PHYS 1441 Section 501 Lecture 11 Wednesday July 7 2004 Dr Jaehoon Yu Collisions Center of Mass CM of a group of particles Fundamentals on Rotation Rotational Kinematics Relationships between linear and angular quantities Today s homework is HW 5 due 6pm next Wednesday Remember the second term exam Monday July 19 Wednesday July 7 2004 PHYS 1441 501 Summer 2004 Dr Jaehoon Yu 1 Announcements Quiz results Class average 57 2 Want to know how you did compared to Quiz 1 Average Quiz 1 36 2 Top score 90 I am impressed of your marked improvement Keep this trend up you will all get 100 soon Wednesday July 7 2004 PHYS 1441 501 Summer 2004 Dr Jaehoon Yu 2 Elastic and Inelastic Collisions Momentum is conserved in any collisions as long as external forces negligible Collisions are classified as elastic or inelastic by the conservation of kinetic energy before and after the collisions Elastic Collision A collision in which the total kinetic energy and momentum are the same before and after the collision Inelastic Collision A collision in which the total kinetic energy is not the same before and after the collision but momentum is Two types of inelastic collisions Perfectly inelastic and inelastic Perfectly Inelastic Two objects stick together after the collision moving at a certain velocity together Inelastic Colliding objects do not stick together after the collision but some kinetic energy is lost Note Momentum is constant in all collisions but kinetic energy is only in elastic collisions Wednesday July 7 2004 PHYS 1441 501 Summer 2004 Dr Jaehoon Yu 3 Example for Collisions A car of mass 1800kg stopped at a traffic light is rear ended by a 900kg car and the two become entangled If the lighter car was moving at 20 0m s before the collision what is the velocity of the entangled cars after the collision The momenta before and after the collision are Before collision p i m1 v1i m2 v 2 i 0 m2 v 2 i m2 20 0m s m1 p f m1 v1 f m2 v 2 f m1 m2 v f After collision Since momentum of the system must be conserved m2 vf m1 v What can we learn from these equations on the direction and magnitude of the velocity before and after the collision Wednesday July 7 2004 m1 m2 v f pi p f f m2 v 2i m 2 v 2i 900 20 0i 6 67 i m s m1 m 2 900 1800 The cars are moving in the same direction as the lighter car s original direction to conserve momentum The magnitude is inversely proportional to its own mass PHYS 1441 501 Summer 2004 Dr Jaehoon Yu 4 Two dimensional Collisions In two dimension one can use components of momentum to apply momentum conservation to solve physical problems m1 m1 v 1i m 2 v 2 i m1 v 1 f m 2 v 2 f v1i m2 v 1f m1 m2 v2f x comp m 1 v1ix m 2 v 2 ix m 1 v1 fx m 2 v 2 fx y comp m 1 v1iy m 2 v 2 iy m 1 v1 fy m 2 v 2 fy Consider a system of two particle collisions and scatters in two dimension as shown in the picture This is the case at fixed target accelerator experiments The momentum conservation tells us m1 v 1i m2 v 2 i m1 v1i m1v1ix m1v1 fx m2v2 fx m1v1 f cos m2v2 f cos m 1 v1 iy 0 m1v1 fy m2v2 fy m1v1 f sin m2v2 f sin And for the elastic conservation the kinetic energy is conserved Wednesday July 7 2004 1 1 1 m1v 12i m1v12f m2 v22 f 2 2 2 PHYS 1441 501 Summer 2004 Dr Jaehoon Yu What do you think we can learn from these relationships 5 Example of Two Dimensional Collisions Proton 1 with a speed 3 50x105 m s collides elastically with proton 2 initially at rest After the collision proton 1 moves at an angle of 37o to the horizontal axis and proton 2 deflects at an angle to the same axis Find the final speeds of the two protons and the scattering angle of proton 2 m1 v1i Since both the particles are protons m1 m2 mp m2 x comp m p v1i m p v1 f cos m p v 2 f cos v 1f m1 y comp m2 v1 f cos 37 v2 f cos 3 50 105 1 v2f 3 50 10 v v 2 1f m p v1 f sin m p v 2 f sin 0 Canceling mp and put in all known quantities one obtains From kinetic energy conservation 5 2 Using momentum conservation one obtains 2 2f Wednesday July 7 2004 v1 f sin 37 v2 f sin Solving Eqs 1 3 3 equations one gets v1 v2 f f 2 2 80 10 5 m s 2 11 10 5 m s 53 0 PHYS 1441 501 Summer 2004 Dr Jaehoon Yu Do this at home 6 Center of Mass We ve been solving physical problems treating objects as sizeless points with masses but in realistic situation objects have shapes with masses distributed throughout the body Center of mass of a system is the average position of the system s mass and represents the motion of the system as if all the mass is on the point What does above statement tell you concerning forces being exerted on the system m2 m1 x1 x2 xCM Wednesday July 7 2004 The total external force exerted on the system of total mass M causes the center of mass to move at an acceleration given by a F M as if all the mass of the system is concentrated on the center of mass Consider a massless rod with two balls attached at either end The position of the center of mass of this system is the mass averaged position of the system m x m 2 x2 CM is closer to the xCM 1 1 m1 m 2 heavier object PHYS 1441 501 Summer 2004 Dr Jaehoon Yu 7 Center of Mass of a Rigid Object The formula for CM can be expanded to Rigid Object or a system of many particles m x m x mn xn xCM 1 1 2 2 m1 m2 mn m x m i i i i y CM i mi ri rCM i z CM i i i mx i i i i mi yi j mi zi k i miri m i i i i M A rigid body an object with shape and size with mass spread throughout the body ordinary objects can be considered as a group of particles with mass mi densely spread throughout the given shape of the object Wednesday July 7 2004 i i i r CM x CM i y CM j z CM k r CM m z m i i i The position vector of the center of mass of a many particle system is m y m x CM m xi i M x CM lim m 0 PHYS 1441 …
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