Mass. The center of mass of an extended system is the point whose dynamics typifies the system as a whole when it is treated as a particle. Sample Problem. Find the location of the center of mass of the system of three particles shown in Fig. 3 below. Each particle has the same mass m. The coordinates (x,y,z) of particle 1 are (0. A special case of the center-of-momentum frame is the center-of-mass frame: an inertial frame in which the center of mass (which is a physical point) remains at the origin. In all COM frames, the center of mass is at rest, but it is not necessarily at the origin of the coordinate system. In special relativity, the COM frame is necessarily unique only when the system is isolated. Contents. 1.

- In the center-of-mass frame, the rapidities of a and b are: and Examples: a) fixed target experiment: b) Collider: K. Reygers, K. Schweda | QGP physics SS2013 | 2. Kinematic Variables 9 Pseudorapidity Special case: Analogous to the relations for the rapidity we find . K. Reygers, K. Schweda | QGP physics SS2013 | 2. Kinematic Variables 10 Example: Beam Rapidities Beam momentum (GeV/c) Beam.
- 9.4 Linear momentum DEFINITION: The system's center of mass is shown in each freeze-frame. The velocity v com of the center of mass is unaffected by the collision. Because the bodies stick together after the collision, their common velocity V must be equal to v com . Example: conservation of momentum The collision within the bullet- block system is so brief. Therefore: (1) During the.
- Momentum and Center of Mass • Next midterm on Thursday (3/15). A sample exam is available on D2L under Content. Chapters 6-9 will be covered. The exam that is posted only included Chapters 5-8, so may want to also look at Exam 3 sample test. • Will cover center of mass today Chap 9.1-5. • Clickers need to be registered each semester - if registered last semester, then it doesn.
- This equality will allow us to calculate the kaon's momentum in the center of mass frame. 4. For two particle systems, the two particles will have equal and opposite momenta in the center-of-mass reference frame, p~ 1 = p~ 2. Letting jp~ 1j p, we have s= (q m2 1 c4 + p 2c2 + q m2 2 c4 + p2c2)2 0 (11) The last term is zero, since p~ 1 + p~ 2 = 0. After a bit of algebra, one can solve for pin.

the lab frame, what is the nal momentum of the combined masses? Since no external forces act on the system, the change in momentum is zero. Thus p~ f = p~ i p~ f = m 1v 1^x: b(10) Taking m 1 and m 2 as our system, what is the velocity of the center of mass of the system (~v cm)? The center-of-mass velcoity is given by the expression ~v cm = 1 m 1 + m 2 (m 1~v 1 + m 2~v 2) : Thus, we have ~v cm. The center of mass follows a straight line along the y-axis. In addition, it will acquire the same velocity as in part b). The translation of the center of mass depends only on the sum of the external forces and not on the point of application of the forces. In this question, the force is applied at the left of the center of mass. As a result, the center of mass of the object translates and.

If we differentiate linear momentum of the center-of-mass with respect to time we obtain. This expression shows that if the net external force acting on a system of particles is zero (F ext = 0 N), the linear momentum of the system is conserved. Example Problem 9-4. A stream of bullets with mass m is fired horizontally with speed v into a large wooden block with mass M that is initially at. centre of **mass** **frame** is just the **frame** in which we are travelling along with the centre of **mass**. This means that all we have to do to go from lab **frame** to CM **frame** velocities is subtract the velocity of the centre of **mass**. u1 = v1 - vCM u2 = v2 vCM. We can determine vCM by recalling that in the CM **frame**, the total **momentum** is zero. The total **momentum** may be written either as the **momentum** **of**.

** Experimental Physics - Mechanics - System of Particles - Conservation of Momentum 3 m 3m Center of mass U = åm i gh i = gåm ih i = Mgh cm F gr! U 0 = åm i gh i U q = U 0 + åm i gDh i U q = U 0 - 2mg sinq = -2 cosq= 0 q mg d dU 2m**. Experimental Physics - Mechanics - System of Particles - Conservation of Momentum 4 R x y Continuous bodies dm ò ò ò ò = = L L L L cm r dr r r dr dm rdm R 0. The length of the energy-momentum 4-vector is given by . The length of this 4-vector is the rest energy of the particle. The invariance is associated with the fact that the rest mass is the same in any inertial frame of reference. Lorentz transformation in 4-vector form: Four-vector sum for momentum-energy: Index Reference Rohlf Sec 4-4 . HyperPhysics***** Relativity : R Nave: Go Back: Lorentz.

A 1kg mass moving through a laboratory at 30m/s in the +x-direction collides with a stationary 2kg mass then rebounds at 16m/s in the -y-directions. (a)What is the final velocity of the 2kg mass? (b)Find the velocity of the center of mass, then find the initial and final velocities of the masses both before and after the collision in the center-of-mass frame Center of Mass for Two Particles beyond One Dimension Now that we have the position, we extend the concept of the center of mass to velocity and acceleration, and thus give ourselves the tools to describe the motion of a system of particles. Taking a simple time derivative of our expression for x cm we see that: v cm = Thus we have a very similar expression for the velocity of the center of. The centre of mass of the frame is, therefore, on the rod CD at a distance L/4 from C. 3. Two charged particles of masses m and 2m are placed a distance d apart on a smooth horizontal table. Because of their mutual attraction, they move towards each other and collide. Where will the collision occur with respect to the initial positions? Sol. As the table is smooth, there is no friction. The. The laboratory frame is a frame whereby positions and velocities are measured with respect to the laboratory. The C.M. frame is the a frame whereby the positions and velocities are measured with respect to the C.M. frame. In lab frame, the total m.. In fact, because the center of mass is stationary in the center-of-mass frame, the total momentum before and after the collision is zero in that frame, like this: Therefore. And after the collision, which means that. Also, if the collision is elastic, kinetic energy is conserved in addition to momentum, so that means the following is true: Substituting . into this equation gives you. Given.

Consider the decay of a particle with mass M to two particles of mass m 1 and m 2 in the rest frame of the parent particle. The two final particles are defined by 8 components (2 energies and 6 momenta). However, for daughter particles of well-defined masses, we have two constraints: p 1 2=m 1 2 and p 2 2=m 2 2, leaving us with 6 variables. Application of 4-momentum conservation leaves us with. Homework Statement On a frictionless table, a glob of clay of mass 0.240kg strikes a bar of mass 1.560kg perpendicularly at a point 0.390m from the center of the bar and sticks to it. a) If the bar is 1.260m long and the clay is moving at 9.300m/s before striking the bar, what is the final.. Center of Mass. The center of mass of an extended system is the point whose dynamics typifies the system as a whole when it is treated as a particle. Sample Problem. Find the location of the center of mass of the system of three particles shown in Fig. 1 below. Each particle has the same mass m. The coordinates (x,y,z) of particle 1 are (0,L,0) of particle 2 are (0,0,0) and of particle 3 (L,0. Other 4-vectors, such as the space-time coordinates of events, of course transform in the same way. The scalar product of two 4-momenta p1 ·p 2= E1E2 −p1 ·p is invariant (frame independent). 46.2. Center-of-mass energy andmomentum In the collision of two particles of masses m1 and m2 the total center-of-mass energ

* A particle or system's 4-momentum *. P Ep =(, ) is a 4-vector, and its norm . PP ⋅ is a Lorentz invariant. This norm defines the system's . rest-mass (or, more simply, mass), m: 2. m PP E p ≡⋅= −22 (A-5) Our systems will always have real . m ≥0 (P. is never space-like). If . m >0, then a reference frame can be found in which the. The center of mass is a point of the rigid object therefore, as any other point of the rod, it rotates about the pivot with angular speed ω. The speed of the center of mass can be expressed as: v cm =ωD/2 (eq. 4) 3. The position and velocity vectors are perpendicular: As a result, (eq. 5

That is, when viewed from the center of mass frame the two objects approach each other with equal and opposite momenta and move away from each other with an equal and opposite momenta. Therefore, the center of mass frame simplifies the analysis since it exhibits a particular symmetry to the problem (see Fig. 6.15) Center of Mass (CM) Frame of Reference: A 1D elastic collision is considered as seen from the CM frame of reference (where the total momentum is zero). Using the velocity of the CM in the Lab frame, you can transfer between the two frames. 4. 1D Inelastic Collision and Internal Energy: A 1D inelastic collision is considered from the laboratory and the CM frame. The kinetic energy is calculated. In addition, we calculate the torque about the center of mass due to all the forces acting the angular momentum of the system with respect to the center of mass in the center of mass reference frame and then apply F. spin . on the particles in the center of mass reference frame. We calculate the time derivative of N . d. L. τext r ∑ cm, i. Learn the definition of center of mass and learn how to calculate it. If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked

result, the total momentum of our system will be conserved, and so we should consider the center of mass, R = m 1r 1 + m 2r 2 m 1 + m 2 = m 1r 1 + m 2r 2 M (4) the time derivative of which is given by the center of mass velocity, v CM = m 1v 1 + m 2v 2 M: (5) Now, in the previous lecture, we found that the acceleration of the center of mass depended on the net external force, F ext = Ma CM: (6. Other articles where Centre-of-mass reference frame is discussed: mechanics: Centre of mass: This is sometimes called the centre-of-mass frame. In this frame, the momentum of the two-body system—i.e., the constant in equation (51)—is equal to zero. Writing each of the v's as the corresponding dr/dt, equation (51) may be expressed in the for

Given, T otal momentum=(18 kgm/sec)i, velocity of Center of mass=(3 m/s)i, Mass of one object=4 kg, Velocity of this object=(1.5 m/s)i Let m be the mass of other object a nd v be the velocity. Now we know total momentum =Total massX velocity of center of mass O all collisions - whenever momentum is conserved only completely inelastic collisions O only elastic collisions QUESTION 2 Consider an elastic collision between a 1-kg block and a 4-kg block. The initial velocity of the 1-kg block is +5 m/s; the 4-kg block is initially at rest. What are the initial velocities of the 1-kg and the 4-kg blocks, respectively, in the center-of-mass reference frame. The linear momentum of a particle with mass m moving with velocity v is deﬁned as p = mv (7.1) center of mass moves as if it were a single particle of mass M moving under the inﬂuence of the sum of the external forces. 7.2 Worked Examples 7.2.1 Linear Momentum 1. A 3.00kg particle has a velocity of (3.0i−4.0j) m s. Find its x and y components of momentum and the magnitude of its. Before the collision the beam particle moves with four-momentum. p b = (p l a b, 0, 0, m b 2 c 4 + p l a b 2 c 2) (10.11) and the target particle m t is at rest, p t = (0, 0, 0, m t c 2). (10.12) We ﬁrst need to determine the velocity v of the Lorentz transformation that bring is to the centre-of-mass frame. We use the Lorentz transformation rules for momenta to ﬁnd that in a Lorentz frame. Centre of mass and Linear momentum . Cross Product of two vectors in Rectangular Coordinate System . Problem on Moment of Inertia . Motion of Center of Mass. Let us talk about centre of mass of a system of particles. Whenever we talk about motion of an object, we usually talk about velocity with which the object is moving or the acceleration with which the object is moving. As we know the.

1.5 Laboratory Frame and the Center-of-Mass Frame. So far, we have discussed collisions of a particle with a fixed center. In reality, however, the target also moves (recoils) as a result of the scattering. In some experiments we may be interested in colliding two beams of particles of comparable energy with each other. Although such situations may appear to be extremely complicated at first. Thus center of momentum means center-of-momentum frame and is a short form of this phrase. [1] A special case of the center-of-momentum frame is the center-of-mass frame: an inertial frame in which the center of mass (which is a physical point) remains at the origin. In all COM frames, the center of mass is at rest, but it is not. Center of mass, spin supplementary conditions, and the momentum of spinning particles L. Filipe O. Costa, Jos e Nat arioy yCAMGSD, Instituto Superior T ecnico, Universidade de Lisboa Lisboa, Portugal Centro de F sica do Porto { CFP, Departamento de F sica e Astronomia Often the physics is best visualized in the center of momentum frame. Details of the calculation: In the rest frame of the m Φ, the 4-vector momentum of the Φ is (E/c,p) = (m Φ c, 0). Each component of the 4-vector is conserved. After the decay we therefore have for each particle: E k 2 = ¼m Φ 2 c 4 = m k 2 c 4 + p k 2 c 2, p k 2 = c 2 (¼m Φ 2 - m k 2). E k = 0.51 GeV, p k = 0.127 GeV/c. IIT JEE Mains and Advanced video lectures for Center of Mass (Finding center of mass for different type of bodies). The notes and videos on Center of Mass have been prepared meticulously by highly qualified and experienced teachers in the field of IIT Preparation having vast experience teaching in Kota. Since, IIT is a problem solving based approach exam, the students not going to coaching.

The basic concept is that for any collision, conservation of momentum holds. For elastic collisions, the relative velocity of each mass in relation to the center of mass will be equal in magnitude but opposite in direction for each mass. (relative to the center of mass) 1. What is the velocity of the center of mass of the system? If we assume that Right is the positive direction. v' = 108(4.7. Let us transform to a new inertial frame of reference--which we shall call the laboratory frame--which is moving with the uniform velocity with respect to the center of mass frame. In the new reference frame, the first particle has initial velocity , and final velocity .Furthermore, the second particle is initially at rest, and has the final velocity An elastic collision of two identical particles must conserve momentum and energy in all inertial frames. In the center of mass frame (which we will consider to be ) (15.88) (15.89) relate the intial and final momenta and energy of the two identical particles. Now, (15.90). In an inertial frame that is not accelerating, the torque is always equal to the rate of change of the angular momentum. However, about an axis through the center of mass of an object which is accelerating, it is still true that the torque is equal to the rate of change of the angular momentum. Even if the center of mass is accelerating, we may still choose one special axis, namely, one.

Conservation of relativistic mass and momentum lead us to the same value of u as the Lorentz Transformation and the principle of relativity. And they give us something more: the combined rest mass of the two objects after the collision. When γ v = 1 this mass is 2m, as we would expect from the classical law of conservation of mass. But when γ. Books by Robert G. Brown Physics Textbooks • Introductory Physics I and II A lecture note style textbook series intended to support the teaching of introductory physics, with calculus, at a level suitable for Duke undergraduates 2.6 Center of mass and gravity For every system and at every instant in time, there is a unique location in space that is the average position of the system's mass. This place is called thecenter of mass, commonly designated by cm, c.o.m., COM, G, c.g., or . One of the routine but important tasks of many real engineers is to ﬁnd the center 1 Nowadays this routine work is often of mass of a. This allows us to define the 4-momentum \( \mathbf{P}\) of a particle in analogy to its 3-momentum, \[\tag{38} \mathbf{P = } {m_{0}\mathbf{U} }, \] \(\mathbf{U}\) being its 4-velocity. Like \(\mathbf{U}, \mathbf{P}\) is timelike and future pointing. And we take as the basic axiom of collision mechanics the conservation of this 4-vector quantity: the sum of the 4-momenta of all the particles. Momentum - Conservation of Momentum - Center of Mass download 204.6M Collisions - Elastic and Inelastic - Center of Mass Frame of Reference downloa

m The center of mass frame is the zero momentum frame so m 1 Îr 1 m 2 Îr 2 0 õ from PHYSICS 9A at University of California, Davi Impulse and Conservation of Linear Momentum. It is best to transform the inertial frame of reference of such that it's at rest ( i.e., ) and accordingly adjust the relative velocity of , which is , in order to accommodate this transformation. § Impulse - Momentum Theorem: § Corollaries: Where there is no net external force acting upon the Center of Mass Frame system, , the following laws.

Momentum and Velocity of Center of Mass Were Conserved (pretty much) Video 2 mass (kg) frames. time (s) delta x (m) vf (m/s) pf (Ns) Kf (J) Girls. 148.4. 65. .270833-.2-.73846-109.588. 40.46315. Boy. 91.4. 41. 0.170833. 0.2. 1.170732. 107.0049. 62.637. System -0.01077-2.58281. 103.1001. Momentum and Velocity of Center of Mass Were Conserved. 4.4 4-momentum To discuss momentum we should rst be explicit concerning what we mean by the symbol m. The rest mass mof any object is the mass of the object as measured in its rest frame. The 4-momentum of a particle (or any other object) with rest mass mis de ned to be mtimes the object's 4-velocity, p= mu: (4.4.1 In the center of mass reference, v cm' = 0 (' means in center of mass reference frame). To switch to this frame of reference every velocity is decreased by v cm (for example v i1' = v i1 - v cm). What is the new expression for the total momentum of the system relative to the center of mass reference frame? Center of Mass III CM CM A. m 1.

6.26 | Center of Mass and the Law of Conservation of Momentum separation speed after collision e approach speed before collision = 12 2 2 1 12 12 12 m em m em v' v v mm mm − + = + ++ 21 1 1 2 21 12 12 m em m em v' v v mm mm −+ = + ++ The C-frame: Total kinetic energy of system in K-frame is related to total kinetic energy in C-frame as: 2 c sys sys/c i Mv K K ;M m 2 =+=∑ For a two-par. Chapter 9 - Centre of Mass, Linear Momentum, Collision solutions from HC Verma Solutions for Class 11 Physics Part 1. Concepts of Physics Part 1, Numerical Problems with their solutions, Short Answer Solutions for Chapter 9 - Centre of Mass, Linear Momentum, Collision from the latest edition of HC Verma Book Utilizing center of mass to make perfectly elastic problems easier 1. v 1i -v 2i = v 2f -v 1f 2. Change the frame of reference to the center of mass by subtracting v cm 3. u 1i -u 2i = u 2f -u 1f but in the frame of reference of the center of mass, total momentum of the system is zero, so m 1u 1i + m 2u 2i = 0 and u 2i = -u 1i m 1 / m 2. ** Angular Momentum with Respect to the Center of Mass Choose center of mass as originabout which angular momentum is calculated (= center of rotation) Use cross-product-equivalent matrix to define the inertia matrix**, I h=−!r!rdm Body ∫ω = − 0−zy z0x −yx0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ 0−zy z0−x −yx0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ z min ⎥ z max ∫ y min y max

Velocity and Momentum l The velocity of the centre of mass of a system of particles is l The momentum can be expressed as l The total linear momentum of the system equals the total mass multiplied by the velocity of the centre of mass ! v CM = d! r CM dt = 1 M m i! v i i ∑ M! v CM =m i! v i i ∑=! p i =! p tot i ∑ l Assume the total mass, M, of the system remains constan (For the answer see the impulse and momentum page) Problem # 9 In the angular momentum page we showed how the angular momentum equations for a rigid body are derived. The figure below shows the set up used for the derivation. Where: r iG is the position vector from point G (the center of mass of the rigid body) to the location of m i We su mmarize by stating that, after the collision, the momenta q 1 and q 2 in the LAB frame (wherem 2 is initially at rest) are q 1 = p 1 ¡ 4® (1+ ®)2 cos2 ' 1=2 (cosµ bx +sinµ by) q 2 = 2p cos' 1+® (cos' xb ¡ sin ' yb) where p 1 = pxb is the initial momentum of particle 1. 2Two-ParticleCollisions in the CM Frame