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Impulse and Momentum. Chapter problems Serway 5,6,10,13,16,17,18,27,29,33,43,44,52,54,59,60 cw.prenhall.com/~bookbind/pubbooks/giancoli. Linear momentum & impulse. Linear momentum is defined as the product of mass and velocity p=mv, p x =mv x , p y = mv y units of momentum are kgm/s
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Impulse and Momentum • Chapter problems Serway • 5,6,10,13,16,17,18,27,29,33,43,44,52,54,59,60 • cw.prenhall.com/~bookbind/pubbooks/giancoli
Linear momentum & impulse • Linear momentum is defined as the product of mass and velocity • p=mv, px=mvx , py= mvy • units of momentum are kgm/s • From Newtons 2nd law • F= ma F=mdv/dt F= dp/dt • The rate of momentum change with respect to time is equal to the resultant force on an object • The product of Force and time is known as IMPULSE • J= Fdt • units of impulse are Ns
Linear momentum & impulse Examples of impulses being applied on everyday objects
Impulse Momentum Theorem Fdt=mdv You apply an impulse on an object and you get an equal change in momentum Area under a Force vs time graph
Linear Momentum and Impulse Example problems 1,2,3 Chapter questions 5,6,10,13,16
Conservation of momentum2 particle system For gravitational or electrostatic force F12 is force of 1 on 2 F21 is force of 2 on 1 m2 m1 F12 F21 F12 =dp1/dt F21 = dp2/dt
Conservation of momentum2 particle system From Newton’s 3rd Law F12 = - F21 or F12 + F21 = 0 F12 is force of 1 on 2 F21 is force of 2 on 1 m2 m1 F12 F21 F12 + F21 =dp1/dt + dp2/dt = 0 d(p1 + p2)/dt= 0 Since this derivative is equal to 0
Conservation of momentum2 particle system Since this derivative is equal to 0 d(p1 + p2)/dt= 0 then integration yields p1 + p2 = a CONSTANT F12 is force of 1 on 2 F21 is force of 2 on 1 m2 m1 F12 F21 Thus the total momentum of the system of 2 particles is a constant.
Conservation of linear momentum Provided the particles are isolated from external forces, the total momentum of the particles will remain constant regards of the interaction between them F12 m1 F21 m2 Simply stated: when two particles collide,their total momentum remains constant. pi = pf p1i + p2i = p1f + p2f (m1v1)i + (m2v2)i = (m1v1)f + (m2v2)f
Conservation of linear momentum Serway problems 9.2 17 & 18
Collisions Event when two particles come together for a short time producing impulsive forces on each other., No external forces acting. Or for the enthusiast: External forces are very small compared to the impulsive forces Types of collisions 1) Elastic- Momentum and Kinetic energy conserved 2) Inelastic- Momentum conserved, some KE lost 3) Perfectly(completely) Inelastic- Objects stick together
Collisions in 1 d Perfectly Elastic 1) Cons. of mom. 2) KE lost in collision 3) KE changes to PE
Collisions - Examples Computer Simulations example 2, problems 5,24,29 Serway Problems 27,29,33,37
Collisions in 2 dimensions After Collision x momentum before collision equals x momentum after the collision mavaf 1 mavafx mavax Before collision mb vel=0 p=0 mbvbxf 2 mbvbf
Collisions in 2 dimensions mavax= mavafx + mbvbxf or mavax= mavaf cos1 + mbvbf cos2
Collisions in 2 dimensions After Collision y momentum before collision equals y momentum after the collision mavaf 1 mavax mavayf Before collision mb vel=0 p=0 Velocity y axis =0 py=o 2 Mbvbyf mbvbf
Collisions in 2 dimensions 0=mavafy - mbvbfy or 0= mavaf sin1-mbvbf sin2
Collisions in 2 dimensions 0= mavaf sin1-mbvbf sin2 mavax= mavaf cos1 + mbvbf cos2 Problems ex 9.9 43,44