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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.

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Presentation on theme: "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."— Presentation transcript:

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2 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

3 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 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

4 Linear momentum & impulse Examples of impulses being applied on everyday objects

5 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

6 Impulse Graph

7 Linear Momentum and Impulse Example problems 1,2,3 Chapter questions 5,6,10,13,16

8 Conservation of momentum 2 particle system For gravitational or electrostatic force m1 m2 F 12 F 21 F 12 is force of 1 on 2 F 21 is force of 2 on 1 F 12 =dp 1 /dt F 21 = dp 2 /dt

9 Conservation of momentum 2 particle system From Newton’s 3rd Law F 12 = - F 21 or F 12 + F 21 = 0 m1 m2 F 12 F 21 F 12 is force of 1 on 2 F 21 is force of 2 on 1 F 12 + F 21 =dp 1 /dt + dp 2 /dt = 0 d(p 1 + p 2 )/dt= 0 Since this derivative is equal to 0

10 Conservation of momentum 2 particle system d(p 1 + p 2 )/dt= 0 then integration yields p 1 + p 2 = a CONSTANT m1 m2 F 12 F 21 F 12 is force of 1 on 2 F 21 is force of 2 on 1 Since this derivative is equal to 0 Thus the total momentum of the system of 2 particles is a constant.

11 Conservation of linear momentum m1 m2 F 12 F 21 Simply stated: when two particles collide,their total momentum remains constant. p i = p f p 1i + p 2i = p 1f + p 2f (m 1 v 1 ) i + (m 2 v 2 ) i = (m 1 v 1 ) f + (m 2 v 2 ) f Provided the particles are isolated from external forces, the total momentum of the particles will remain constant regards of the interaction between them

12 Conservation of linear momentum Serway problems 9.2 17 & 18

13 Collisions

14 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

15 Collisions in 1 d Perfectly Elastic 1) Cons. of mom. 2) KE lost in collision 3) KE changes to PE

16 Elastic Collision Calculation 2 objects

17 Collisions - Examples Computer Simulations example 2, problems 5,24,29 Serway Problems 27,29,33,37

18 Collisions in 2 dimensions m a v ax m b vel=0 p=0 Before collision After Collision m a v af m b v bf m a v afx m b v bxf x momentum before collision equals x momentum after the collision 11 22

19 Collisions in 2 dimensions m a v ax= m a v afx + m b v bxf or m a v ax= m a v af cos  1 + m b v bf cos  2

20 Collisions in 2 dimensions m a v ax m b vel=0 p=0 Before collision After Collision m a v af m b v bf m a v ayf M b v byf y momentum before collision equals y momentum after the collision Velocity y axis =0 p y =o 22 11

21 Collisions in 2 dimensions 0= m a v afy - m b v bfy or 0= m a v af sin  1 -m b v bf sin  2

22 Collisions in 2 dimensions 0= m a v af sin  1 -m b v bf sin  2 m a v ax= m a v af cos  1 + m b v bf cos  2 Problems ex 9.9 43,44


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