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C HAPTER-9

C HAPTER-9. Center of Mass and Linear Momentum. Ch 9-2 The Center of Mass. Center of mass (com) of a system of particles is required to describe the position and motion of the system

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C HAPTER-9

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  1. CHAPTER-9 Center of Mass and Linear Momentum

  2. Ch 9-2 The Center of Mass • Center of mass (com) of a system of particles is required to describe the position and motion of the system • The com of a system of particles is the point that moves as if the entire mass of the system were concentrated there and all external forces were applied there

  3. Ch 9-2 The Center of Mass com is defined with reference to origin of an axis For Fig. b Xcom=(m1x1+m2x2)/(m1+m2) =(m1x1+m2x2)/M where M= m1+m2 =mi xcom =mixi / M= mixi/mi ycom=miyi /M; zcom=miyi /M rcom= xcom i +ycom j +zcom k = miri / M

  4. Ch 9-2 The Center of Mass(solid bodies) For a solid body having continuous distribution of matter, the particle becomes differential mass element dm xcom=(1/M) x dm ; ycom=(1/M) y dm ; zcom=(1/M) z dm M is mass of the object and its density  are related to its volume through  = M/V =dm/dV Then xcom=(1/V) x dV ; ycom=(1/V) y dV ; zcom=(1/V) z dV

  5. The figure shows a uniform square plate from which four identical squares at the corners will be removed a) where is com of plate originally? b) where it is after removal of square 1 c) after removal of square 1 and 2 d) after removal of square 1 and 3 e) after removal of square 1, 2 and 3 f) All four square Answers in term of Quadrant, axes or points (a) origin; (b) fourth quadrant; (c) on y axis below origin; (d) origin; (e) third quadrant; (f) origin Ch-9 Check Point 1

  6. Ch 9-3 Newton’s Second Law for a System of Particles For a system of particles with com defined by: rcom = miri / M , Newton’s Second law : Fnet=M acom Mrcom = miri ; differentiating w.r.t time Mvcom= mivi ; differentiating w.r.t time Macom= miai = Fi= Fnet Fnet-x=Macom-x; Fnet-y=Macom-y; Fnet-z=Macom-z

  7. Ch 9-4,5 Linear Momentum • Linear momentum of a particle p- a vector quantity p = mv (linear momentum of a particle) • Newton’s second law of motion: Time rate change of the momentum of a particle is equal to the net force acting on the particle and is in the direction of force Fnet = d/dt (p) = d/dt (mv) = m dv/dt = ma • Linear momentum P of a system of particles is vector sum of individuals particle’s linear momenta p P = pi = mivi =Mvcom(System of particles) Fnet = d/dt (P)= Macom(System of particles)

  8. The figure gives the magnitude of the linear momentum versus time t for a particle moving along an axis. A force directed along the axis acts on the particle. Rank the four regions indicated according to the magnitude of the force, greatest first b) In which region particle is slowing F= dp/dt Region 1: largest slope 2: Zero slope 3: Negative slope 4: Zero slopes Ans: 1, 3, then 2 and 4 tie (zero force); (b) 3 Ch-9 Check Point 3

  9. Ch 9-6 Collision and Impulse • Momentum p of any point- like object can be changed by application of an external force • Single collision of a moving particle-like object (projectile) with another body ( target) • Ball (Projectile-R) – bat (target-L) system • Change in momentum of ball in time dt , dp=F(t) dt. • Net change  dp=  F(t) dt. • Impulse J=  F(t) dt. = Favgt • Change in momentum P =Pf-Pi =J

  10. Ch 9-7 Conservation of Linear Momentum • If Fnet-external = 0 then J=  F(t) dt =0 • P =Pf-Pi =J=0; then Pf = Pi • Law of conservation of linear momentum: • If no net external force acts on a system of particles, the total linear momentum of the system of particles cannot change. • Pfx = Pix ; Pfy = Piy

  11. A paratrooper whose chute fails to open lands in snow, he slightly hurt. Had he landed on bare ground, the stoppong time would be 10 times shorter and the collision lethal. Does the presence of snow increases, decreases or leaves unchanged the values of (a) the paratrooper change in momentum (b) the impulse stopping the paratrooper C) the force stopping the paratrooper Answer: (a)p =m(vf-vi) unchanged; (b) J= p ; unchanged; (c) J=F.dt ; t increase, F decrease Ch-9 Check Point 4

  12. The figure shows an overhead view of a ball bouncing from a vertical wall without any change in its speed. Consider the change p in the balls’ linear momentum a) Is px positive, negative, or zero b) Is px positive, negative, or zero b) What is direction of p? px=pxf-pxi = 0 py=pyf-pyi=pyf-(-pyi) =positive Direction of P towards y-axis Ch-9 Check Point 5 • q q x

  13. An initially stationary device lying on a frictionless floor explodes into two pieces, which then slides across the floor. One piece slides in the positive direction of an x axis. a) What is the sum of the momenta of the two pieces after the explosion? b) Can the second piece move at an angle to the x-axis? c) What is the direction of the momentum of the second piece? a) Pi= Pf =0 b) No because Pf = 0=P1fx+P2 Then P2 =-P1fx c) Negative x-axis Ch-9 Check Point 6

  14. Ch 9-8 Momentum and Kinetic Energy in collisions • Elastic collision: Momentum and kinetic energy of the system is conserved • then Pf = Pi and Kf = Ki • Inelastic collision: Momentum of the system isconserved but kinetic energy of the system is not conserved • then Pf = Pi and Kf  Ki • Completely inelastic collision: Momentum of the system is conserved but kinetic energy of the system is not conserved. After the collision the colliding bodies stick together and moves as a one body.

  15. Ch-9-9 Inelastic Collision in one Dimension • One-dimensional inelastic collision • For a two body system Total momentum Pi before collision = Total momentum Pf after collision p1i+p2i=p1f+p2f m1v1i+m2v2i=m1v1f+m2v2f • One-dimensional completelyinelastic collision (v2i=0) m1v1i=(m1+m2)V m1v1i/(m1+m2) Hence Vv1i [m1 (m1+m2)]

  16. Ch-9-9: Velocity of the Center of Mass • For One-dimensional completely inelastic collision of a two body system • P=(m1+m2)vcom • Pi=Pf and m1v1i=(m1+m2)V Pi=(m1+m2)vcom= m1v1i and Pf=(m1+m2)vcom =(m1+m2)V • vcom = m1v1i /(m1+m2)= V • Vcom has a constant speed

  17. Sample-Problem-9-8Ballistic Pedulum • One dimensional completely inelastic collision Conservation of linear momentum mv =(m+M)V; V=mv/(m+M) • Conservation of mechanical energy (K+PE)initial= (K+PE)final (m+M)V2/2= (m+M)gh V= (2gh); V=mv/(m+M) v= [(m+M)(2gh)]/m

  18. Ch-9-10: Elastic Collision in One Dimensions • Elastic collision: Momentum and kinetic energy of the system is conserved then Pf = Pi and Kf = Ki Two classes: • Stationary target (v2i=0) m1v1i=m1v1f+m2v2f m1v1i2/2=m1v1f2/2+m2v2f2/2 • v1f=v1i(m1-m2)/(m1+m2) • v2f=2v1im1/(m1+m2)

  19. Ch-9-10: Elastic Collision in One Dimensions • Stationary target (v2i=0) • v1f=v1i(m1-m2)/(m1+m2) • v2f=2v1im1/(m1+m2) • Three cases: • Equal masses: m1=m2 • v1f=0 and v2f=v1i • Massive target : m2>>m1 • v1f=-v1i and v2f  (2m1/m2)v1i • Massive projectile : m1>>m2 • v1f  v1i and v2f 2v1i

  20. Ch-9-10: Elastic Collision in One Dimensions Moving target m1v1i+ m2v2i = m1v1f+m2v2f m1v1i2/2+ m2v2i2/2 = m1v1f2/2+m2v2f2/2 v1f=[v1i (m1-m2)/(m1+m2)] +2m2v2i/(m1+m2) v2f=2v1i m1/(m1+m2) + [v2i (m2-m1) / (m1+m2)]

  21. What is the final linear momentum of the target if the initial linear momentum of the projectile is 6 kg.m/s and final linear momentum of the projectile is: a) 2 kg.m/s b) -2 kg.m/s c) what is the final kinetic energy of the target if the initial and final kinetic energies of the projectile are , respectively 5 J and 2 J? Pi=Pf Pf2=Pi-Pf1 =6-2= 4 kg.m/s =6-(-2)=8 kg.m/s c) Ki1= Kf1+Kf2 Then Kf2=Ki1-Kf1 = 5 – 2=3 J Ch-9 Check Point 10

  22. Ch-9-11: Elastic Collision in Two Dimensions • Solve equations: P1i + P2i = P1f + P2f K1i + K2i = K1f + K2f • Along x-axis m1v1i= m1v1f cos1+m2v2f cos2 • Along y-axis 0= - m1v1f sin1+m2v2f sin2 • m1v1i2/2 = m1v1f2/2+m2v2f2/2

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